A Stefan-type stochastic moving boundary problem

Abstract

Motivated by applications in economics and finance, in particular to the modeling of limit order books, we study a class of stochastic second-order PDEs with non-linear Stefan-type boundary interaction. To solve the equation we transform the problem from a moving boundary problem into a stochastic evolution equation with fixed boundary conditions. Using results from interpolation theory we obtain existence and uniqueness of local strong solutions, extending results of Kim, Zheng and Sowers. In addition, we formulate conditions for existence of global solutions and provide a refined analysis of possible blow-up behavior in finite time.

Appendices

Appendix 1: The Nemytskii operator on Sobolev spaces

In this section we prove some regularity results on the Nemytskii operator N,

$$\begin{aligned} N(u)(x) = \mu (x,u(x)), \end{aligned}$$

on the Sobolev spaces $H^k(\mathbb {R}_+)$. Here, $\mu :\mathbb {R}_+ \times \mathbb {R}^d \rightarrow \mathbb {R}$ and $x\in \mathbb {R}_+$. Note that these results are well-known, even for more general spaces, in the case of bounded domains, see e.g. [1, 32]. However, in the case of unbounded domains several additional conditions on $\mu$ are necessary to make them work. First, we state a result which guarantees, that under certain assumptions on $\mu$, N maps $H^k$ into $H^k$. For a proof we refer to [31, Theorem 1], of which it is a special case.

Lemma 6.1

For each integer $k\ge 1$ the space $H^k(\mathbb {R}_+)$ is a Banach algebra. In particular, there exists a constant c such that for all u, $v\in H^k(\mathbb {R}_+)$ it holds that $uv\in H^k(\mathbb {R}_+)$ and

$$\begin{aligned} \left|\left|uv\right|\right|_{H^k} \le c \left|\left|u\right|\right|_{H^k}\left|\left|v\right|\right|_{H^k}. \end{aligned}$$

Next, we adapt [31, Theorem 2] to our setting. For notational reasons we also introduce the Nemytskii operators

$$\begin{aligned} N_x(u)(x) := \left( \frac{\partial }{\partial x}\mu \right) (x,u(x)),\quad N_{y_j}(u)(x):=\left( \frac{\partial }{\partial y_j} \mu \right) (x,u(x)),\;j=1,...,d, \end{aligned}$$

for $\quad u\in H^k(\mathbb {R}_+; \mathbb {R}^d),\;x\in \mathbb {R}_+$. In order for N to map $H^k$ into $H^k$ again, we need certain growths restrictions, which is not the case on bounded domains.

Assumption 6.2

Assume $\mu \in C^m(\mathbb {R}_{\ge 0}\times \mathbb {R}^d, \mathbb {R})$ and

• (a) For each integer l, $0\le l\le m$ there exists an $a_l\in L^2(\mathbb {R}_+)$ and some $b_l:\mathbb {R}^d\rightarrow \mathbb {R}_+$ locally bounded, such that

  $$\begin{aligned} \left|D^{(l,0,...,0)} \mu (x,y)\right| \le a_l(x) + b_l(y)\left|y\right|,\quad \forall \,x\in \mathbb {R}_+,\,y\in \mathbb {R}^d \end{aligned}$$

• (b) For each multiindex $\alpha$ with $\alpha _1 \ne \left|\alpha \right| \le m$, the functions $\sup _{x\in \mathbb {R}_+} \left|D^\alpha \mu (x,.)\right|$ are locally bounded.

Assumption 6.3

Assume that $\mu \in C^{m}(\mathbb {R}_{\ge 0} \times \mathbb {R}^d, \mathbb {R})$ and $D^{\alpha }\mu (x,.)$ is locally Lipschitz for all multi-indices $\alpha$, $\left|\alpha \right|\le m$ with Lipschitz constants uniform in $x\in \mathbb {R}_{\ge 0}$, i. e. we assume that for all $r\ge 0$ there exists $L_r\ge 0$ such that

$$\begin{aligned} \left|D^{\alpha }\mu (x,y) - D^{\alpha }\mu (x,z)\right| \le L_r \left|y-z\right|. \end{aligned}$$

holds for all $x\in \mathbb {R}_{\ge 0}$ and y, $z\in \mathbb {R}^d$ with $\left|y\right|$, $\left|z\right|\le r$ and $\alpha$, $\left|\alpha \right|\le m$.

Remark 6.4

If $\mu$ satisfies Assumption 6.2 for some integer $m\ge 1$, then $\mu$ satisfies Assumption 6.3 for $m-1$.

Remark 6.5

Recall the Sobolev embeddings

$$\begin{aligned} H^{m+1}(\mathbb {R}_+) \hookrightarrow BUC^m(\mathbb {R}_+), \end{aligned}$$

where $BUC^m(\mathbb {R}_+)$ denotes the Banach space of functions with bounded and uniformly continuous derivatives up to order m. As usual $BUC^m(\mathbb {R}_+)$ is equipped with the $C^m$-norm. In the following, we will work with the $BUC^m$ representative of the elements in $H^{m+1}$ without further comment.

Theorem 6.6

If Assumption 6.2 holds for some integer $m\ge 1$, then the operator N is continuous from $(H^m(\mathbb {R}_+))^d$ into $H^m(\mathbb {R}_+)$.

Proof

We adapt the proof of [31, Theorem 2] for the domain $\mathbb {R}_+$ and the spaces $H^k$ by incorporating the additional growths assumptions. We proceed by induction and consider $m=1$, first. Since $\mu \in C^1(\mathbb {R}_{\ge 0}\times \mathbb {R}^d)$ we get immediately that N(u), $N_x(u)$ and $N_{y_j}(u)$, $j=1,...,d$ are bounded and continuous functions for $u\in H^1(\mathbb {R}_+; \mathbb {R}^d)$ fixed. Let now $(u^n) \subset H^1(\mathbb {R}_+; \mathbb {R}^d) \cap C^\infty (\mathbb {R}_{\ge 0};\mathbb {R}^d)$ such that $u^n \longrightarrow u$ in $H^1$. Then the convergence also takes place in $\left|\left|.\right|\right|_{\infty }$ and by the chain rule we can write

$$\begin{aligned} \frac{\,{\text {d}}}{\,{\text {d}}x} N(u^n) = N_x(u^n) + \sum _{j=1}^n N_{y_j}(u^n) \nabla u^n_j. \end{aligned}$$

(6.1)

By assumption,

$$\begin{aligned} \left|N(u^n)\right| \le a_0 + (b_0 (u^n)) \left|u^n\right|,\text { and }\left|N_x(u^n)\right| \le a_1 + (b_1(u^n)) \left|u^n\right|. \end{aligned}$$

Recall that $u^n$ are globally bounded by Sobolev embeddings and $b_0$, $b_1$, are locally bounded by assumption. Hence, $b_i\circ u^n$ are globally bounded for $n\in \mathbb {N}$, $i\in \{0,1\}$. Since $u^n\rightarrow u$ uniformly and in $L^2$ we get for each estimate that

$$\begin{aligned} \sup _{n\in \mathbb {N}} \left|\left|b_i(u^n)\right|\right|_{\infty } < \infty ,\quad \text { for }i \in \{0,1\}, \end{aligned}$$

and hence $N(u^n)$ and $N_x(u^n)$ are bounded by $L^2$-converging sequences. Hence, we can apply a version of Lebesgue’s dominated convergence [14, Theorem 1.21] and obtain the $L^2$ convergence of

$$\begin{aligned} N(u^n) \longrightarrow N(u),\quad \text {and}\quad N_x(u^n) \longrightarrow N_x(u),\;\text { as }n\rightarrow \infty . \end{aligned}$$

For the remaining summands we get

$$\begin{aligned} \left|\left|N_{y_j} (u^n) \nabla u^n_j - N_{y_j} (u) \nabla u_j\right|\right|_{L^2}\le & {} \sup _{n\in \mathbb {N}}\left|\left|N_{y_j}(u^n)\right|\right|_{\infty } \left|\left|\nabla u^n_j - \nabla u_j\right|\right|_{L^2} \\&+ \left|\left|\left( N_{y_j}(u^n) - N_{y_j}(u)\right) \nabla u_j\right|\right|_{L^2} \end{aligned}$$

which goes to 0 as $n\rightarrow \infty$. Indeed, uniform convergence of $(u^n)$ and dominated convergence yield $L^2$-convergence of $(N_{y_j}(u^n)\nabla u)$. Hence,

$$\begin{aligned} N(u^n) \xrightarrow []{H^1}N(u),\quad \text { as }n\rightarrow \infty . \end{aligned}$$

By completeness of $H^1$ this implies $N(u)\in H^1(\mathbb {R}_+)$ and also shows the continuity of N for the case $m=1$.

For the induction step from m to $m+1$ we may assume that the claim holds true for $m \ge 1$ and that Assumption 6.2 holds for $(m+1)$. Clearly, the assumption also holds for m and so N maps $H^{m+1}$ continuously into $H^m$ by induction hypothesis. It thus remains to show that also $\frac{\,{\text{ d }}}{\,{\text{ d }}x}N$ maps $H^{m+1}$ into $H^m$. We decompose $\frac{\,{\text{ d }}}{\,{\text{ d }}x}N$ as in (6.1) and note that Assumption 6.2 for m is also satisfied by $\frac{\partial }{\partial x}\mu$ and by

$$\begin{aligned} (x,y,z) \mapsto {\tilde{\mu }}_j(x,y,z) := \frac{\partial }{\partial y_j} \mu (x,y)z. \end{aligned}$$

(6.2)

Hence, by induction hypothesis the operators $N_x$ and ${\tilde{N}}_j$ are continuous from $H^{m}(\mathbb {R}_+)^d$, resp. $H^m (\mathbb {R}_+)^{d+1}$, into $H^m (\mathbb {R}_+)$, where ${\tilde{N}}_j$ is the Nemytskii operator defined by ${\tilde{\mu }}_j$, $j=1,...,d$. Since also $u\mapsto \nabla u$ is continuous from $H^{m+1}$ into $H^m$ and Lemma 6.1 shows continuity of multiplication, (6.1) yields continuity of $\frac{\,{\text{ d }}}{\,{\text{ d }}x}N$ from $H^{m+1}$ into $H^m$, as claimed. $\square$

Theorem 6.7

Let $\mu$ satisfy Assumptions 6.2 and 6.3 for some positive integer m. Then, N is Lipschitz continuous from bounded subsets of $(H^m(\mathbb {R}_+))^d$ into $H^m(\mathbb {R}_+)$.

Proof

We proceed as above, by induction on $m\in \mathbb {N}$. First, let $m=1$ and u, $v\in H^1(\mathbb {R}_+;\mathbb {R}^d)$ with $\left|\left|u\right|\right|_{H^1}$, $\left|\left|v\right|\right|_{H^1} \le r$. By continuity of N we can assume, w. l. o. g. u, $v\in H^1(\mathbb {R}_+;\mathbb {R}^d)\cap C^\infty (\mathbb {R}_{\ge 0}; \mathbb {R}^d)$. By Sobolev embeddings there exists a constant c s. t. $\left|\left|u\right|\right|_{\infty }$,$\left|\left|v\right|\right|_{\infty } \le cr$. By Assumption 6.3,

$$\begin{aligned} \left|\left|N(u) - N(v)\right|\right|_{L^2}&\le L_{cr} \left|\left|u-v\right|\right|_{L^2(\mathbb {R}_+;\mathbb {R}^d)},\\ \left|\left|N_x(u) - N_x(v)\right|\right|_{L^2}&\le L_{cr} \left|\left|u-v\right|\right|_{L^2(\mathbb {R}_+;\mathbb {R}^d)}, \end{aligned}$$

and for $j=1,...,d$,

$$\begin{aligned} \begin{aligned}&\left|\left|N_{y_j}(u) \nabla u_j - N_{y_j}(v) \nabla v_j\right|\right|_{L^2}\\&\quad \le \left|\left|N_{y_j}(u)\right|\right|_{\infty } \left|\left|\nabla u_j - \nabla v_j\right|\right|_{L^2} + \left|\left|N_{y_j}(u) - N_{y_j}(v)\right|\right|_{\infty } \left|\left|\nabla v_j\right|\right|_{L^2}\\&\quad \le K_{j,cr} \left|\left|\nabla u_j - \nabla v_j\right|\right|_{L^2} + L_{cr}\left|\left|u_j - v_j\right|\right|_{L^2}, \end{aligned} \end{aligned}$$

(6.3)

for $K_{j,cr} := \sup _{\left|y\right|\le cr} \sup _{x\in \mathbb {R}}\left|\frac{\partial }{\partial y}_j \mu (x,y)\right|<\infty .$ Chain rule (6.1) then yields the assertion for $m=1$.

For the induction step we may assume that the theorem holds for fixed m and that Assumptions 6.2 and 6.3 are satisfied for $m+1$. By induction hypothesis, N is Lipschitz on bounded sets from $H^m(\mathbb {R}_+)^d$ into $H^m(\mathbb {R}_+)$ and thus, also from $H^{m+1}(\mathbb {R}_+)^d$ into $H^m(\mathbb {R}_+)$. Hence, it suffices to show that $\frac{\,{\text{ d }}}{\,{\text{ d }}x}N$ is Lipschitz on bounded sets from $H^m(\mathbb {R}_+)^d$ into $H^m(\mathbb {R}_+)$. To this end note that $\frac{\partial }{\partial x}\mu$ satisfies Assumptions 6.2 and 6.3 as well as ${\tilde{\mu }}_j$, $j=1,...,d$, defined in (6.2). By induction hypothesis, the operators $N_x$ and ${\tilde{N}}_j$, $j=1,...,d$, defined in the proof of Theorem 6.6, are Lipschitz on bounded sets. Again, approximation by elements in $H^m\cap C^m$ and (6.1) then show that the same holds true for $\frac{\,{\text{ d }}}{\,{\text{ d }}x}N$. $\square$

Appendix 2: The noise operator

In this section we will study the operator-valued map $\mathcal {C}$, defined in (2.13) by

$$\begin{aligned} (\mathcal {C}(u)w)(x) = \begin{pmatrix} \sigma _+(x,u_1(x)) (T_\zeta w)(u_3+x)\\ \sigma _-(-x,u_2(x)) (T_\zeta w)(u_3-x) \\ 0 \end{pmatrix} \end{aligned}$$

for $u\in \mathcal {D}(\mathcal {A})$, $w\in U$ and $x\in \mathbb {R}$. We can reduce the problem to the operator

$$\begin{aligned} (u,x_*)\mapsto \sigma (.,u(.)) T_\zeta (.+x_*) \end{aligned}$$

(7.1)

for $\sigma$ satisfying Assumption 2.3 and $\zeta$ as in Assumption 2.6. Define the Nemytskii operator

$$\begin{aligned} N_\sigma : H^2(\mathbb {R}_+) \rightarrow H^2(\mathbb {R}_+),\; u\mapsto \sigma (.,u(.)), \end{aligned}$$

which is Lipschitz on bounded sets by Theorem 6.7.

Lemma 7.1

Multiplication is bilinear continuous from $H^2(\mathbb {R}_+) \times BUC^2(\mathbb {R}_{\ge 0})$ into $H^2(\mathbb {R}_+)$.

Proof

By density of $C^n \cap H^n$ in $H^n$, $n\in \mathbb {N}$, one can check that Leibniz formula holds for multiplication on $H^n\times BUC^n$, so that

$$\begin{aligned} D^k(uf) = \sum _{j=0}^k \left( {\begin{array}{c}k\\ j\end{array}}\right) D^{j}u f^{(k-j)} \end{aligned}$$

which is clearly square integrable for $u\in H^n(\mathbb {R}_+)$, $f\in BUC^n(\mathbb {R}_{\ge 0})$ and $k\le n$. In particular, for $n=2$,

$$\begin{aligned} \left|D^k(uf)(x)\right| \le K \left|\left|f\right|\right|_{C^2} \sum _{j=0}^k D^{j}u(x),\quad k\le 2 \end{aligned}$$

for some constant K, so that for some ${\tilde{K}}$,

$$\begin{aligned} \left|\left|uf\right|\right|_{H^2} \le {\tilde{K}} \left|\left|f\right|\right|_{C^2}\left|\left|u\right|\right|_{H^2}. \end{aligned}$$

(7.2)

$\square$

Lemma 7.2

$T_\zeta$ maps U into $BUC^2(\mathbb {R})$. Moreover, $T_\zeta w$ and its first two derivatives are Lipschitz continuous for all $w\in U$.

Proof

First note that for all $x\in \mathbb {R}$ and $i\in \{0,1,2\}$ it holds that

$$\begin{aligned} \frac{\,{\text {d}}}{\,{\text {d}}x} T_{\zeta ^{(i)}}w(x) = T_{\zeta ^{(i+1)}}w(x),\quad x\in \mathbb {R},\;w\in U. \end{aligned}$$

(7.3)

Indeed, for all $z\ne 0$, the fundamental theorem of calculus yields

$$\begin{aligned}&\frac{1}{z}\left|\int _\mathbb {R}\left( \zeta ^{(i)}(x+z,y) - \zeta ^{(i)}(x,y) - z\zeta ^{(i+1)}(x,y)\right) w(y)\,{\text {d}}y\right|\\&\quad \le \int _\mathbb {R}\int _0^1 \left|\zeta ^{(i+1)}(x+\epsilon z,y) - \zeta ^{(i+1)}(x,y)\right|\,{\text {d}}\epsilon \,\, d w(y) \,{\text {d}}y \\&\quad \le \left|\left|w\right|\right|_{L^2} \int _0^1 \left|\left|\zeta ^{(i+1)}(x+\epsilon z,.) - \zeta ^{(i+1)}(x,.)\right|\right|_{L^2} \,{\text {d}}\epsilon . \end{aligned}$$

By strong continuity of the shift group on $L^2$, the integrand on the right hand side converges to 0, as $z\rightarrow 0$. Since (2.6) holds true for $i\in \{0,..,3\}$, (7.3) follows by dominated convergence.

Now, it suffices to show that $T_\zeta w\in BUC(\mathbb {R})$ provided that (2.6) holds for $i=0$ and $i=1$. For fixed $w\in U$ and $x_1$, $x_2\in \mathbb {R}$ we directly get

$$\begin{aligned} \left|T_\zeta w(x_1) - T_\zeta w(x_2)\right|\le & {} \int _\mathbb {R}\int _0^1 \left| \zeta '(x_2 + \epsilon (x_1-x_2), y)\right| \left|w(y)\right| \,{\text {d}}\epsilon \,{\text {d}}y \left|x_1-x_2\right| \nonumber \\\le & {} \sup _{x\in \mathbb {R}} \left|\left|\zeta '(x,.)\right|\right|_{L^2(\mathbb {R})} \left|\left|w\right|\right|_{L^2} \left|x_1-x_2\right|. \end{aligned}$$

(7.4)

Here, we used fundamental theorem of calculus, Tonelli’s theorem and the Cauchy–Schwartz inequality. Hence, $T_\zeta w$ is globally Lipschitz and particularly uniformly continuous. Analogously, Cauchy–Schwartz inequality yields

$$\begin{aligned} \sup _{x\in \mathbb {R}}\left|T_\zeta w(x)\right| \le \sup _{x\in \mathbb {R}} \left|\left|\zeta (x,.)\right|\right|_{L^2(\mathbb {R})} \left|\left|w\right|\right|_{U} <\infty . \end{aligned}$$

(7.5)

$\square$

Remark 7.3

From (7.5) we immediatly get $T_\zeta \in L(U, BUC^2(\mathbb {R}))$. However, in general $T_\zeta$ itself is not Hilbert–Schmidt. To get the Hilbert–Schmidt property we need the multiplication with $N_\sigma$ as we will show in the next lemma.

Lemma 7.4

Let $\Delta$ be the Dirichlet Laplacian on $L^2(\mathbb {R}_+)$. For $u\in \mathcal {D}(\Delta )$ and $x_*\in \mathbb {R}$ it holds that

$$\begin{aligned} \left|\left|u \cdot (\theta _{x_*}\circ T_\zeta (.))\right|\right|_{{\text {HS}}(U; \mathcal {D}(\Delta ))} \le K \left|\left|u\right|\right|_{H^2} \sup _{x\in \mathbb {R}}\sum _{i=0}^2\left|\left|\zeta ^{(i)}(x,.)\right|\right|_{L^2} \end{aligned}$$

Remark 7.5

This result immediately extends to $\mathcal {C}(u)$, because Assumption 2.3 and Lemma 7.1 assure $N_\sigma (u)\in \mathcal {D}(\Delta )$ for all $u\in \mathcal {D}(\Delta )$. Moreover, note that

$$\begin{aligned} \theta _x\circ T_{\zeta } = T_{\zeta _x}, \end{aligned}$$

where $\zeta _x := \zeta (x+.,.)$ satisfies Assumption 2.6, too.

Proof

Linearity and continuity in w follow directly from the construction and Remark 7.3 and we are now interested in the Hilbert Schmidt norm. Without loss of generality, we can choose $x_* = 0$. So denote by $(e_k)$ an arbitrary CONS of U, then

$$\begin{aligned} \left|\left|u\cdot (T_\zeta (.))\right|\right|_{{\text {HS}}(U,\mathcal {D}(\Delta ))}^2 = \sum _{k=1}^\infty \left|\left|u\cdot (T_\zeta e_k)\right|\right|_{L^2}^2 + \left|\left|\Delta (u \cdot (T_\zeta e_k))\right|\right|_{L^2}^2, \end{aligned}$$

(7.6)

and the first sum equals

$$\begin{aligned} \begin{aligned}&\sum _{k=1}^\infty \int _{\mathbb {R}_+} u(x)^2 \langle \zeta (x, .), e_k\rangle _{L^2(\mathbb {R})}^2\\&\quad = \int _{\mathbb {R}_+} u(x)^2 \left|\left|\zeta (x,.)\right|\right|_{L^2(\mathbb {R})}^2 \le \left|\left|u\right|\right|_{L^2(\mathbb {R}_+)}^2 \sup _{x\in \mathbb {R}}\left|\left|\zeta (x,.)\right|\right|_{L^2(\mathbb {R})}^2, \end{aligned} \end{aligned}$$

(7.7)

where we used Tonelli’s theorem and Parseval’s identity for the first equality. To bound the second sum we proceed on exactly the same way but first apply Leibnitz rule to get the second (weak) derivative

$$\begin{aligned} \begin{aligned}&\sum _{k=1}^\infty \sum _{i=0}^2 \left( {\begin{array}{c}2\\ i\end{array}}\right) \int _{\mathbb {R}_+}\left|\frac{\partial ^{i}}{\partial x^i}u(x)\right|^2\ \langle \zeta ^{(2-i)}(x,.),e_k\rangle _{L^2(\mathbb {R})}^2\\&\quad \le \sum _{i=0}^2 \left( {\begin{array}{c}2\\ i\end{array}}\right) \left|\left|\frac{\partial ^{i}}{\partial x^i} u\right|\right|_{L^2(\mathbb {R}_+)}^2 \sup _{x\in \mathbb {R}}\left|\left|\zeta ^{(2-i)}(x,.)\right|\right|_{L^2(\mathbb {R})}^2\\&\quad \le 4 \left|\left|u\right|\right|_{H^2(\mathbb {R}_+)}^2 \sup _{x\in \mathbb {R}}\sum _{i=0}^2\left|\left|\zeta ^{(i)}(x,.)\right|\right|_{L^2(\mathbb {R})}^2. \end{aligned} \end{aligned}$$

(7.8)

$\square$

To show the main result of this appendix, we just need to combine the previous lemmas.

Theorem 7.6

The map $\mathcal {C}:\mathcal {D}(\mathcal {A})\rightarrow {\text {HS}}(U,\mathcal {D}(\mathcal {A}))$ is Lipschitz continuous on bounded sets.

Proof

By the structure of $\mathcal {A}$ and $\mathfrak L^2$, it suffices to show the property for the operator defined in (7.1). Assumption 2.3 yields $N_\sigma (\mathcal {D}(\Delta ))\subset \mathcal {D}(\Delta )$ and we can apply Lemma 7.4. For u, ${\tilde{u}}\in \mathcal {D}(\Delta )$, $x_*$, $y_*\in \mathbb {R}$ and writing $\zeta _{z,\tilde{z}}(x,y) := \zeta (z+x,y) - \zeta (\tilde{z} + x)$, it holds that

$$\begin{aligned} \begin{aligned}&\left|\left|N_\sigma (u)(\theta _{x}T_\zeta ) - N_\sigma ({\tilde{u}}) (\theta _{y}T_\zeta ) \right|\right|_{{\text {HS}}} \\&\quad \le \left|\left|(N_\sigma (u) - N_\sigma ({\tilde{u}}))\cdot (\theta _{x} T_\zeta .)\right|\right|_{{\text {HS}}} + \left|\left|N_\sigma ({\tilde{u}}) \cdot (T_{ \zeta _{x,y}} .)\right|\right|_{{\text {HS}}}\\&\quad \le K_\zeta \left|\left|N_\sigma (u) - N_\sigma ({\tilde{u}})\right|\right|_{H^2} + K\left|\left|N_\sigma ({\tilde{u}})\right|\right|_{H^2} \sup _{z\in \mathbb {R}} \sum _{i=0}^2 \left|\left| \zeta _{x,y}^{(i)}(z,.)\right|\right|_{L^2}. \end{aligned} \end{aligned}$$

A computation similar to (7.4) shows

$$\begin{aligned} \sup _{z\in \mathbb {R}}\left|\left|\zeta _{x,y}^{(i)}(z,.)\right|\right|_{L^2}^2 \le \left|x-y\right|^2 \sup _{z\in \mathbb {R}}\left|\left|\zeta ^{(i+1)}(z,.)\right|\right|_{L^2}^2. \end{aligned}$$

Finally, we put everything together and use that on bounded sets $N_\sigma$ is Lipschitz, and thus bounded, to get the assertion. $\square$