Traveling Waves Near Couette Flow for the 2D Euler Equation

Abstract

In this paper we reveal the existence of a large family of new, nontrivial and smooth traveling waves for the 2D Euler equation at an arbitrarily small distance from the Couette flow in $H^s$, with $s<3/2$, at the level of the vorticity. The speed of these waves is of order 1 with respect to this distance. This result strongly contrasts with the setting of very high regularity in Gevrey spaces [see Bedrossian and Masmoudi (Publ Math Inst Hautes Etudes Sci 122:195–300, 2015)], where the problem exhibits an inviscid damping mechanism that leads to relaxation of perturbations back to nearby shear flows. It also complements the fact that there not exist nontrivial traveling waves in the $H^{\frac{3}{2}+}$ neighborhoods of Couette flow [see Lin and Zeng (Arch Ration Mech Anal 200:1075–1097, 2011)].

Appendix

In order to facilitate the presentation of the manuscript, we collect in this section all the technical lemmas used previously. We start recalling the definition of the kernel given by

$$\begin{aligned} K[g](\textbf{x},{\bar{\textbf{x}}}):=\log \left[ \cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[g](\textbf{x}))-\cos ({\bar{x}})\right] , \end{aligned}$$

(125)

where

$$\begin{aligned} \Delta _{{\bar{\textbf{x}}}}[g](\textbf{x}):=g(\textbf{x})-g(\textbf{x}-{\bar{\textbf{x}}}). \end{aligned}$$

In order to take derivatives into the kernel (125) we have introduced the following two functions

$$\begin{aligned} \Psi _1[g](\textbf{x},{\bar{\textbf{x}}})&:=\frac{\sinh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[g](\textbf{x}))}{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[g](\textbf{x}))-\cos ({\bar{x}})},\\ \Psi _2[g](\textbf{x},{\bar{\textbf{x}}})&:=\frac{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[g](\textbf{x}))}{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[g](\textbf{x}))-\cos ({\bar{x}})}, \end{aligned}$$

with derivatives given by the expressions

$$\begin{aligned} \partial _{\textbf{x}}\Psi _1[g](\textbf{x},{\bar{\textbf{x}}})&=[\Psi _2[g]-\Psi _1^2[g]](\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x}), \end{aligned}$$

(126)

$$\begin{aligned} \partial _{\textbf{x}}\Psi _2[g](\textbf{x},{\bar{\textbf{x}}})&=[\Psi _1[g]\left( 1-\Psi _2[g]\right) ](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x}). \end{aligned}$$

(127)

Combining (125) with (126) and (127) it is just a matter of algebra to obtain

$$\begin{aligned} \partial _{\textbf{x}}K[g](\textbf{x},{\bar{\textbf{x}}})&=\Psi _1[g](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x}),\\ \partial _{\textbf{x}}^2 K[g](\textbf{x},{\bar{\textbf{x}}})&=[\Psi _2[g]-\Psi _1^2[g]](\textbf{x},{\bar{\textbf{x}}})\left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x})\right) ^2+\Psi _1[g](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2 g](\textbf{x}),\\ \partial _{\textbf{x}}^3 K[g](\textbf{x},{\bar{\textbf{x}}})&=\Psi _1[g](\textbf{x},{\bar{\textbf{x}}})[ 1-3\Psi _2[g]+ 2 \Psi _1^2[g]](\textbf{x},{\bar{\textbf{x}}}) \left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x})\right) ^3\\&\quad +3[\Psi _2[g]- \Psi _1^2[g]](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2 g](\textbf{x})\\&\quad + \Psi _1[g](\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^3 g](\textbf{x}). \end{aligned}$$

Now, before starting with the proof of the lemmas, we remember the Taylor expansion of the trigonometric functions involved in the definition of $\Psi _{i}(\textbf{x},{\bar{\textbf{x}}})$ for $i=1,2.$

$$\begin{aligned} \sinh (z):=\sum _{n=0}^{+\infty }\frac{z^{2n+1}}{(2n+1)!},\qquad \cosh (z):=\sum _{n=0}^{+\infty }\frac{z^{2n}}{(2n)!},\qquad \text {and} \qquad \cos (z):=\sum _{n=0}^{+\infty }\frac{(-1)^n z^{2n}}{(2n)!}. \end{aligned}$$

Using the above expressions, it is simple to check that there exists some constant $C(\Vert g\Vert _{H^{3}(D_{\varepsilon })})$ such that for any ${\bar{\textbf{x}}}\in {\mathbb {T}}\times {\mathbb {R}}$ with $|{\bar{\textbf{x}}}|\ll 1$, we get

$$\begin{aligned} |\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[g](\textbf{x}))|&\le C(\Vert g\Vert _{H^{3}(D_{\varepsilon })}), \nonumber \\ |\sinh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[g](\textbf{x}))|&\le C(\Vert g\Vert _{H^{3}(D_{\varepsilon })}) |{\bar{\textbf{x}}}|, \nonumber \\ |\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[g](\textbf{x}))-\cos ({\bar{x}})|&\ge \left( \tfrac{1}{2}-\Vert g\Vert _{H^{3}(D_{\varepsilon })}\right) |{\bar{\textbf{x}}}|^2-C(\Vert g\Vert _{H^{3}(D_{\varepsilon })}) |{\bar{\textbf{x}}}|^4, \end{aligned}$$

(128)

and as an immediate consequence we obtain the following result.

Corollary 7.1

Let $g\in {\mathbb {B}}_{\delta }(H^3(D_{\varepsilon }))$ with $0< \delta (\varepsilon )\ll 1$. For ${\bar{\textbf{x}}}\in {\mathbb {T}}\times {\mathbb {R}}$ such that $|{\bar{\textbf{x}}}|\ll 1$ we have

$$\begin{aligned} |\Psi _i[g](\textbf{x},{\bar{\textbf{x}}})|\le \frac{C(\Vert g \Vert _{H^{3}(D_{\varepsilon })})}{ 1-C(\Vert g \Vert _{H^{3}(D_{\varepsilon })}) }\frac{1}{|{\bar{\textbf{x}}}|^i} \qquad \text {for} \quad i=1,2. \end{aligned}$$

Now, we focus our attention in the higher order derivatives $\partial _{\textbf{x}}^2 K[g](\textbf{x},{\bar{\textbf{x}}})$ and $\partial _{\textbf{x}}^3 K[g](\textbf{x},{\bar{\textbf{x}}})$. It will be convenient to introduce some notation in order to handle these terms. To be more specific, we will use the following expressions:

$$\begin{aligned} \partial _{\textbf{x}}^2 K[g](\textbf{x},{\bar{\textbf{x}}})&={\textsf{I}}_1[g](\textbf{x},{\bar{\textbf{x}}})+{\textsf{I}}_2[g](\textbf{x},{\bar{\textbf{x}}}), \end{aligned}$$

(129)

$$\begin{aligned} \partial _{\textbf{x}}^3 K[g](\textbf{x},{\bar{\textbf{x}}})&={\textsf{J}}_1[g](\textbf{x},{\bar{\textbf{x}}})+{\textsf{J}}_2[g](\textbf{x},{\bar{\textbf{x}}})+{\textsf{J}}_3[g](\textbf{x},{\bar{\textbf{x}}}) , \end{aligned}$$

(130)

where

$$\begin{aligned} \begin{aligned} {\textsf{I}}_1[g](\textbf{x},{\bar{\textbf{x}}})&:={\tilde{{\textsf{I}}}}_1[g](\textbf{x},{\bar{\textbf{x}}})\left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x})\right) ^2,\\ {\textsf{I}}_2[g](\textbf{x},{\bar{\textbf{x}}})&:={\tilde{{\textsf{I}}}}_2[g](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2 g](\textbf{x}),\\ {\textsf{J}}_1[g](\textbf{x},{\bar{\textbf{x}}})&:={\tilde{{\textsf{J}}}}_1[g](\textbf{x},{\bar{\textbf{x}}})\left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x})\right) ^3,\\ {\textsf{J}}_2[g](\textbf{x},{\bar{\textbf{x}}})&:={\tilde{{\textsf{J}}}}_2[g](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2 g](\textbf{x}),\\ {\textsf{J}}_3[g](\textbf{x},{\bar{\textbf{x}}})&:={\tilde{{\textsf{J}}}}_3[g](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^3 g](\textbf{x}), \end{aligned} \quad \quad \begin{aligned} {\tilde{{\textsf{I}}}}_1[g]&:=\Psi _2[g]-\Psi _1^2[g],\\ {\tilde{{\textsf{I}}}}_2[g]&:=\Psi _1[g],\\ {\tilde{{\textsf{J}}}}_1[g]&:=\Psi _1[g]\left( 1- 3\Psi _2[g]+2 \Psi _1^2[g]\right) ,\\ {\tilde{{\textsf{J}}}}_2[g]&:=3\Psi _2[g]-\Psi _1^2[g],\\ {\tilde{{\textsf{J}}}}_3[g]&:=\Psi _1[g]. \end{aligned} \end{aligned}$$

As an immediate consequence of Corollary 7.1 we get the following bounds for the above expressions.

Corollary 7.2

Let $g\in {\mathbb {B}}_{\delta }(H^3(D_{\varepsilon }))$ with $0< \delta (\varepsilon )\ll 1$. For ${\bar{\textbf{x}}}\in {\mathbb {T}}\times {\mathbb {R}}$ such that $|{\bar{\textbf{x}}}|\ll 1$ we have

$$\begin{aligned} |{\tilde{{\textsf{I}}}}_2[g](\textbf{x},{\bar{\textbf{x}}})|, |{\tilde{{\textsf{J}}}}_3[g](\textbf{x},{\bar{\textbf{x}}})|\le C(\Vert g\Vert _{H^{3}(D_{\varepsilon })}) |{\bar{\textbf{x}}}|^{-1},\\ |{\tilde{{\textsf{I}}}}_1[g](\textbf{x},{\bar{\textbf{x}}})|, |{\tilde{{\textsf{J}}}}_2[g](\textbf{x},{\bar{\textbf{x}}})|\le C(\Vert g\Vert _{H^{3}(D_{\varepsilon })}) |{\bar{\textbf{x}}}|^{-2},\\ |{\tilde{{\textsf{J}}}}_1[g](\textbf{x},{\bar{\textbf{x}}})|\le C(\Vert g \Vert _{H^{3}(D_{\varepsilon })}) |{\bar{\textbf{x}}}|^{-3}. \end{aligned}$$

Before starting with the proof of the main lemmas of the Appendix, we will collect a couple of auxiliary results. In first place, we note that for any $g\in H^{4,3}(D_{\varepsilon }),$ an inequality that we will repeatedly apply in our arguments is the following:

$$\begin{aligned} |\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^ig](\textbf{x})|\le |{\bar{\textbf{x}}}|^{2-i} \Vert g\Vert _{H^{4,3}(D_{\varepsilon })}, \qquad i=1,2. \end{aligned}$$

(131)

Note that the above reduces to the Sobolev embedding $H^{3,2}(D_\varepsilon )\subset C^{1}(D_{\varepsilon })$ and $H^{2,1}(D_\varepsilon )\subset C(D_{\varepsilon })$. Secondly, we obtain an uniform bound for the difference of the auxiliary functions $\Psi _i[{\textsf{f}}']-\Psi _i[{\textsf{f}}'']$ and their derivatives in terms of the $H^{4,3}(D_{\varepsilon })$-norm of the difference ${\textsf{f}}'-{\textsf{f}}''.$

Lemma 7.3

Let ${\textsf{f}}', {\textsf{f}}'' \in {\mathbb {B}}_{\delta }(H^{4,3}(D_{\varepsilon }))$ with $0<\delta (\varepsilon )\ll 1$ small enough. The following bounds hold

• i)
  $$\begin{aligned} |(\Psi _i[{\textsf{f}}']-\Psi _i[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})|^2 |{\bar{\textbf{x}}}|^{2i}\le C(\varepsilon ) \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{3}(D_{\varepsilon })}^2, \qquad i=1,2. \end{aligned}$$
• ii)
  $$\begin{aligned} |(\partial _{\textbf{x}}\Psi _i[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _i[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})|^2 |{\bar{\textbf{x}}}|^{2i}\le C(\varepsilon ) \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}^2, \qquad i=1,2. \end{aligned}$$
• iii)
  $$\begin{aligned} |(\partial _{\textbf{x}}^2\Psi _i[{\textsf{f}}']-\partial _{\textbf{x}}^2\Psi _i[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})|^2 |{\bar{\textbf{x}}}|^{2(i+1)}\le C(\varepsilon ) \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}^2, \qquad i=1,2. \end{aligned}$$

Proof of (i)

Using some trigonometric identities for hyperbolic functions $\sinh (\cdot ), \cosh (\cdot )$ we get

$$\begin{aligned}&\left( \Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}'']\right) (\textbf{x},{\bar{\textbf{x}}})\\&\quad =\frac{\sinh (\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'-{\textsf{f}}''](\textbf{x}))-\cos ({\bar{x}})\cosh ({\bar{y}})\left[ \sinh (\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x}))-\sinh (\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x}))\right] }{\left[ \cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x}))-\cos ({\bar{x}})\right] \left[ \cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x}))-\cos ({\bar{x}})\right] }\nonumber \\&\qquad -\frac{\cos ({\bar{x}})\sinh ({\bar{y}})\left[ \cosh (\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x}))-\cosh (\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x}))\right] }{\left[ \cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x}))-\cos ({\bar{x}})\right] \left[ \cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x}))-\cos ({\bar{x}})\right] }, \end{aligned}$$

and

$$\begin{aligned}&\left( \Psi _2[{\textsf{f}}']-\Psi _2[{\textsf{f}}'']\right) (\textbf{x},{\bar{\textbf{x}}})\\&\quad =\frac{\cos ({\bar{x}})\left[ \cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x}))-\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x}))\right] }{\left[ \cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x}))-\cos ({\bar{x}})\right] \left[ \cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x}))-\cos ({\bar{x}})\right] }. \end{aligned}$$

Notice that both denominators vanish if and only if ${\bar{\textbf{x}}}=0$. Now, since ${\textsf{f}}',{\textsf{f}}''\in H^{4,3}(D_{\varepsilon })\subset C^2(D_{\varepsilon })$ are continuous functions on a bounded domain, the outer region is easily bounded by the required term. Then, we just focus our attention on the inner region $|{\bar{\textbf{x}}}|\ll 1$.

Applying Taylor expansion of each trigonometric function involved we obtain that each numerator of the previous expression can be bound as follows

$$\begin{aligned}{} & {} \left| \sinh (\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'-{\textsf{f}}''](\textbf{x}))-\cos ({\bar{x}})\cosh ({\bar{y}})\left[ \sinh (\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x}))-\sinh (\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x}))\right] \right| \\{} & {} \quad \le C(\varepsilon ) |{\bar{\textbf{x}}}|^3 \frac{1}{1-\max \{\Vert {\textsf{f}}' - {\textsf{f}}''\Vert _{H^{3}},\Vert {\textsf{f}}'\Vert _{H^{3}},\Vert {\textsf{f}}''\Vert _{H^{3}}\}\text {diam}(D_{\varepsilon })} \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{3}},\\{} & {} \left| \cos ({\bar{x}})\sinh ({\bar{y}})\left[ \cosh (\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x}))-\cosh (\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x}))\right] \right| \\{} & {} \quad \le C(\varepsilon ) |{\bar{\textbf{x}}}|^3 \frac{\max \{\Vert {\textsf{f}}'\Vert _{H^{3}},\Vert {\textsf{f}}''\Vert _{H^{3}}\}}{1-\max \{\Vert {\textsf{f}}'\Vert _{H^{3}},\Vert {\textsf{f}}''\Vert _{H^{3}}\}\text {diam}(D_{\varepsilon })} \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{3}}, \end{aligned}$$

and

$$\begin{aligned}{} & {} \left| \cos ({\bar{x}})\left[ \cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x}))-\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x}))\right] \right| \\{} & {} \quad \le C(\varepsilon ) |{\bar{\textbf{x}}}|^2 \frac{\max \{\Vert {\textsf{f}}'\Vert _{H^{3}},\Vert {\textsf{f}}''\Vert _{H^{3}}\}}{1-\max \{\Vert {\textsf{f}}'\Vert _{H^{3}},\Vert {\textsf{f}}''\Vert _{H^{3}}\}\text {diam}(D_{\varepsilon })} \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{3}}. \end{aligned}$$

Combining all with the usual lower bound for the denominator (see (128)), we obtain

$$\begin{aligned} |(\Psi _i[{\textsf{f}}']-\Psi _i[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})|^2 |{\bar{\textbf{x}}}|^{2j} \le \left( \frac{C(\varepsilon ,\Vert {\textsf{f}}'\Vert _{H^{3}(D_{\varepsilon })},\Vert {\textsf{f}}''\Vert _{H^{3}(D_{\varepsilon })})}{1-\max \{\Vert {\textsf{f}}'\Vert _{H^{3}},\Vert {\textsf{f}}''\Vert _{H^{3}}\}\text {diam}(D_{\varepsilon })}\right) \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^3(D_{\varepsilon })}^2. \end{aligned}$$

Finally, as ${\textsf{f}}', {\textsf{f}}'' \in {\mathbb {B}}_{\delta }(H^{4,3}(D_{\varepsilon })),$ taking $0<\delta <\text {diam}(D_{\varepsilon })^{-1}$ we have proved our goal. $\square$

Proof of (ii). Due to the relations

$$\begin{aligned} \partial _{\textbf{x}}\Psi _1[g](\textbf{x},{\bar{\textbf{x}}})&=(\Psi _2[g]-\Psi _1^2[g])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x}),\\ \partial _{\textbf{x}}\Psi _2[g](\textbf{x},{\bar{\textbf{x}}})&=(\Psi _1[g]-\Psi _1[g]\Psi _2[g])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x}), \end{aligned}$$

it is clear that adding and subtracting some appropriate term we obtain the expressions

$$\begin{aligned} (\partial _{\textbf{x}}\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})&=(\Psi _2[{\textsf{f}}']-(\Psi _1[{\textsf{f}}'])^2)(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}({\textsf{f}}'-{\textsf{f}}'')](\textbf{x}) \nonumber \\&\quad +(\Psi _2[{\textsf{f}}']-\Psi _2[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})\nonumber \\&\quad - (\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})(\Psi _1[{\textsf{f}}']+\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}'](\textbf{x}), \end{aligned}$$

(132)

and

$$\begin{aligned} (\partial _{\textbf{x}}\Psi _2[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _2[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})&=(\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}']\Psi _2[{\textsf{f}}'])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}({\textsf{f}}'-{\textsf{f}}'')](\textbf{x})\nonumber \\&\quad + (\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''] )(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})\nonumber \\&\quad - \Psi _1[{\textsf{f}}'](\textbf{x},{\bar{\textbf{x}}})(\Psi _2[{\textsf{f}}']-\Psi _2[{\textsf{f}}''] )(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})\nonumber \\&\quad - \Psi _2[{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})(\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''] )(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x}). \end{aligned}$$

(133)

As before, we just focus on the inner region $|{\bar{\textbf{x}}}|\ll 1,$ which is the most singular part. Now, remembering (131) and applying repeatedly Corollary 7.1, we get for $1\le i \le 2$ and $|{\bar{\textbf{x}}}|\ll 1$ that

$$\begin{aligned}{} & {} |(\partial _{\textbf{x}}\Psi _i[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _i[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})||{\bar{\textbf{x}}}|^{i}\\{} & {} \quad \lesssim \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}+ |{\bar{\textbf{x}}}||(\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})| +|{\bar{\textbf{x}}}|^2|(\Psi _2[{\textsf{f}}']-\Psi _2[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})|. \end{aligned}$$

Finally, applying i) in each of the last two terms of the above expressions we have proved ii). $\square$

Proof of (iii)

Taking one derivative on (132), (133) and adding and subtracting appropriate terms, we obtain the expressions

$$\begin{aligned}&(\partial _{\textbf{x}}^2\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}^2\Psi _1[{\textsf{f}}'])(\textbf{x},{\bar{\textbf{x}}})\\&\quad =(\Psi _2[{\textsf{f}}']-\Psi _1^2[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2({\textsf{f}}'-{\textsf{f}}'')](\textbf{x})\\&\qquad + (\Psi _1[{\textsf{f}}']-3\Psi _1[{\textsf{f}}'] \Psi _2[{\textsf{f}}']+2\Psi _1^3[{\textsf{f}}'])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _x{\textsf{f}}'](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}({\textsf{f}}'-{\textsf{f}}'')](\textbf{x})\\&\qquad + (\Psi _2[{\textsf{f}}']-\Psi _2[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}''](\textbf{x})\\&\qquad + (\partial _\textbf{x}\Psi _2[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _2[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})\\&\qquad - (\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\Psi _1[{\textsf{f}}']+\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}'](\textbf{x}),\\&\qquad - (\partial _{\textbf{x}}\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}''])(\Psi _1[{\textsf{f}}']+\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}'](\textbf{x}),\\&\qquad - (\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\Psi _2[{\textsf{f}}']-\Psi _1^2[{\textsf{f}}'])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}'](\textbf{x}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}'](\textbf{x}),\\&\qquad - (\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\Psi _2[{\textsf{f}}'']-\Psi _1^2[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}'](\textbf{x}),\\ \end{aligned}$$

and

$$\begin{aligned}&(\partial _{\textbf{x}}^2\Psi _2[{\textsf{f}}']-\partial _{\textbf{x}}^2\Psi _2[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})\\&\quad =(\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}']\Psi _2[{\textsf{f}}'])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2({\textsf{f}}'-{\textsf{f}}'')](\textbf{x})\\&\qquad +(\Psi _2[{\textsf{f}}']-2\Psi _1^2[{\textsf{f}}'])(1-\Psi _2[{\textsf{f}}'])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}'](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}({\textsf{f}}'-{\textsf{f}}'')](\textbf{x})\\&\qquad + (\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''] )(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}''](\textbf{x})\\&\qquad + (\partial _{\textbf{x}}\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}''] )(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})\\&\qquad - \Psi _1[{\textsf{f}}'](\Psi _2[{\textsf{f}}']-\Psi _2[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}''](\textbf{x})\\&\qquad - \Psi _1[{\textsf{f}}'](\partial _{\textbf{x}}\Psi _2[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _2[{\textsf{f}}''] )(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})\\&\qquad - (\Psi _2[{\textsf{f}}']-\Psi _1^2[{\textsf{f}}'])(\Psi _2[{\textsf{f}}']-\Psi _2[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}'](\textbf{x}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})\\&\qquad - \Psi _2[{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})\left( \Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''] \right) (\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}''](\textbf{x})\\&\qquad - \Psi _2[{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})(\partial _{\textbf{x}}\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}''] )(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})\\&\qquad - (\Psi _1[{\textsf{f}}'']-\Psi _1[{\textsf{f}}'']\Psi _2[{\textsf{f}}''])\left( \Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''] \right) (\textbf{x},{\bar{\textbf{x}}})\left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})\right) ^2. \end{aligned}$$

The computations are very long and tedious but share lot of similarities. For this reason we shall focus only on $(\partial _{\textbf{x}}^2\Psi _2[{\textsf{f}}']-\partial _{\textbf{x}}^2\Psi _2[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})$ to illustrate how the estimates work. The ideas is to write the above long expressions in groups of terms as

• $(1-\Psi _2[{\textsf{f}}'])(\Psi _1[{\textsf{f}}'] \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2({\textsf{f}}'-{\textsf{f}}'')](\textbf{x}) +(\Psi _2[{\textsf{f}}']-2\Psi _1^2[{\textsf{f}}'])\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}'](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}({\textsf{f}}'-{\textsf{f}}'')](\textbf{x})),$

• $(\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''] )(\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}''](\textbf{x})- (\Psi _1[{\textsf{f}}'']-\Psi _1[{\textsf{f}}'']\Psi _2[{\textsf{f}}''])\left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})\right) ^2 - \Psi _2[{\textsf{f}}''] \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}''](\textbf{x})),$

• $(\Psi _2[{\textsf{f}}']-\Psi _2[{\textsf{f}}''])\left( -\Psi _1[{\textsf{f}}']\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}''](\textbf{x})- (\Psi _2[{\textsf{f}}']-\Psi _1^2[{\textsf{f}}'])\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}'](\textbf{x}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})\right) ,$

• $(\partial _{\textbf{x}}\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}''] )(1-\Psi _2[{\textsf{f}}''])\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x}),$

• $(\partial _{\textbf{x}}\Psi _2[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _2[{\textsf{f}}''] ) (- \Psi _1[{\textsf{f}}'] \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})).$

Remembering (131) and applying repeatedly Corollary 7.1, we get for $1\le i \le 2$ and $|{\bar{\textbf{x}}}|\ll 1$ that

$$\begin{aligned} |(\partial _{\textbf{x}}^2\Psi _i[{\textsf{f}}']-\partial _{\textbf{x}}^2\Psi _i[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})||{\bar{\textbf{x}}}|^{i}&\lesssim |{\bar{\textbf{x}}}|^{-1}\Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}+ |(\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})|\\&\quad +|{\bar{\textbf{x}}}||(\Psi _2[{\textsf{f}}']-\Psi _2[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})|\\&\quad +|{\bar{\textbf{x}}}||(\partial _{\textbf{x}}\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})|\\&\quad +|{\bar{\textbf{x}}}|^2|(\partial _{\textbf{x}}\Psi _2[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _2[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})|. \end{aligned}$$

Finally, applying i) and ii) in each of the terms of the above expressions we have proved iii). $\square$

At this point, we have all the ingredients to prove the main lemmas of the Appendix. The first one allows us to take derivatives into the kernel $\partial _{\textbf{x}}^i K[g]$ as follows:

Lemma 7.4

Let $g \in {\mathbb {B}}_{\delta }(H^{4,3}(D_{\varepsilon }))$ with $0<\delta (\varepsilon )\ll 1$ small enough. The following bounds hold

• i)
  $$\begin{aligned} \sup _{\textbf{x}\in D_{\varepsilon }}\left( \sup _{{\bar{\textbf{x}}}\in D_\varepsilon (y)} |K[g](\textbf{x},{\bar{\textbf{x}}})|^2\right) \le C(\varepsilon ,\Vert g\Vert _{H^{4,3}(D_{\varepsilon })}), \end{aligned}$$
• ii)
  $$\begin{aligned} \sup _{\textbf{x}\in D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}|\partial _{\textbf{x}} K[g](\textbf{x},{\bar{\textbf{x}}})|^2\hbox {d}{\bar{\textbf{x}}} \right) \le C(\varepsilon ,\Vert g\Vert _{H^{4,3}(D_{\varepsilon })}), \end{aligned}$$
• iii)
  $$\begin{aligned} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| \partial _{\textbf{x}}^2 K[g](\textbf{x},{\bar{\textbf{x}}}) \right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\le C(\varepsilon ,\Vert g\Vert _{H^{4,3}(D_{\varepsilon })}), \end{aligned}$$
• iv)
  $$\begin{aligned} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| \partial _{\textbf{x}}^3 K[g](\textbf{x},{\bar{\textbf{x}}}) \right| ^2 |{\bar{\textbf{x}}}|^{2}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\le C(\varepsilon ,\Vert g\Vert _{H^{4,3}(D_{\varepsilon })}). \end{aligned}$$

Proof

As we can see the computations share lot of similarities. For this reason we shall focus only on one significant term iii) to illustrate how the estimates work. Remembering (129) we get

$$\begin{aligned} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| \partial _{\textbf{x}}^2 K[g](\textbf{x},{\bar{\textbf{x}}}) \right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\le \sum _{i=1}^{2} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| {\textsf{I}}_i[g](\textbf{x},{\bar{\textbf{x}}}) \right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\end{aligned}$$

Proceeding as usual, we split the integral into inner and outer regions. As $g\in H^{4,3}(D_{\varepsilon })\subset C^{2}(D_{\varepsilon })$ is a continuous function on a bounded domain the outer integral is trivially bounded by some universal constant. To handle the remaining inner integral $\{(\textbf{x},{\bar{\textbf{x}}})\in D_{\varepsilon }\times D_{\varepsilon }(y):|{\bar{\textbf{x}}}|\ll 1\}$ of each term, we use Corollary 7.2 as follows:

${\textsf{I}}_1)$:

As $g\in H^{4,3}(D_{\varepsilon })$ we have that $\partial _{\textbf{x}}g \in H^{3,2}(D_{\varepsilon })\subset C^1(D_{\varepsilon })$ and consequently we get

$$\begin{aligned}&\int _{D_{\varepsilon }}\left( \int _{|{\bar{\textbf{x}}}|\ll 1}\left| {\textsf{I}}_1[g](\textbf{x},{\bar{\textbf{x}}}) \right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\&\quad = \int _{D_{\varepsilon }}\left( \int _{|{\bar{\textbf{x}}}|\ll 1}\left| {\tilde{{\textsf{I}}}}_1[g](\textbf{x},{\bar{\textbf{x}}})\right| ^2 \left| \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x})\right| ^4 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\&\quad \le C(\varepsilon ,\Vert g \Vert _{H^{4,3}(D_{\varepsilon })})\left( \int _{|{\bar{\textbf{x}}}|\ll 1} |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}}\right) , \end{aligned}$$

${\textsf{I}}_2)$:

As $g\in H^{4,3}(D_{\varepsilon })$ we have that $\partial _{\textbf{x}}^2g \in H^{2,1}(D_{\varepsilon })\subset C(D_{\varepsilon })$ and consequently we get

$$\begin{aligned}&\int _{D_{\varepsilon }}\left( \int _{|{\bar{\textbf{x}}}|\ll 1}\left| {\textsf{I}}_2[g](\textbf{x},{\bar{\textbf{x}}}) \right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\&\quad = \int _{D_{\varepsilon }}\left( \int _{|{\bar{\textbf{x}}}|\ll 1}\left| {\tilde{{\textsf{I}}}}_2[g](\textbf{x},{\bar{\textbf{x}}})\right| ^2 \left| \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2 g](\textbf{x})\right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\&\le C(\delta ,\Vert g \Vert _{H^{4,3}(D_{\varepsilon })})\left( \int _{|{\bar{\textbf{x}}}|\ll 1} \frac{1}{|{\bar{\textbf{x}}}|^{2(1-\gamma )}}\hbox {d}{\bar{\textbf{x}}}\right) . \end{aligned}$$

As $0<\gamma <1,$ the last term is integrable and we have proved our goal. The rest of the terms i), ii) and iv) work in a similar way and we omit the details. $\square$

Now, we continue with an analog result for the difference $\partial _{\textbf{x}}^i K[{\textsf{f}}']-\partial _{\textbf{x}}^iK[{\textsf{f}}'']$.

Lemma 7.5

Let ${\textsf{f}}', {\textsf{f}}'' \in {\mathbb {B}}_{\delta }(H^{4,3}(D_{\varepsilon }))$ with $0<\delta (\varepsilon )\ll 1$ small enough. The following bounds hold

• i)
  $$\begin{aligned} \sup _{\textbf{x}\in D_{\varepsilon }}\left( \sup _{{\bar{\textbf{x}}}\in D_\varepsilon (y)} |{\mathcal {K}}[{\textsf{f}}',{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})|^2\right) \le C(\varepsilon ) \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}^2. \end{aligned}$$
• ii)
  $$\begin{aligned} \sup _{\textbf{x}\in D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}|\partial _{\textbf{x}}{\mathcal {K}}[{\textsf{f}}',{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})|^2\hbox {d}{\bar{\textbf{x}}} \right) \le C(\varepsilon ) \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}^2. \end{aligned}$$
• iii)
  $$\begin{aligned} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| \partial _{\textbf{x}}^2 {\mathcal {K}}[{\textsf{f}}',{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}}) \right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\le C(\varepsilon ) \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}^2. \end{aligned}$$
• iv)
  $$\begin{aligned} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| \partial _{\textbf{x}}^3 {\mathcal {K}}[{\textsf{f}}',{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}}) \right| ^2 |{\bar{\textbf{x}}}|^{2}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\le C(\varepsilon ) \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}^2. \end{aligned}$$

Proof of i). We start by remembering the kernel definition ${\mathcal {K}}[{\textsf{f}}',{\textsf{f}}'']=K[{\textsf{f}}']-K[{\textsf{f}}'']$ given by

$$\begin{aligned} {\mathcal {K}}[{\textsf{f}}',{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})=\log \left[ \frac{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\cos ({\bar{x}})}{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) -\cos ({\bar{x}})}\right] . \end{aligned}$$

Adding and subtracting some appropriate term, one finds

$$\begin{aligned} \frac{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\cos ({\bar{x}})}{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) -\cos ({\bar{x}})}=1+\frac{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) }{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) -\cos ({\bar{x}})}, \end{aligned}$$

and using the standard logarithmic inequality $1-x^{-1}\le \log x\le x-1$ for all $x>0$, we get the lower and upper bounds

$$\begin{aligned} {\mathcal {K}}[{\textsf{f}}',{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})&\le \frac{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) }{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) -\cos ({\bar{x}})},\\ {\mathcal {K}}[{\textsf{f}}',{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})&\ge \frac{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) }{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\cos ({\bar{x}})}. \end{aligned}$$

As ${\textsf{f}}', {\textsf{f}}'' \in {\mathbb {B}}_{\delta }(H^{4,3}(D_{\varepsilon }))$ are arbitrary functions we can assume without loss of generality that

$$\begin{aligned} |{\mathcal {K}}[{\textsf{f}}',{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})|\le \left| \frac{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) }{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) -\cos ({\bar{x}})}\right| , \end{aligned}$$

and using the trigonometric identity $\cosh (a+b)=\cosh (a)\cosh (b)+\sinh (a)\sinh (b)$ we finally get

$$\begin{aligned}&\frac{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) }{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) -\cos ({\bar{x}})}\\&\quad =\cosh ({\bar{y}})\frac{\cosh \left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\cosh \left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) }{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) -\cos ({\bar{x}})}\\&\qquad +\sinh ({\bar{y}})\frac{\sinh \left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\sinh \left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) }{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) -\cos ({\bar{x}})}. \end{aligned}$$

Now, we will use the Taylor expansion of the trigonometric functions $\cosh (\cdot )$ and $\sinh (\cdot )$ together with the algebraic identity $a^n-b^n=(a-b)\sum _{k=0}^{n-1}a^k b^{n-1-k}$ to obtain

$$\begin{aligned}&\cosh \left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\cosh \left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) \\&\quad =\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'-{\textsf{f}}''](\textbf{x})\sum _{n=1}^{\infty }\sum _{k=0}^{2n-1} \frac{\left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) ^{k}\left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) ^{2n-1-k}}{(2n)!},\\&\sinh \left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\sinh \left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) \\&\quad =\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'-{\textsf{f}}''](\textbf{x})\sum _{n=0}^{\infty }\sum _{k=0}^{2n} \frac{\left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) ^{k}\left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) ^{2n-k}}{(2n+1)!}. \end{aligned}$$

Computing the infinite sum of the geometric series $\sum _{n=0}^{\infty }r^{2n}$ with

$$\begin{aligned} r:=\max \{\Vert {\textsf{f}}'\Vert _{H^{4,3}(D_{\varepsilon })},\Vert {\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}\}\text {diam}(D_{\varepsilon }), \end{aligned}$$

we finally get (under condition $\delta <\text {diam}(D_{\varepsilon })^{-1}$) that

$$\begin{aligned}&\left| \cosh \left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\cosh \left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) \right| \\&\quad \le C(\varepsilon ) |{\bar{\textbf{x}}}|^2 \frac{ \max \{\Vert {\textsf{f}}'\Vert _{H^{4,3}(D_{\varepsilon })},\Vert {\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}\} \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}}{1-\max \{\Vert {\textsf{f}}'\Vert _{H^{4,3}(D_{\varepsilon })},\Vert {\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}\} \text {diam}(D_{\varepsilon })} ,\\&\left| \sinh \left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}'](\textbf{x})\right) -\sinh \left( \Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x})\right) \right| \\&\quad \le C(\varepsilon ) |{\bar{\textbf{x}}}| \frac{ \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}}{1-\max \{\Vert {\textsf{f}}'\Vert _{H^{4,3}(D_{\varepsilon })},\Vert {\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}\}\text {diam}(D_{\varepsilon })}, \end{aligned}$$

where we have applied the Sobolev embedding $L^{\infty }(D_{\varepsilon })\hookrightarrow H^2(D_{\varepsilon }).$

Now, it is not difficult to see that the following uniform bound holds

$$\begin{aligned} \sup _{\textbf{x}\in D_{\varepsilon }}\left( \sup _{{\bar{\textbf{x}}}\in D_\varepsilon (y)} \left| \frac{\cosh ({\bar{y}})|{\bar{\textbf{x}}}|^2+\sinh ({\bar{y}})|{\bar{\textbf{x}}}|}{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x}))-\cos ({\bar{x}})}\right| ^2\right) \le C(\varepsilon ) \end{aligned}$$

thanks to the fact (see (128)) that for $|{\bar{\textbf{x}}}|\ll 1$ we have

$$\begin{aligned} |\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}''](\textbf{x}))-\cos ({\bar{x}})|\ge \left( \tfrac{1}{2}-\Vert {\textsf{f}}''\Vert _{H^{3}(D_{\varepsilon })}\right) |{\bar{\textbf{x}}}|^2-C(\Vert {\textsf{f}}''\Vert _{H^{3}(D_{\varepsilon })}) |{\bar{\textbf{x}}}|^4. \end{aligned}$$

Then, taking $0<\delta < \text {diam}(D_{\varepsilon })^{-1}$ small enough and combining all we have proved our goal. $\square$

Proof of ii). As ${\mathcal {K}}[{\textsf{f}}',{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})=( K[{\textsf{f}}']-K[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})$ and $\partial _{\textbf{x}}K[g](\textbf{x},{\bar{\textbf{x}}})=\Psi _1[g](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x})$, adding and subtracting some appropriate term we obtain

$$\begin{aligned} \partial _{\textbf{x}}{\mathcal {K}}[{\textsf{f}}',{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})=\Psi _1[{\textsf{f}}'](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}({\textsf{f}}'-{\textsf{f}}'')](\textbf{x})+(\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x}). \nonumber \\ \end{aligned}$$

(134)

To bound the first term we use ${\textsf{f}}',{\textsf{f}}''\in {\mathbb {B}}_{\delta }(H^{4,3}(D_{\varepsilon }))$, which implies $\partial _{\textbf{x}}({\textsf{f}}'-{\textsf{f}}'')\in H^{3,2}(D_\varepsilon )\subset C^{1}(D_{\varepsilon })$. As usual, splitting the integral and using Corollary 7.1 in the inner region we obtain directly that

$$\begin{aligned} \int _{D_\varepsilon (y)}|\Psi _1[{\textsf{f}}'](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}({\textsf{f}}'-{\textsf{f}}'')](\textbf{x})|^2\hbox {d}{\bar{\textbf{x}}} \le C(\varepsilon ,\Vert {\textsf{f}}'\Vert _{H^{4,3}(D_{\varepsilon })})\Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}^2. \end{aligned}$$

For the second term of (134), using that $\partial _{\textbf{x}}{\textsf{f}}''\in H^{3,2}(D_{\varepsilon })\subset C^{1}(D_{\varepsilon })$ and applying Lemma 7.3 we get

$$\begin{aligned}&\int _{D_\varepsilon (y)}|(\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})|^2\hbox {d}{\bar{\textbf{x}}} \\&\quad \le \Vert {\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })} \int _{D_\varepsilon (y)}|(\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}})|^2 |{\bar{\textbf{x}}}|^2\hbox {d}{\bar{\textbf{x}}}\\&\quad \le C(\varepsilon ,\Vert {\textsf{f}}'\Vert _{H^{4,3}(D_{\varepsilon })},\Vert {\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}) \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}^2. \end{aligned}$$

Finally, as the above holds for any $\textbf{x}\in D_{\varepsilon }$, combining the above and taking the supremum over all the domain we have proved the desired inequality. $\square$

Proof of (iii)

Taking a derivative of (134) we obtain

$$\begin{aligned} \partial _{\textbf{x}}^2{\mathcal {K}}[{\textsf{f}}',{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})&=\partial _{\textbf{x}}\Psi _1[{\textsf{f}}'](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}({\textsf{f}}'-{\textsf{f}}'')](\textbf{x})+\Psi _1[{\textsf{f}}'](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2({\textsf{f}}'-{\textsf{f}}'')](\textbf{x}) \nonumber \\&\quad +(\partial _{\textbf{x}}\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x})\nonumber \\&\quad +(\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}''](\textbf{x}). \end{aligned}$$

(135)

To bound the first term of (135), as ${\textsf{f}}',{\textsf{f}}''\in {\mathbb {B}}_{\delta }(H^{4,3}(D_{\varepsilon }))$ we get $\partial _{\textbf{x}}{\textsf{f}}', \partial _{\textbf{x}}({\textsf{f}}'-{\textsf{f}}'')\in H^{3,2}(D_\varepsilon )\subset C^{1}(D_{\varepsilon })$. Now, splitting the integral and using (126) together with Corollary 7.1 in the inner region we obtain

$$\begin{aligned}{} & {} \int _{D_{\varepsilon }}\left( \int _{|{\bar{\textbf{x}}}|\ll 1}\left| \partial _{\textbf{x}}\Psi _1[{\textsf{f}}'](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}({\textsf{f}}'-{\textsf{f}}'')](\textbf{x}) \right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\{} & {} \quad \le \frac{C(\Vert {\textsf{f}}' \Vert _{H^{4,3}(D_{\varepsilon })})}{ 1-C(\Vert {\textsf{f}}' \Vert _{H^{4,3}(D_{\varepsilon })}) } \Vert {\textsf{f}}'-{\textsf{f}}'' \Vert _{H^{4,3}(D_{\varepsilon })}^2. \end{aligned}$$

To bound the second term of (135) we use the same type of ideas. As ${\textsf{f}}',{\textsf{f}}''\in {\mathbb {B}}_{\delta }(H^{4,3}(D_{\varepsilon }))$ we get $\partial _{\textbf{x}}^2({\textsf{f}}'-{\textsf{f}}'')\in H^{2,1}(D_\varepsilon )\subset C(D_{\varepsilon })$. Now, splitting the integral and applying Corollary 7.1 in the inner region we obtain

$$\begin{aligned}{} & {} \int _{D_{\varepsilon }}\left( \int _{|{\bar{\textbf{x}}}|\ll 1}\left| \Psi _1[{\textsf{f}}'](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2({\textsf{f}}'-{\textsf{f}}'')](\textbf{x}) \right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\{} & {} \quad \le \frac{C(\Vert {\textsf{f}}' \Vert _{H^{4,3}(D_{\varepsilon })})}{ 1-C(\Vert {\textsf{f}}' \Vert _{H^{4,3}(D_{\varepsilon })}) } \Vert {\textsf{f}}'-{\textsf{f}}'' \Vert _{H^{4,3}(D_{\varepsilon })}^2\left( \int _{|{\bar{\textbf{x}}}|\ll 1}\frac{1}{|{\bar{\textbf{x}}}|^{2(1-\gamma )}}\hbox {d}{\bar{\textbf{x}}}\right) , \end{aligned}$$

where the last integral is bounded thanks to the fact that $0<\gamma <1.$ Finally, to bound the last two terms of (135) we use the fact that $\partial _{\textbf{x}}{\textsf{f}}'' \in H^{3,2}(D_{\varepsilon })\subset C^1(D_{\varepsilon })$ and $\partial _{\textbf{x}}^2{\textsf{f}}'' \in H^{2,1}(D_{\varepsilon })\subset C(D_{\varepsilon })$ to get

$$\begin{aligned}{} & {} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| (\partial _{\textbf{x}}\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x}) \right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\{} & {} \quad \le \Vert {\textsf{f}}'' \Vert _{H^{4,3}(D_{\varepsilon })} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| (\partial _{\textbf{x}}\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \right| ^2 |{\bar{\textbf{x}}}|^{2(1+\gamma )}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\, \end{aligned}$$

and

$$\begin{aligned}{} & {} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| (\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}''](\textbf{x}) \right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\{} & {} \quad \le \Vert {\textsf{f}}'' \Vert _{H^{4,3}(D_{\varepsilon })}\int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| (\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}. \end{aligned}$$

Using auxiliary Lemma 7.3 on the last terms of the above expressions we have proved our goal. $\square$

Proof of (iv)

As before, taking a derivative of (135) we obtain

$$\begin{aligned} \partial _{\textbf{x}}^3{\mathcal {K}}[{\textsf{f}}',{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})&=\partial _{\textbf{x}}^2\Psi _1[{\textsf{f}}'](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}({\textsf{f}}'-{\textsf{f}}'')](\textbf{x})+2\partial _{\textbf{x}}\Psi _1[{\textsf{f}}'](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2({\textsf{f}}'-{\textsf{f}}'')](\textbf{x}) \nonumber \\&\quad +\Psi _1[{\textsf{f}}'](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^3({\textsf{f}}'-{\textsf{f}}'')](\textbf{x})\nonumber \\&\quad +2\left( \partial _{\textbf{x}}\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}'']\right) (\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}''](\textbf{x}) \nonumber \\&\quad +\left( \Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}'']\right) (\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^3{\textsf{f}}''](\textbf{x})\nonumber \\&\quad +\left( \partial _{\textbf{x}}^2\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}^2\Psi _1[{\textsf{f}}'']\right) (\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x}). \end{aligned}$$

(136)

In first place, due to the relations (126) and (127), we have

$$\begin{aligned} \partial _{\textbf{x}}^2\Psi _1[{\textsf{f}}']&=[\Psi _1[{\textsf{f}}'](1-\Psi _2[{\textsf{f}}'])-2\Psi _1[{\textsf{f}}'](\Psi _2[{\textsf{f}}']-\Psi _1^2[{\textsf{f}}'])]\left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}'](\textbf{x})\right) ^2\\&\quad +(\Psi _2[{\textsf{f}}']-\Psi _1^2[{\textsf{f}}'])\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}']. \end{aligned}$$

Remembering (131) and applying repeatedly Corollary 7.1, we get that the first three terms of (136) can be handled in the following way:

$$\begin{aligned}&\sum _{i=1}^{3}\int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| \partial _{\textbf{x}}^{3-i}\Psi _1[{\textsf{f}}'](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^{i}({\textsf{f}}'-{\textsf{f}}'')](\textbf{x}) \right| ^2 |{\bar{\textbf{x}}}|^{2}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\&\quad \le C(\varepsilon ) \Vert {\textsf{f}}'-{\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })}^2. \end{aligned}$$

The remaining three terms of (136) follows by auxiliary Lemma 7.3. Notice that using (131) we get

$$\begin{aligned}{} & {} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| (\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^3{\textsf{f}}''](\textbf{x}) \right| ^2 |{\bar{\textbf{x}}}|^{2}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\{} & {} \quad \le \Vert {\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })} \int _{D_{\varepsilon }}\left( \sup _{{\bar{\textbf{x}}}\in D_\varepsilon (y)}\left| (\Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \right| ^2 |{\bar{\textbf{x}}}|^{2} \right) \hbox {d}\textbf{x},\\{} & {} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| (\partial _{\textbf{x}}\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}''](\textbf{x}) \right| ^2 |{\bar{\textbf{x}}}|^{2}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\{} & {} \quad \le \Vert {\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| (\partial _{\textbf{x}}\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \right| ^2 |{\bar{\textbf{x}}}|^{2}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}, \end{aligned}$$

and

$$\begin{aligned}{} & {} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| (\partial _{\textbf{x}}^2\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}^2\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x}) \right| ^2 |{\bar{\textbf{x}}}|^{2}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\{} & {} \quad \le \Vert {\textsf{f}}''\Vert _{H^{4,3}(D_{\varepsilon })} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| (\partial _{\textbf{x}}^2\Psi _1[{\textsf{f}}']-\partial _{\textbf{x}}^2\Psi _1[{\textsf{f}}''])(\textbf{x},{\bar{\textbf{x}}}) \right| ^2 |{\bar{\textbf{x}}}|^{4}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}. \end{aligned}$$

Using auxiliary Lemma 7.3 on the last terms of the above expressions we have proved our goal. $\square$

Lemma 7.6

Let ${\textsf{f}}\in {\mathbb {B}}_{\delta }(H^{4,3}(D_{\varepsilon }))$ with $0<\delta (\varepsilon )\ll 1$ small enough and $h\in H^{4,3}(D_{\varepsilon })$ with $\Vert h\Vert _{H^{4,3}(D_{\varepsilon })}=1.$ For $0<\tau \ll 1$, the following bound holds

• i)
  $$\begin{aligned} \sup _{\textbf{x}\in D_{\varepsilon }}\left( \sup _{{\bar{\textbf{x}}}\in D_{\varepsilon }(y)}\left| \frac{{\mathcal {K}}[{\textsf{f}}+\tau h,{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})}{\tau }-\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right| ^2 \right) \le C(\varepsilon )\tau ^2. \end{aligned}$$
• ii)
  $$\begin{aligned} \sup _{\textbf{x}\in D_{\varepsilon }}\left( \int _{D_{\varepsilon }(y)}\left| \partial _{\textbf{x}}\left\{ \frac{{\mathcal {K}}[{\textsf{f}}+\tau h,{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})}{\tau }-\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right\} \right| ^2\hbox {d}{\bar{\textbf{x}}} \right) \le C(\varepsilon )\tau ^2. \end{aligned}$$
• iii)
  $$\begin{aligned} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| \partial _{\textbf{x}}^2\left\{ \frac{{\mathcal {K}}[{\textsf{f}}+\tau h,{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})}{\tau }-\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right\} \right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\le C(\varepsilon )\tau ^2. \end{aligned}$$
• iv)
  $$\begin{aligned} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| \partial _{\textbf{x}}^3\left\{ \frac{{\mathcal {K}}[{\textsf{f}}+\tau h,{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})}{\tau }-\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right\} \right| ^2 |{\bar{\textbf{x}}}|^{2}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\le C(\varepsilon )\tau ^2. \end{aligned}$$

Proof of (i)

Just applying the standard logarithmic inequality $1-x^{-1}\le \log x\le x-1$ for all $x>0$, we get the lower and upper bounds

$$\begin{aligned} {\mathcal {K}}[{\textsf{f}}+\tau h,{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})&\le \frac{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x})\right) -\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x})\right) }{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x})\right) -\cos ({\bar{x}})},\\ {\mathcal {K}}[{\textsf{f}}+\tau h,{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})&\ge \frac{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x})\right) -\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x})\right) }{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x})\right) -\cos ({\bar{x}})}. \end{aligned}$$

Therefore, we have

$$\begin{aligned}{} & {} \left| \frac{{\mathcal {K}}[{\textsf{f}}+\tau h,{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})}{\tau }-\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right| \\{} & {} \quad \le \max _{0\le \xi \le \tau }\left| \frac{1}{\tau }\left( \frac{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x})\right) -\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x})\right) }{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\xi h](\textbf{x})\right) -\cos ({\bar{x}})}\right) -\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right| . \end{aligned}$$

Now, using the trigonometric identity $\cosh (a+b)=\cosh (a)\cosh (b)+\sinh (a)\sinh (b)$ on the above expression, we get

$$\begin{aligned}{} & {} \frac{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x})\right) -\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x})\right) }{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\xi h](\textbf{x})\right) -\cos ({\bar{x}})}\\{} & {} \quad =\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](x))\frac{\cosh \left( \Delta _{{\bar{\textbf{x}}}}[\tau h](\textbf{x})\right) -1}{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\xi h](\textbf{x})\right) -\cos ({\bar{x}})}\\{} & {} \qquad +\sinh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](x))\frac{\sinh \left( \Delta _{{\bar{\textbf{x}}}}[\tau h](\textbf{x})\right) }{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\xi h](\textbf{x})\right) -\cos ({\bar{x}})}, \end{aligned}$$

and consequently

$$\begin{aligned}{} & {} \left| \frac{{\mathcal {K}}[{\textsf{f}}+\tau h,{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})}{\tau }-\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right| \\{} & {} \quad \le \left| \frac{\cosh \left( \Delta _{{\bar{\textbf{x}}}}[\tau h](\textbf{x})\right) -1}{\tau }\right| \max _{0\le \xi \le \tau }\left| \frac{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](x))}{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\xi h](\textbf{x})\right) -\cos ({\bar{x}})}\right| \\{} & {} \qquad +\left| \frac{\sinh \left( \Delta _{{\bar{\textbf{x}}}}[\tau h](\textbf{x})\right) }{\tau }-\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right| \max _{0\le \xi \le \tau }\left| \frac{\sinh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](x))}{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\xi h](\textbf{x})\right) -\cos ({\bar{x}})}\right| . \end{aligned}$$

Applying the Taylor expansion of the trigonometric functions $\sinh (\cdot ), \cosh (\cdot )$ together with the fact that $\Vert h\Vert _{H^3(D_{\varepsilon })}=1$ we get the bounds

$$\begin{aligned} \left| \frac{\cosh \left( \Delta _{{\bar{\textbf{x}}}}[\tau h](\textbf{x})\right) -1}{\tau }\right| \le C(\varepsilon )\tau |{\bar{\textbf{x}}}|^2,\\ \left| \frac{\sinh \left( \Delta _{{\bar{\textbf{x}}}}[\tau h](\textbf{x})\right) }{\tau }-\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right| \le C(\varepsilon )\tau ^2 |{\bar{\textbf{x}}}|^3. \end{aligned}$$

In addition, we have

$$\begin{aligned}{} & {} \max \left\{ \left| \frac{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](x))}{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\xi h](\textbf{x})\right) -\cos ({\bar{x}})}\right| ,\left| \frac{\sinh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](x))}{\cosh \left( {\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\xi h](\textbf{x})\right) -\cos ({\bar{x}})}\right| \right\} \\{} & {} \quad \le \frac{C(\Vert {\textsf{f}}\Vert _{H^{3}(D_{\varepsilon })})}{\left( \tfrac{1}{2}-\Vert {\textsf{f}}+\xi h\Vert _{H^{3}(D_{\varepsilon })}\right) |{\bar{\textbf{x}}}|^2-C(\Vert {\textsf{f}}+\xi h\Vert _{H^{3}(D_{\varepsilon })}) |{\bar{\textbf{x}}}|^4}, \end{aligned}$$

which give us

$$\begin{aligned}{} & {} \left| \frac{{\mathcal {K}}[{\textsf{f}}+\tau h,{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})}{\tau }-\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right| \\{} & {} \quad \le \max _{0\le \xi \le \tau }\frac{C(\varepsilon ,\Vert {\textsf{f}}\Vert _{H^{3}(D_{\varepsilon })}) \tau }{\left( \tfrac{1}{2}-\Vert {\textsf{f}}+\xi h\Vert _{H^{3}(D_{\varepsilon })}\right) -C(\Vert {\textsf{f}}+\xi h\Vert _{H^{3}(D_{\varepsilon })}) |{\bar{\textbf{x}}}|^2}. \end{aligned}$$

Coming back to the starting point, as ${\textsf{f}}\in {\mathbb {B}}_{\delta }(X(D_{\varepsilon }))$ and $\Vert h\Vert _{H^{4,3}(D_{\varepsilon })}=1,$ taking $0<\delta (\varepsilon ),\tau \ll 1$ small enough we have proved our desired inequality. $\square$

Proof of (ii)

As ${\mathcal {K}}({\textsf{f}}',{\textsf{f}}'')(\textbf{x},{\bar{\textbf{x}}})=\left( K[{\textsf{f}}']-K[{\textsf{f}}'']\right) (\textbf{x},{\bar{\textbf{x}}})$ and $\partial _{\textbf{x}}K[g](\textbf{x},{\bar{\textbf{x}}})=\Psi _1[g](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x})$, adding and subtracting some appropriate term we obtain

$$\begin{aligned} \partial _{\textbf{x}}{\mathcal {K}}[{\textsf{f}}',{\textsf{f}}''](\textbf{x},{\bar{\textbf{x}}})=\Psi _1[{\textsf{f}}'](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}({\textsf{f}}'-{\textsf{f}}'')](\textbf{x})+\left( \Psi _1[{\textsf{f}}']-\Psi _1[{\textsf{f}}'']\right) (\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}''](\textbf{x}). \end{aligned}$$

So, it is just a matter of algebra to get

$$\begin{aligned} \partial _{\textbf{x}}\left\{ \frac{{\mathcal {K}}[{\textsf{f}}+\tau h,{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})}{\tau }-\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right\} = {\textsf{A}}[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{B}}[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}), \nonumber \\ \end{aligned}$$

(137)

where both new functionals are given respectively by

$$\begin{aligned} {\textsf{A}}[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&:=\left[ \frac{(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau }-(\Psi _2[{\textsf{f}}]-\Psi _1^2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right] \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x}),\nonumber \\ {\textsf{B}}[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&:=(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}h](\textbf{x}). \end{aligned}$$

(138)

Directly from auxiliary Lemma 7.3, as $h\in H^{4,3}(D_{\varepsilon })$ implies $\partial _{\textbf{x}}h\in H^{3,2}(D_{\varepsilon })\subset C^{1}(D_{\varepsilon })$ and by hypothesis $\Vert h\Vert _{H^{4,3}(D_{\varepsilon })}=1,$ we get

$$\begin{aligned} \sup _{\textbf{x}\in D_{\varepsilon }} \int _{D_{\varepsilon }(y)}\left| {\textsf{B}}[{\textsf{f}},h]\right| ^2\hbox {d}{\bar{\textbf{x}}}\le C(\varepsilon )\tau ^2. \end{aligned}$$

To obtain something similar for ${\textsf{A}}[{\textsf{f}},h]$ we need to work a little bit more. In first place, thanks to the definition of $\Psi _1[\cdot ](\textbf{x},{\bar{\textbf{x}}})$, adding and subtracting some appropriate term we have

$$\begin{aligned}{} & {} \left( \Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}]\right) (\textbf{x},{\bar{\textbf{x}}})\\{} & {} \quad =\frac{\sinh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x}))-\sinh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x}))}{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x}))-\cos ({\bar{x}})} -\sinh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x}))\\{} & {} \qquad \frac{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x}))-\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x}))}{\left[ \cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x}))-\cos ({\bar{x}})\right] \left[ \cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x}))-\cos ({\bar{x}})\right] }, \end{aligned}$$

and using on it some trigonometric identities, we obtain

$$\begin{aligned}{} & {} \left( \Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}]\right) (\textbf{x},{\bar{\textbf{x}}})\\{} & {} \quad =\left[ \Psi _2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})-\Psi _1^2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\cosh \left( \tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) \right] \sinh \left( \tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) +\Psi _1^2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\\{} & {} \qquad \frac{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x})-\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x})}{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x})-\cos ({\bar{x}})}\sinh \left( \tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) \cosh \left( \tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) . \end{aligned}$$

Now, dividing the above expression by $\tau$ and subtracting the term $(\Psi _2[{\textsf{f}}]-\Psi _1^2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})$ we obtain that

$$\begin{aligned}{} & {} \frac{(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau }-(\Psi _2[{\textsf{f}}]-\Psi _1^2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\\{} & {} \quad =\Psi _2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\left( \frac{\sinh (\tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x}))}{\tau } - \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) \\{} & {} \qquad -\Psi _1^2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\left( \cosh (\tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x}))\frac{\sinh (\tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x}))}{\tau } - \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) \\{} & {} \qquad +\Psi _1^2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}}) \frac{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x})-\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x})}{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x})-\cos ({\bar{x}})}\\{} & {} \qquad \sinh \left( \tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) \cosh \left( \tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) . \end{aligned}$$

Finally, using the Taylor expansion of the trigonometric functions $\sinh (\cdot ), \cosh (\cdot )$ and (131) give us the desired inequality

$$\begin{aligned} \sup _{\textbf{x}\in D_{\varepsilon }} \int _{D_{\varepsilon }(y)}\left| {\textsf{A}}[{\textsf{f}},h]\right| ^2\hbox {d}{\bar{\textbf{x}}}\le C(\varepsilon )\tau ^2. \end{aligned}$$

(139)

$\square$

Proof of (iii)

Recalling (137), we have

$$\begin{aligned} \partial _{\textbf{x}}^2\left\{ \frac{{\mathcal {K}}[{\textsf{f}}+\tau h,{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})}{\tau }-\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right\} = \partial _{\textbf{x}}{\textsf{A}}[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+\partial _{\textbf{x}}{\textsf{B}}[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}), \nonumber \\ \end{aligned}$$

(140)

where, using relation $\partial _{\textbf{x}}\Psi _1[g](\textbf{x},{\bar{\textbf{x}}})={\tilde{{\textsf{I}}}}_1[g](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}g](\textbf{x})$ whit ${\tilde{{\textsf{I}}}}_1[g]=\left( \Psi _2[g]\right. \left. -(\Psi _1[g])^2\right)$, we get

$$\begin{aligned} \partial _{\textbf{x}}{\textsf{A}}[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})= & {} ({\tilde{{\textsf{I}}}}_1[{\textsf{f}}+\tau h]-{\tilde{{\textsf{I}}}}_1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}h](\textbf{x})\\{} & {} \quad +\frac{(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau }\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}](\textbf{x})\\{} & {} \quad -{\tilde{{\textsf{I}}}}_1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}]\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\\{} & {} \quad +\frac{({\tilde{{\textsf{I}}}}_1[{\textsf{f}}+\tau h]-{\tilde{{\textsf{I}}}}_1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau }\left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\right) ^2\\{} & {} \quad -\left( \partial _{\textbf{x}}\Psi _2[{\textsf{f}}]-2\Psi _1[{\textsf{f}}]\partial _{\textbf{x}}\Psi _1[{\textsf{f}}]\right) (\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x}), \end{aligned}$$

and

$$\begin{aligned} \partial _{\textbf{x}}{\textsf{B}}[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})= & {} ({\tilde{{\textsf{I}}}}_1[{\textsf{f}}+\tau h]-{\tilde{{\textsf{I}}}}_1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}h](\textbf{x})\\{} & {} +(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2 h](\textbf{x})\\{} & {} + \tau {\tilde{{\textsf{I}}}}_1[{\textsf{f}}+\tau h](\textbf{x},{\bar{\textbf{x}}})\left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}} h](\textbf{x})\right) ^2. \end{aligned}$$

By definition of ${\tilde{{\textsf{I}}}}_1[\cdot ]$ and relations (126) and (127), the previous expressions can be rewritten in a more manageable way as

$$\begin{aligned} \partial _{\textbf{x}}{\textsf{A}}[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&=({\textsf{A}}_1'+{\textsf{A}}_2'+{\textsf{A}}_3'+{\textsf{A}}_4'+{\textsf{A}}_5')[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}), \end{aligned}$$

(141)

$$\begin{aligned} \partial _{\textbf{x}}{\textsf{B}}[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&=({\textsf{B}}_1'+{\textsf{B}}_2'+{\textsf{B}}_3'+{\textsf{B}}_4')[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}), \end{aligned}$$

(142)

with

$$\begin{aligned} {\textsf{A}}_1'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&:=\left[ \frac{(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau }-(\Psi _2[{\textsf{f}}]-\Psi _1^2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right] \\&\qquad \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}](\textbf{x}),\\ {\textsf{A}}_2'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&:=(\Psi _2[{\textsf{f}}+\tau h]-\Psi _2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}h](\textbf{x}),\\ {\textsf{A}}_3'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&:=-(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\Psi _1[{\textsf{f}}+\tau h]+\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}h](\textbf{x}),\\ {\textsf{A}}_4'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&:=\left[ \frac{(\Psi _2[{\textsf{f}}+\tau h]-\Psi _2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau } -\left( \Psi _1[{\textsf{f}}]-\Psi _1[{\textsf{f}}]\Psi _2[{\textsf{f}}]\right) (\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right] \\&\qquad \left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\right) ^2,\\ {\textsf{A}}_5'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&:=\bigg [-(\Psi _1[{\textsf{f}}+\tau h]+\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\frac{(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau }\\&\qquad +2\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}}) (\Psi _2[{\textsf{f}}]-\Psi _1^2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\bigg ]\left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\right) ^2, \end{aligned}$$

and

$$\begin{aligned} {\textsf{B}}_1'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&:=(\Psi _2[{\textsf{f}}+\tau h]-\Psi _2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}h](\textbf{x}),\\ {\textsf{B}}_2'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&:=-(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\Psi _1[{\textsf{f}}+\tau h]\\&\qquad +\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}h](\textbf{x}),\\ {\textsf{B}}_3'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&:= \tau (\Psi _2[{\textsf{f}}+\tau h]-\Psi _1^2[{\textsf{f}}+\tau h])(\textbf{x},{\bar{\textbf{x}}})\left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}} h](\textbf{x})\right) ^2,\\ {\textsf{B}}_4'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&:=(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2 h](\textbf{x}). \end{aligned}$$

Note that the only new type of term that appear is ${\textsf{A}}_4'[{\textsf{f}},h].$ The rest of terms can be easily handle using auxiliary Lemma 7.3, Corollary 7.1 and (131). To sum up, our proof reduces to check that

$$\begin{aligned} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| {\textsf{A}}_4'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})\right| ^2 |{\bar{\textbf{x}}}|^{2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\le C(\varepsilon ,\Vert {\textsf{f}}\Vert _{H^{3}(D_{\varepsilon })})\tau ^2. \end{aligned}$$

(143)

By (131), as ${\textsf{f}}\in H^{4,3}(D_{\varepsilon })$ we have that $\partial _{\textbf{x}}{\textsf{f}}\in H^{3,2}(D_{\varepsilon })\subset C^{1}(D_{\varepsilon })$ and we just need to prove the same type of estimate for the term

$$\begin{aligned}{} & {} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| \frac{(\Psi _2[{\textsf{f}}+\tau h]-\Psi _2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau }-\left( \Psi _1[{\textsf{f}}]-\Psi _1[{\textsf{f}}]\Psi _2[{\textsf{f}}]\right) (\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right| ^2 |{\bar{\textbf{x}}}|^{4+2\gamma }\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}. \end{aligned}$$

Thanks to the definition of $\Psi _2[\cdot ](\textbf{x},{\bar{\textbf{x}}})$, adding and subtracting some appropriate term we have

$$\begin{aligned}{} & {} \left( \Psi _2[{\textsf{f}}+\tau h]-\Psi _2[{\textsf{f}}]\right) (\textbf{x},{\bar{\textbf{x}}})\\{} & {} \quad =\frac{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x}))-\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x}))}{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x}))-\cos ({\bar{x}})} -\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x}))\\{} & {} \qquad \frac{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x}))-\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x}))}{\left[ \cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x}))-\cos ({\bar{x}})\right] \left[ \cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x}))-\cos ({\bar{x}})\right] }, \end{aligned}$$

and using on it some trigonometric identities, we obtain

$$\begin{aligned}{} & {} \left( \Psi _2[{\textsf{f}}+\tau h]-\Psi _2[{\textsf{f}}]\right) (\textbf{x},{\bar{\textbf{x}}})\\{} & {} \quad =\left[ \Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})-\Psi _1[{\textsf{f}}]\Psi _2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\cosh \left( \tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) \right] \sinh \left( \tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) \\{} & {} \qquad -\Psi _1^2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\sinh ^2(\tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})) +2\Psi _1[{\textsf{f}}]\Psi _2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\sinh (\tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})) \\{} & {} \qquad \quad \sinh ^2\left( \tfrac{\tau }{2}\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) \\{} & {} \qquad +2\Psi _2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\left[ 1-\Psi _2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\cosh (\tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x}))\right] \sinh ^2\left( \tfrac{\tau }{2}\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) \\{} & {} \qquad +\Psi _1[{\textsf{f}}+\tau h](\textbf{x},{\bar{\textbf{x}}})\left[ \frac{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x}))-\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x}))}{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x}))-\cos ({\bar{x}})}\right] ^2. \end{aligned}$$

Now, dividing the above expression by $\tau$ and subtracting the term $(\Psi _1[{\textsf{f}}]-\Psi _1[{\textsf{f}}]\Psi _2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})$ we obtain that

$$\begin{aligned}{} & {} \frac{(\Psi _2[{\textsf{f}}+\tau h]-\Psi _2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau } -\left( \Psi _1[{\textsf{f}}]-\Psi _1[{\textsf{f}}]\Psi _2[{\textsf{f}}]\right) (\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\\{} & {} \quad =\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\left( \frac{\sinh \left( \tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) }{\tau }-\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x}) \right) \\{} & {} \qquad -\Psi _1[{\textsf{f}}]\Psi _2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\left( \cosh \left( \tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) \frac{\sinh \left( \tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) }{\tau }-\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x}) \right) \\{} & {} \qquad -\Psi _1^2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\sinh ^2(\tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})) \\{} & {} \qquad +2\Psi _1[{\textsf{f}}]\Psi _2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\sinh (\tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})) \sinh ^2\left( \tfrac{\tau }{2}\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) \\{} & {} \qquad +2\Psi _2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\left[ 1-\Psi _2[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\cosh (\tau \Delta _{{\bar{\textbf{x}}}}[h](\textbf{x}))\right] \sinh ^2\left( \tfrac{\tau }{2}\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right) \\{} & {} \qquad +\Psi _1[{\textsf{f}}+\tau h](\textbf{x},{\bar{\textbf{x}}})\left[ \frac{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}+\tau h](\textbf{x}))-\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x}))}{\cosh ({\bar{y}}+\Delta _{{\bar{\textbf{x}}}}[{\textsf{f}}](\textbf{x}))-\cos ({\bar{x}})}\right] ^2. \end{aligned}$$

Finally, using the Taylor expansion of the trigonometric functions $\sinh (\cdot ), \cosh (\cdot )$ and (131) give us the desired inequality (143). $\square$

Proof of (iv)

Recalling (140) together with (141) and (142), we have

$$\begin{aligned}{} & {} \partial _{\textbf{x}}^3\left\{ \frac{{\mathcal {K}}[{\textsf{f}}+\tau h,{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})}{\tau }-\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right\} \nonumber \\{} & {} \quad = \sum _{i=1}^{5}\partial _{\textbf{x}}{\textsf{A}}_i'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+\sum _{i=1}^{4}\partial _{\textbf{x}}{\textsf{B}}_i'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}), \end{aligned}$$

(144)

where,

and

$$\begin{aligned}{} & {} \partial _{\textbf{x}}{\textsf{B}}_1'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})=(\partial _{\textbf{x}}\Psi _2[{\textsf{f}}+\tau h]-\partial _{\textbf{x}}\Psi _2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}h](\textbf{x})\\{} & {} \qquad +(\Psi _2[{\textsf{f}}+\tau h]-\Psi _2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\left[ \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}h](\textbf{x})+\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2h](\textbf{x})\right] \\{} & {} \quad =:{\textsf{B}}_{11}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{B}}_{12}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}),\\{} & {} \partial _{\textbf{x}}{\textsf{B}}_2'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})=-(\partial _{\textbf{x}}\Psi _1[{\textsf{f}}+\tau h]-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}])(\Psi _1[{\textsf{f}}+\tau h]+\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\\{} & {} \qquad \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}h](\textbf{x})\\{} & {} \qquad \qquad \qquad \qquad \qquad -(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\partial _{\textbf{x}}\Psi _1[{\textsf{f}}+\tau h]\\{} & {} \qquad \qquad \qquad \qquad \qquad +\partial _{\textbf{x}}\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}h](\textbf{x})\\{} & {} \qquad \qquad \qquad \qquad \qquad -(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\Psi _1[{\textsf{f}}+\tau h]\\{} & {} \qquad \qquad \qquad \qquad \qquad +\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}h](\textbf{x})\\{} & {} \qquad \qquad \qquad \qquad \qquad -(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\Psi _1[{\textsf{f}}+\tau h]\\{} & {} \qquad \qquad \qquad \qquad \qquad +\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2h](\textbf{x})\\{} & {} \quad =:{\textsf{B}}_{21}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{B}}_{22}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})\\{} & {} \qquad \qquad \qquad \qquad \qquad \quad +{\textsf{B}}_{23}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{B}}_{24}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}),\\{} & {} \partial _{\textbf{x}}{\textsf{B}}_3'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})= \tau (\partial _{\textbf{x}}\Psi _2[{\textsf{f}}+\tau h]\\{} & {} \qquad \qquad \qquad \qquad \qquad -2\Psi _1[{\textsf{f}}+\tau h]\partial _{\textbf{x}}\Psi _1[{\textsf{f}}+\tau h])(\textbf{x},{\bar{\textbf{x}}})\left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}} h](\textbf{x})\right) ^2\\{} & {} \qquad \qquad \qquad \qquad \qquad +\tau (\Psi _2[{\textsf{f}}+\tau h]-\Psi _1^2[{\textsf{f}}+\tau h])(\textbf{x},{\bar{\textbf{x}}})2 \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}} h](\textbf{x})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2 h](\textbf{x})\\{} & {} \quad =:{\textsf{B}}_{31}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{B}}_{32}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}),\\{} & {} \partial _{\textbf{x}}{\textsf{B}}_4'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})= (\partial _{\textbf{x}}\Psi _1[{\textsf{f}}+\tau h]-\partial _{\textbf{x}}\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2 h](\textbf{x})\\{} & {} \qquad \qquad \qquad \qquad \qquad +(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^3 h](\textbf{x})\\{} & {} \quad =:{\textsf{B}}_{41}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{B}}_{42}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}). \end{aligned}$$

Taking into account all the above we have proved that

$$\begin{aligned}{} & {} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| \partial _{\textbf{x}}^3\left\{ \frac{{\mathcal {K}}[{\textsf{f}}+\tau h,{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})}{\tau }-\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right\} \right| ^2 |{\bar{\textbf{x}}}|^{2}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\{} & {} \quad \lesssim \sum _{i=1}^{5} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| \partial _{\textbf{x}}{\textsf{A}}_i'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})\right| ^2 |{\bar{\textbf{x}}}|^{2}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\\{} & {} \qquad + \sum _{i=1}^{4} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| \partial _{\textbf{x}}{\textsf{B}}_i'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})\right| ^2 |{\bar{\textbf{x}}}|^{2}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}, \end{aligned}$$

where each of the terms that appear can be decomposed as follows:

$$\begin{aligned} \partial _{\textbf{x}}{\textsf{A}}_1'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&={\textsf{A}}_{11}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{A}}_{12}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}),\\ \partial _{\textbf{x}}{\textsf{A}}_2'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&={\textsf{A}}_{21}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{A}}_{22}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{A}}_{23}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}),\\ \partial _{\textbf{x}}{\textsf{A}}_3'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&={\textsf{A}}_{31}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{A}}_{32}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{A}}_{33}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{A}}_{34}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}),\\ \partial _{\textbf{x}}{\textsf{A}}_4'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&={\textsf{A}}_{41}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{A}}_{42}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}),\\ \partial _{\textbf{x}}{\textsf{A}}_5'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&={\textsf{A}}_{51}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{A}}_{52}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}), \end{aligned}$$

and

$$\begin{aligned} \partial _{\textbf{x}}{\textsf{B}}_1'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&={\textsf{B}}_{11}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{B}}_{12}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}),\\ \partial _{\textbf{x}}{\textsf{B}}_2'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&={\textsf{B}}_{21}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{B}}_{22}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{B}}_{23}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{B}}_{24}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}),\\ \partial _{\textbf{x}}{\textsf{B}}_3'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&={\textsf{B}}_{31}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{B}}_{32}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}),\\ \partial _{\textbf{x}}{\textsf{B}}_4'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})&={\textsf{B}}_{41}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{B}}_{42}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}). \end{aligned}$$

For the sake of brevity, we shall present here the complete details for the term ${\textsf{A}}_{51}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})$ and the other terms can be dealt using the previous lemmas via straightforward variations. To sum up, our goal reduces to check that

$$\begin{aligned} \int _{D_{\varepsilon }}\left( \int _{D_\varepsilon (y)}\left| {\textsf{A}}_{51}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})\right| ^2 |{\bar{\textbf{x}}}|^{2}\hbox {d}{\bar{\textbf{x}}} \right) \hbox {d}\textbf{x}\le C(\varepsilon )\tau ^2. \end{aligned}$$

(145)

Notice that ${\textsf{A}}_{51}''[{\textsf{f}},h]$ can be written in a more manageable way (adding and subtracting terms) as

$$\begin{aligned} {\textsf{A}}_{51}''[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})= & {} 2(\Psi _1[{\textsf{f}}]-\Psi _1[{\textsf{f}}+\tau h])(\textbf{x},{\bar{\textbf{x}}}) \partial _{\textbf{x}}\\{} & {} \left\{ \frac{(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau } \right\} \left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\right) ^2\\{} & {} -2 \partial _{\textbf{x}}\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}}) \left\{ \frac{(\Psi _1[{\textsf{f}}+\tau h] -\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau } \right. \\{} & {} \left. -(\Psi _2[{\textsf{f}}]-\Psi _1^2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right\} \left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\right) ^2\\{} & {} -2 \Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\partial _{\textbf{x}}\left\{ \frac{(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau }-(\Psi _2[{\textsf{f}}]\right. \\{} & {} \left. -\Psi _1^2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right\} \left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\right) ^2. \end{aligned}$$

Now, each one of the above terms can be easily handled. Using (131) and applying repeatedly Lemma 7.3 we get that the first term is bounded as required. For the second term, we only need to note that it can be written, remembering (138), as

$$\begin{aligned}{} & {} 2 \partial _{\textbf{x}}\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}}) \left\{ \frac{(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau } \right. \\{} & {} \quad \left. - (\Psi _2[{\textsf{f}}]-\Psi _1^2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right\} \left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\right) ^2\\{} & {} \quad =2 \partial _{\textbf{x}}\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}}) {\textsf{A}}[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x}). \end{aligned}$$

Combining (126) with Corollary 7.2 and (131) we obtain the bound $2 |\partial _{\textbf{x}}\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})|\le C(\varepsilon ).$ After that, as an immediate consequence of (139) we get the required bound for the second term. For the latter, we proceed in the same spirit as before, trying to split that term into previously studied terms. Notice that

$$\begin{aligned}{} & {} 2 \Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\partial _{\textbf{x}}\left\{ \frac{(\Psi _1[{\textsf{f}}+\tau h]-\Psi _1[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})}{\tau }-(\Psi _2[{\textsf{f}}]-\Psi _1^2[{\textsf{f}}])(\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[h](\textbf{x})\right\} \\{} & {} \qquad \left( \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})\right) ^2\\{} & {} \quad =2 \Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\partial _{\textbf{x}}{\textsf{A}}[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})-2 \Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}}){\textsf{A}}[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}}) \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}^2{\textsf{f}}](\textbf{x})\\{} & {} \quad =2 \Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}}) \left\{ {\textsf{A}}_1'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{A}}_4'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})+{\textsf{A}}_5'[{\textsf{f}},h](\textbf{x},{\bar{\textbf{x}}})\right\} \Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x}). \end{aligned}$$

Using Corollary 7.1 and (131) we obtain the bound $2 |\Psi _1[{\textsf{f}}](\textbf{x},{\bar{\textbf{x}}})\Delta _{{\bar{\textbf{x}}}}[\partial _{\textbf{x}}{\textsf{f}}](\textbf{x})|\le C(\varepsilon )$. Finally, working as we did to get inequality (143) we get the required bound for each of the above terms. Therefore, combining all we have proved our goal. $\square$