The Morse Property of Limit Functions Appearing in Mean Field Equations on Surfaces with Boundary

Abstract

In this paper, we study the Morse property for functions related to limit functions of mean field equations on a smooth, compact surface $\Sigma$ with boundary $\partial \Sigma$. Given a Riemannian metric g on $\Sigma$ we consider functions of the form

where $\sigma _i \ne 0$ for $i=1,\ldots ,m$, $G^g$ is the Green function of the Laplace-Beltrami operator on $(\Sigma ,g)$ with Neumann boundary conditions, $R^g$ is the corresponding Robin function, and $h \in {{\mathcal {C}}}^{2}(\Sigma ^m,\mathbb {R})$ is arbitrary. We prove that for any Riemannian metric g, there exists a metric $\widetilde{g}$ which is arbitrarily close to g and in the conformal class of g such that $f_{\widetilde{g}}$ is a Morse function. Furthermore we show that, if all $\sigma _i>0$, then the set of Riemannian metrics for which $f_g$ is a Morse function is open and dense in the set of all Riemannian metrics.

1 Introduction and Main Results

Let $\Sigma$ be a smooth, compact surface with boundary $\partial \Sigma$. For a Riemannian metric g on $\Sigma$ let $G^g:\Sigma \times \Sigma \rightarrow \mathbb {R}$ be the Green function for the Laplace-Beltrami operator $-\Delta _g$ with Neumann boundary conditions and mean value 0; i.e. for each $\xi \in \Sigma$ the function $G^g(\cdot ,\xi )$ is the unique solution of

$$\begin{aligned}\left\{ \begin{aligned} -\Delta _g G^g(\cdot ,\xi )&= \delta _\xi -\frac{1}{|\Sigma |_g}{} & {} \qquad \text {in }{\mathring{\Sigma }}:= \Sigma \setminus \partial \Sigma ,\\ \partial _{\nu _g} G^g(\cdot ,\xi )&= 0{} & {} \qquad \text {on }\partial \Sigma ,\\ \int _\Sigma G^g(\cdot ,\xi )\,dv_g&= 0. \end{aligned} \right. \end{aligned}$$

Here $\nu _g$ is the unit outward normal on $\partial \Sigma$, $dv_g$ is the area element of $\Sigma$, $|\Sigma |_g=\int _{\Sigma } dv_g$ and $\delta _{\xi }$ is the Dirac distribution on $\Sigma$ concentrated at $\xi \in \Sigma$. Setting $\kappa (\xi )=2\pi$ for $\xi \in {\mathring{\Sigma }}$ and $\kappa (\xi )=\pi$ for $\xi \in \partial \Sigma$, the Robin function $R^g:\Sigma \rightarrow \mathbb {R}$ is defined by

$$\begin{aligned} R^g(\xi ):= \lim _{x\rightarrow \xi }\left( G^g(x,\xi ) + \frac{1}{\kappa (\xi )}\log d_g(x,\xi )\right) , \end{aligned}$$

where $d_g$ denotes the distance with respect to the metric g. Given integers $m\ge l\ge 0$ with $m\ge 1$ and real numbers $\sigma _i\ne 0$, $i=1,\dots ,m$, we set

$$\begin{aligned} X:= {\mathring{\Sigma }}^l\times (\partial \Sigma )^{m-l}\qquad \text {and}\qquad \Delta _X:= \{x\in X:\hbox {} x_i=x_j\; \hbox {for some}\; i\ne j\}, \end{aligned}$$

and define, for an arbitrary given function $h\in {{\mathcal {C}}}^2(\Sigma ^m, \mathbb {R})$:

(1.1)

We will prove that $f_g$ is a Morse function for a “generic” metric g on $\Sigma$, in a sense that will be made precise below.

Functions of the form (1.1) play a crucial role in understanding the blow-up behavior of solutions of mean field equations like

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _g u= \lambda \left( \frac{V e^u}{\int _{\Sigma } Ve^u\, dv_g} - \frac{1}{|\Sigma |_g}\right) &{} \text { in } {\mathring{\Sigma }}, \\ \partial _{\nu _g} u = 0 &{} \text { on } \partial \Sigma , \\ \int _{\Sigma } u\, dv_g = 0. &{} \end{array}\right. } \end{aligned}$$

(1.2)

For a compact surface without boundary it has been proven in [7] that a nondegenerate or more generally, an isolated stable critical point $x=(x_1,\dots ,x_m)$ of $f_g$, with $l=m$ and $h(x) = 2 \sum _{i=1}^m\sigma _i\log V(x_i)$, gives rise to solutions $(\lambda ,u)$ of (1.2) with $\lambda$ close to $8\pi m$ and such that u blows up as $\lambda \rightarrow 8\pi m$ precisely at $x_1,\dots ,x_m\in \Sigma$. Similar results have been obtained in the case of mean field equations on a bounded domain in $\mathbb {R}^2$ with Dirichlet boundary conditions in [8, 13], and for solutions of Gelfand’s problem and steady states of the Keller–Segel system (see [2, 5, 6, 10] for instance).

Now we state our main results. We fix an integer $k\ge 0$ and $0<\alpha <1$, and let $\textrm{Riem}^{k+2,\alpha }(\Sigma )$ be the space of Riemannian metrics of class ${{\mathcal {C}}}^{k+2,\alpha }$ on $\Sigma$, i.e. the space of symmetric and positive definite sections $\Sigma \rightarrow (T\Sigma \otimes T\Sigma )^*$ of class ${{\mathcal {C}}}^{k+2,\alpha }$.

Theorem 1.1

If $\sigma _i\ne 0$ for $i=1,\dots ,m$ and $\sigma _1+\dots +\sigma _m\ne 0$ then for any $g\in \textrm{Riem}^{k+2,\alpha }(\Sigma )$ the set

$$\begin{aligned} {{\mathcal {M}}}_g^{k+2,\alpha }(\Sigma ):= \{\psi \in {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+):{f_{\psi g}\text { is a Morse function }} \} \end{aligned}$$

is a residual subset of ${{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)$.

As a consequence of Theorem 1.1 for any Riemannian metric g on $\Sigma$, there exists a metric $\widetilde{g}$ which is arbitrarily close to g and conformal to g, and such that $f_{\widetilde{g}}$ is a Morse function.

Theorem 1.2

If $\sigma _i>0$ for all $i=1,\dots ,m$ then the set

$$\begin{aligned} \textrm{Riem}_{Morse}^{k+2,\alpha }(\Sigma ):= \{g\in \textrm{Riem}^{k+2,\alpha }(\Sigma ):f_g \text { is a Morse function} \} \end{aligned}$$

is an open and dense subset of $\textrm{Riem}^{k+2,\alpha }(\Sigma )$.

Remark 1.3

In [1] the authors considered functions $f_g$ of the form (1.1) on a surface $\Sigma$ without boundary and with $h(x_1,\dots ,x_m)=\sum _{i=1}^m\log V(x_i)$ where $V\in {{\mathcal {C}}}^2(\Sigma ,\mathbb {R}_+)$. This is motivated by the mean field equation (1.2). They proved that $f_g$ is a Morse function for V in an open and dense subset of ${{\mathcal {C}}}^2(\Sigma ,\mathbb {R}_+)$. Also related is the paper [3], which deals with functions like $f_g$ where G is the Green function of the Dirichlet Laplace operator on a bounded smooth domain $\Omega \subset \mathbb {R}^n$. In [3], it is proved that $f_g$ is a Morse function for a generic domain $\Omega$. The present paper seems to be the first investigating the Morse property of $f_g$ as in (1.1) as a function of the Riemannian metric.

2 Preliminaries

Riemann surfaces are locally conformally flat, so there exist isothermal coordinates where the metric is conformal to the Euclidean metric (see [4, 11, 15]). We modify the isothermal coordinates applied in [7, 9, 16] to a Riemann surface with boundary. For $\xi \in \Sigma$ there exists a local chart $y_{\xi }:\Sigma \supset U(\xi )\rightarrow B^\xi \subset \mathbb {R}^2$ transforming g to $e^{\varphi _\xi \circ y_\xi }\langle \cdot ,\cdot \rangle _{\mathbb {R}^2}$. We may assume that $y_\xi (\xi )=0$. For $\xi \in {\mathring{\Sigma }}$ we may also assume that $\overline{U(\xi )}\subset {\mathring{\Sigma }}$ and that the image of $y_{\xi }$ is a disc $B^\xi :=\{ y\in \mathbb {R}^2: |y|< 2r_{\xi }\}$ of radius $2r_\xi >0$. For $\xi \in \partial \Sigma$, by Lemma 4 in [16], there exists an isothermal coordinate system $\left( U(\xi ), y_{\xi }\right)$ near $\xi$ such that the image of $y_{\xi }$ is a half disk ${B}_{2r_{\xi }}^{+}:=\{y\in \mathbb {R}^2: |y|<2r_{\xi }, y_2\ge 0\}$ of radius $2r_{\xi }>0$ and $y_{\xi }\left( U(\xi )\cap \partial \Sigma \right) = {{B}_{2r_{\xi }}^{+}} \cap \partial \mathbb {R}^{2}_+$. For any $x \in$ $y_{\xi }^{-1}\left( {{B}_{2r_{\xi }}^{+}} \cap \partial \mathbb {R}^{2}_+\right)$, we have

$$\begin{aligned} \left( y_{\xi }\right) _*(\nu _g(x))=\left. -\exp \left( -\frac{\varphi _{\xi }(y)}{2}\right) \frac{\partial }{ \partial y_2 }\right| _{ y=y_{\xi }(x)}. \end{aligned}$$

(2.1)

In this case, we take $B^\xi = {B}_{2r_{\xi }}^{+}$. For $\xi \in \Sigma$ and $0<r\le 2r_\xi$ we set

$$\begin{aligned} B_r^\xi := B^\xi \cap \{ y\in \mathbb {R}^2: |y|< r\}\quad \text {and}\quad U_{r}(\xi ):=y_\xi ^{-1}(B_{r}^{\xi }). \end{aligned}$$

Recall that $\varphi _\xi :B^\xi \rightarrow \mathbb {R}$ is related to K, the Gaussian curvature of $\Sigma$, by the equation

$$\begin{aligned} -\Delta \varphi _\xi (y) = 2K\big (y^{-1}_\xi (y)\big ) e^{\varphi _\xi (y)} \quad \text {for all }y\in B^\xi . \end{aligned}$$

We can assume that $y_\xi$ and $\varphi _\xi$ depend smoothly on $\xi$, and that $\varphi _\xi (0)=0$ and $\nabla \varphi _\xi (0)=0$.

Observe that we have for $\zeta \in U(\xi )$,

$$\begin{aligned} \lim _{x\rightarrow \zeta }\frac{d_g(x,\zeta )}{\big |\varphi _\xi (x)-\varphi _\xi (\zeta )\big |} = e^{\frac{1}{2}\varphi _\xi (\zeta )}, \end{aligned}$$

which implies

$$\begin{aligned} R^g(\zeta ) = \lim _{x\rightarrow \zeta }\left( G^g(x,\zeta )+\frac{1}{\kappa (\zeta )}\log \big |y_\xi (x)-y_\xi (\zeta )\big |\right) + \frac{1}{2\kappa (\zeta )}\varphi _\xi \big (y_\xi (\zeta )\big ), \end{aligned}$$

(2.2)

and in particular, using $\varphi _\xi \big (y_\xi (\xi )\big )=\varphi _\xi (0)=0$:

$$\begin{aligned} R^g(\xi ) = \lim _{x\rightarrow \xi }\left( G^g(x,\xi )+\frac{1}{\kappa (\xi )}\log \big |y_\xi (x)\big |\right) . \end{aligned}$$

Now we construct a regular part $H^g(x,\xi )$ of the Green function $G^g(x,\xi )$ such that $H^g(\xi ,\xi )=R^g(\xi )$. Let $\chi \in {{\mathcal {C}}}^\infty (\mathbb {R},[0,1])$ be such that

$$\begin{aligned} \chi (s) = {\left\{ \begin{array}{ll} 1 &{}\quad \text {if }|s|\le 1,\\ 0 &{}\quad \text {if }|s|\ge 2. \end{array}\right. } \end{aligned}$$

For $\xi \in {\mathring{\Sigma }}$ we choose $\delta _\xi = \min \{\frac{1}{2} r_\xi ,\frac{1}{2}{{\,\textrm{dist}\,}}(x,\partial \Sigma )\}$, and for $\xi \in \partial \Sigma$ we set $\delta _\xi :=\frac{1}{2} r_\xi$. Next we define the cut-off function $\chi _\xi \in {{\mathcal {C}}}^\infty (\Sigma ,[0,1])$ by

$$\begin{aligned} \chi _\xi (x):= {\left\{ \begin{array}{ll} \chi \big (|y_\xi (x)|/\delta _\xi \big ) &{}\quad \text {if }x\in U(\xi )\\ 0&{}\quad \text {if }x\in \Sigma \setminus U(\xi ) \end{array}\right. }. \end{aligned}$$

Then, for $\xi \in \Sigma$ the function $H^g_{\xi }:=H^g(\cdot ,\xi ):\Sigma \rightarrow \mathbb {R}$ is defined to be the unique solution of the Neumann problem

$$\begin{aligned} \left\{ \begin{aligned}&\Delta _g H^g_{\xi } = \frac{1}{\kappa (\xi )} (\Delta _g\chi _\xi )\log \big |y_\xi \big |+\frac{2}{\kappa (\xi )} \left\langle \nabla ^g\chi _\xi ,\nabla ^g\log |y_\xi |\right\rangle _g+\frac{1}{|\Sigma |_g}&\quad \text {in }{\mathring{\Sigma }},\\&\partial _{ \nu _g} H^g_{\xi } = \frac{1}{\kappa (\xi )}\big (\partial _{\nu _g }\chi _\xi \big ) \log |y_\xi | + \frac{1}{\kappa (\xi )}\chi _\xi \partial _{\nu _g} \log |y_\xi |&\quad \text {on }\partial \Sigma ,\\&\int _{\Sigma } H^g_{\xi } dv_g = \frac{1}{\kappa (\xi )}\int _{\Sigma } \chi _{\xi }\log |y_\xi |\, dv_g. \end{aligned} \right. \end{aligned}$$

(2.3)

Lemma 2.1

For $g\in \textrm{Riem}^{k+2,\alpha }(\Sigma )$ the function $H^g$ is of class ${{\mathcal {C}}}^{k+3,\alpha }$ in any compact subsets of $\Sigma \times {\mathring{\Sigma }}$ and in $\Sigma \times \partial \Sigma$. Moreover, it satisfies

$$\begin{aligned} G^g(x,\xi ) = -\frac{1}{\kappa (\xi )}\chi _\xi (x)\cdot \log \big |y_\xi (x)\big | + H^g(x,\xi ) \end{aligned}$$

and $H^g(\xi ,\xi )=R^g(\xi )$. Consequently $R^g$ is of class ${{\mathcal {C}}}^{k+3,\alpha }$ in any compact subsets of ${\mathring{\Sigma }}$ and in $\partial \Sigma$.

Proof

First we observe that $e^{\varphi _{\xi }}\in {{\mathcal {C}}}^{k+2,\alpha }(B^{\xi }, \mathbb {R})$ because $g\in \textrm{Riem}^{k+2,\alpha }(\Sigma )$. For $\xi \in {\mathring{\Sigma }}$, by the choice of $\delta _{\xi }$ we have that $-\Delta _g H^g_\xi$ is of class ${{\mathcal {C}}}^{k+2,\alpha }$ in $\Sigma$ and $\partial _{\nu _g } H^g(x,\xi ) \equiv 0$ on $\partial \Sigma$. Now the Schauder estimate for the Neumann problem (see [14, 16], for instance) implies that the solution of (2.3) uniquely exists in ${{\mathcal {C}}}^{k+4,\alpha }(\Sigma ,\mathbb {R})$.

For $x\in U(\xi )\cap \partial \Sigma$, setting $y=y_{\xi }(x)$ we have:

$$\begin{aligned} \partial _{ \nu _g } \log |y_{\xi }(x)| {\mathop {=}\limits ^{(2.1)}} - e^{-\frac{1}{2} {\varphi }_{\xi }(y)}\frac{\partial }{\partial y_2} \log |y| = - e^{-\frac{1}{2} {\varphi }_{\xi }(y)}\frac{y_2}{|y|^2}\equiv 0. \end{aligned}$$

Clearly, $\partial _{\nu _g} \chi (|y_{\xi }|(x))=0$ for $x \in \partial \Sigma \cap U_{\delta _{\xi }}(\xi )$. It follows that $\partial _{\nu _g} H^g_\xi$ is of class ${{\mathcal {C}}}^{k+2,\alpha }$ on $\partial \Sigma$. Moreover $\Delta _{g}H^g_\xi$ is of class ${{\mathcal {C}}}^{k+2,\alpha }$ in $\Sigma$. Consequently (2.3) has a unique solution $H^g_\xi \in {{\mathcal {C}}}^{k+3,\alpha }(\Sigma ,\mathbb {R})$ by the Schauder estimates.

Finally $H^g_\xi$ is uniformly bounded in ${{\mathcal {C}}}^{k+3, \alpha }$ for $\xi$ in any compact subsets of ${\mathring{\Sigma }}$ and in $\partial \Sigma$. Therefore $H^g(\xi ,\xi )$ is in ${{\mathcal {C}}}^{k+3, \alpha }$ in any compact subsets of ${\mathring{\Sigma }}$ and in $\partial \Sigma$, and $H^g(\xi ,\xi )=R^g(\xi )$ by (2.2). $\square$

For $g\in \textrm{Riem}^{k+2,\alpha }(\Sigma )$ we now consider the map

$$\begin{aligned} {{\mathcal {H}}}^g: \Sigma \times \Sigma \times {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+) \rightarrow \mathbb {R},\quad {{\mathcal {H}}}^g(x,\xi ,\psi ):= H^{\psi g}(x,\xi ). \end{aligned}$$

Proposition 1

The map ${{\mathcal {H}}}^g$ is ${{\mathcal {C}}}^1$ in $\Sigma \times {\mathring{\Sigma }}\times {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)$ and in $\Sigma \times \partial \Sigma \times {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)$. Moreover, we have

$$\begin{aligned} ~ D_{\psi } {{\mathcal {H}}}^g(x,\xi , 1)[\theta ] = - \frac{1}{|\Sigma |_g}\int _{\Sigma } (G^{g}(z,x) + G^g(z,\xi )) \theta (z) dv_g(z), \end{aligned}$$

(2.4)

for any $\theta \in {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R})$.

Proof

For $g\in \textrm{Riem}^{k+2,\alpha }(\Sigma )$ and $\psi \in {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)$ we clearly have $\psi g\in \textrm{Riem}^{k+2,\alpha }(\Sigma )$. By a direct calculation we obtain the following equations for $H^{\psi g}_\xi - H^g_\xi$:

$$\begin{aligned} \left\{ \begin{aligned} -\Delta _{g} (H^{\psi g}_\xi - H^g_\xi )&= \frac{1}{|\Sigma |_g} - \frac{\psi }{ |\Sigma |_{\psi g}}{} & {} \text {in }{\mathring{\Sigma }},\\ \partial _{ \nu _g} (H^{\psi g}_\xi - H^g_\xi )&= 0{} & {} \text {on }\partial \Sigma ,\\ \int _{\Sigma } (H^{\psi g}_\xi -H^{{g}}_\xi )\, dv_g&= -\int _{\Sigma } G^{\psi g}_\xi (\psi -1) dv_g. \end{aligned} \right. \end{aligned}$$

An expansion of $|\Sigma |_{\psi g}= \int _{\Sigma } dv_{\psi g}$ yields:

$$\begin{aligned} -\Delta _g(H^{\psi g}_\xi - H^g_\xi ) = \frac{1-\psi }{|\Sigma |_g}+ \frac{\int _{\Sigma }(\psi -1) dv_g}{|\Sigma |^2_{g}} + {\mathcal {O}}(\Vert \psi -1\Vert _{{{\mathcal {C}}}^{k+2,\alpha }}^2) \quad \text {as } \Vert \psi -1\Vert _{{{\mathcal {C}}}^{k+2,\alpha }} \rightarrow 0. \end{aligned}$$

Recall that $G^{\psi g}_\xi = - \frac{4}{\sigma (\xi )}\chi (|y_{\xi }|)\log {|y_{\xi }|}+H^{\psi g}_\xi$, so the representation formula gives:

$$\begin{aligned} H^{\psi g}_\xi (x) - H^g_\xi (x)= & {} \frac{1}{|\Sigma |_g} \int _{\Sigma } G^{\psi g}(z,\xi )(\psi (z)-1)\,dv_g(z) \nonumber \\ {}{} & {} -\frac{1}{|\Sigma |_{\psi g}} \int _{\Sigma } G^g(z,x)(\psi (z)-1)\,dv_g(z). \end{aligned}$$

(2.5)

By standard elliptic estimates (see [14, 16]) there exists a constant C such that

$$\begin{aligned} \big \Vert H^{\psi g}_\xi - H^g_\xi \big \Vert _{{{\mathcal {C}}}^{k+4,\alpha }} \le C\cdot \Vert \psi -1\Vert _{{{\mathcal {C}}}^{k+2,\alpha }}, \end{aligned}$$

thus

$$\begin{aligned} H^{\psi g}_\xi \rightarrow H^g_\xi \quad \text { in }{{\mathcal {C}}}^{k+4,\alpha }\text { as }\psi \rightarrow 1 \text { in }{{\mathcal {C}}}^{k+2,\alpha }. \end{aligned}$$

(2.6)

According to the construction of $H^g(x,\xi )$, the convergence in (2.6) is uniform for $\xi$ in any compact subset of ${\mathring{\Sigma }}$, and in $\partial \Sigma$. It follows that ${{\mathcal {H}}}_g(x,\xi ,\cdot )$ is continuous at 1, uniformly for $x\in \Sigma$ and $\xi$ in any compact subsets of ${\mathring{\Sigma }}$ or $\xi \in \partial \Sigma$. Using ${{\mathcal {H}}}_{g}(x,\xi ,\psi )= {{\mathcal {H}}}_{\psi g} (x,\xi ,1)$ we see that ${{\mathcal {H}}}_g(x,\xi , \cdot )$ is continuous at every $\psi \in {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)$.

Next we prove that ${{\mathcal {H}}}^g(x,\xi ,\psi )$ is ${{\mathcal {C}}}^1$ with respect to $\psi$. We fix $\theta \in {{\mathcal {C}}}^{2+k,\alpha }(\Sigma , \mathbb {R})$ and consider the metric $(1+t\theta )g$ with t sufficiently small so that $1+t\theta > 0$. Then

$$\begin{aligned} w_\xi ^t(x):= & {} \frac{1}{t}\big (H^{(1+t\theta )g}(x,\xi )-H^g(x,\xi )\big )\nonumber \\ {}= & {} \frac{1}{t}\big ({{\mathcal {H}}}^g(x,\xi ,1+t\theta )-{{\mathcal {H}}}^g(x,\xi ,1)\big ) \end{aligned}$$

(2.7)

satisfies the following equations as $t\rightarrow 0$:

$$\begin{aligned} ~ \left\{ \begin{aligned} -\Delta _{g}w_\xi ^t&= \frac{1}{|\Sigma |^2_g}\int _{\Sigma } \theta \, dv_{g} -\frac{\theta }{|\Sigma |_g} + {{\mathcal {O}}}(t){} & {} \quad \text {in }{\mathring{\Sigma }},\\ \partial _{\nu _g} w_t(x,\xi )&= 0{} & {} \quad x\in \partial \Sigma ,\\ \int _{\Sigma } w_\xi ^t\, dv_{g}&=- \int _{\Sigma } G^{(1+t\theta )g}_\xi \cdot \theta \, dv_g \end{aligned} \right. \end{aligned}$$

(2.8)

where ${{\mathcal {O}}}(t)$ is defined with respect to the ${{\mathcal {C}}}^{k+2,\alpha }$-norm. Applying the standard elliptic estimates, $w_\xi ^t$ converges as $t\rightarrow 0$ to some function $w_\xi ^0(x) = D_\psi {{\mathcal {H}}}^g(x,\xi ,1)[\theta ]$ in ${{\mathcal {C}}}^{k+4,\alpha }(\Sigma , \mathbb {R})$. Moreover, $w_\xi ^0$ satisfies the equations

$$\begin{aligned} ~ \left\{ \begin{aligned} -\Delta _{g}w_\xi ^0&= \frac{1}{|\Sigma |^2_g}\int _{\Sigma } \theta \, dv_{g} -\frac{\theta }{|\Sigma |_g}{} & {} \quad \text { in }{\mathring{\Sigma }}, \\ \partial _{\nu _g} w_\xi ^0&= 0{} & {} \quad \text { on }\partial \Sigma ,\\ \int _{\Sigma } w_\xi ^0\, dv_{g}&= -\int _{\Sigma } G^{g}_\xi \cdot \theta \, dv_g. \end{aligned} \right. \end{aligned}$$

(2.9)

The representation formula now implies (2.4):

$$\begin{aligned} D_\psi {{\mathcal {H}}}^g(x,\xi ,1)[\theta ] = w_\xi ^0(x) = -\frac{1}{|\Sigma |_{g}}\int _{\Sigma } \big (G^{g}(z,x) +G^{g}(z,\xi )\big ) \cdot \theta (z)\, dv_g(z). \end{aligned}$$

Replacing g by $\psi g$ and $\theta$ by $\frac{\theta }{\psi }$, we obtain the derivative of ${{\mathcal {H}}}_g(x,\xi ,\psi )$ for arbitrary $\psi \in {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)$:

$$\begin{aligned} D_{\psi }{{\mathcal {H}}}^g(x,\xi ,\psi )[\theta ]= & {} \lim _{t\rightarrow 0}\frac{1}{t}\big (H^{(\psi +t\theta ) g}(x,\xi )-H^{\psi g}(x,\xi )\big ) \nonumber \\= & {} -\frac{1}{|\Sigma |_{\psi g}}\int _{\Sigma } \left( G^{\psi g} (z,x)+G^{\psi g}(z,\xi ) \right) \cdot \frac{\theta (z) }{\psi (z)}dv_{\psi g}(z) \nonumber \\= & {} -\frac{1}{|\Sigma |_{\psi g}}\int _{\Sigma } \left( G^{\psi g} (z,x)+G^{\psi g}(z,\xi ) \right) \cdot \theta (z) dv_{g}(z). \qquad \end{aligned}$$

(2.10)

In order to see that ${{\mathcal {H}}}_g(x,\xi ,\psi )$ is continuously Fréchet differentiable with respect to $\psi$ it is sufficient to prove that $D_{\psi }{{\mathcal {H}}}_g(x,\xi ,\psi )$ is continuous in $\psi$ as a linear operator on ${{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)$. For $\psi _1,\psi _2\in {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)$ there holds

$$\begin{aligned} \begin{aligned}&\big | D_{\psi }{{\mathcal {H}}}_g(x,\xi ,\psi _1)[\theta ]-D_{\psi }{{\mathcal {H}}}_g(x,\xi ,\psi _2)[\theta ] \big |\\ {}&= \left| \frac{1}{|\Sigma |_{\psi _1 g}}\int _{\Sigma } \left( G^{\psi _1 g} (z,x) +G^{\psi _1 g}(z,\xi ) \right) \theta (z) dv_{g}(z)\right. \\&\quad \left. - \frac{1}{|\Sigma |_{\psi _2 g}}\int _{\Sigma } \left( G^{\psi _2 g} (z,x)+G^{\psi _2 g}(z,\xi ) \right) \theta (z) dv_{g}(z)\right| \\&\le C\cdot \Vert \theta \Vert _{{{\mathcal {C}}}^{k+2,\alpha }} \cdot \Vert \psi _1-\psi _2\Vert _{{{\mathcal {C}}}^{k+2,\alpha }} \end{aligned} \end{aligned}$$

where we applied (2.10); here $C>0$ is a constant. Therefore ${{\mathcal {H}}}^g$ is ${{\mathcal {C}}}^1$ in $\Sigma \times {\mathring{\Sigma }}\times {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)$ and in $\Sigma \times \partial \Sigma \times {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+).$ $\square$

3 Proof of Theorem 1.1

The proof is based on the following transversality theorem [12, Theorem 5.4].

Theorem 3.1

  Let $M,\Psi ,N$ be Banach manifolds of class ${{\mathcal {C}}}^r$ for some $r\in \mathbb {N}$, let ${{\mathcal {D}}}\subset M\times \Psi$ be open, let ${{\mathcal {F}}}: {{\mathcal {D}}}\rightarrow N$ be a ${{\mathcal {C}}}^r$ map, and fix a point $z\in N$. Assume for each $(y,\psi )\in {{\mathcal {F}}}^{-1}(z)$ that:

• (1) $D_y{{\mathcal {F}}}(y,\psi ): T_y M\rightarrow T_{z} N$ is semi-Fredholm with index $<r$;

• (2) $D{{\mathcal {F}}}(y,\psi ): T_y M \times T_\psi \Psi \rightarrow T_{z} N$ is surjective;

• (3) ${{\mathcal {F}}}^{-1}(z)\rightarrow \Psi$, $(y,\psi )\mapsto \psi$, is $\sigma$-proper.

Then

$$\begin{aligned} \Psi _{reg}:=\{ \psi \in \Psi : z \text { is a regular value of } {{\mathcal {F}}}(\cdot , \psi )\} \end{aligned}$$

is a residual subset of $\Psi$.

Proof of Theorem 1.1

We set $\Psi := {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)$ and consider the functions $f_{\psi g}:X \setminus \Delta _X \rightarrow \mathbb {R}$ for $\psi \in {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)$ first in a local isothermal chart. Let $y_{\xi _i}:\Sigma \supset U(\xi _i) \rightarrow B^{\xi _i} \subset \mathbb {R}^2$ be isothermal charts of $\Sigma$ as in Sect. 2 with $\xi _1,\dots ,\xi _l\in {\mathring{\Sigma }}$, $\xi _{l+1},\dots ,\xi _m\in \partial \Sigma$. For simplicity of notation we set $Y_i:=y_{\xi _i}:U_i:=U(\xi _i) \rightarrow B_i:=B^{\xi _{i}}\subset \mathbb {R}^2$ for $i=1,\dots ,l$, and $Y_i:=\pi _1\circ y_{\xi _i}:U_i:=U(\xi _i)\cap \partial \Sigma \rightarrow B_i \subset \mathbb {R}$ for $i=l+1,\dots ,m$; here $\pi _1:\mathbb {R}^2\rightarrow \mathbb {R}$ is the projection onto the first component. For $i=l+1,\dots ,m$ we thus have $y_{\xi _i}(x) = (Y_i(x),0)$. Then

$$\begin{aligned} Y= & {} Y_1\times \dots \times Y_m: X \supset U:=U_1\times \cdots \times U_m \rightarrow \mathbb {R}^{l+m},\ \ (x_1,\dots ,x_m) \\ {}\mapsto & {} \big (Y_1(x_1),\dots ,Y_m(x_m)\big ), \end{aligned}$$

is a chart of X. Set $M=N:=\mathbb {R}^{l+m}$ and $V:= Y(U\cap X{\setminus } \Delta _X) \subset \mathbb {R}^{l+m}=M$ so that ${{\mathcal {D}}}:= V \times {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)$ is an open subset of $\mathbb {R}^{l+m} \times {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)$.

We will apply Theorem 3.1 to prove that $0\in \mathbb {R}^{l+m}$ is a regular value of

$$\begin{aligned} \nabla \big (f_{\psi g}\circ Y^{-1}\big ): \mathbb {R}^{l+m}\supset V=Y(U\cap X\setminus \Delta _X) \rightarrow \mathbb {R}^{l+m} \end{aligned}$$

for $\psi$ in a residual subset $\Psi _U \subset \Psi$. This implies that the restriction of $f_{\psi g}$ to $U\cap X\setminus \Delta _X$ is a Morse function for $\psi \in \Psi _U$. Then Theorem 1.1 follows because X is covered by finitely many neighborhoods U as above and because the intersection of finitely many residual sets is a residual set.

It remains to prove that the map

$$\begin{aligned} {{\mathcal {F}}}_g: \mathbb {R}^{l+m}\times {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+) \supset {{\mathcal {D}}}\rightarrow \mathbb {R}^{l+m},\quad (y,\psi ) \mapsto \nabla \big (f_{\psi g}\circ Y^{-1}\big )(y), \end{aligned}$$

satisfies the assumptions of Theorem 3.1 with $r=1$. Concerning the differentiability it is clear that ${{\mathcal {F}}}_g$ is ${{\mathcal {C}}}^1$ as a function of $y\in V$. In order to see that ${{\mathcal {F}}}_g$ is also ${{\mathcal {C}}}^1$ in $\psi$, by Lemma 2.1 it is sufficient to prove that $\nabla _y \left( H^{\psi g}\left( Y_i^{-1}(\cdot ),Y_j^{-1}(\cdot )\right) \right)$ is ${{\mathcal {C}}}^1$ in $\psi$. We recall that $w^t_{\xi }$ is defined by (2.7). Applying the representation formula of $w^t_{\xi }$ and Lebesgue’s dominated convergence theorem, we have for $(y,\psi )\in {{\mathcal {D}}}$ and $\theta \in {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R})$:

$$\begin{aligned}{} & {} D_{\psi }\big |_{\psi =1}\nabla _y H^{\psi g}(Y_i^{-1}(y_i), Y_j^{-1}(y_j) )[\theta ]\\ {}{} & {} \quad =\lim _{t\rightarrow 0}\nabla _y \left( \left. w^t_{\xi }(x) \right| _{x= Y^{-1}_i(y_i),\xi =Y^{-1}_{j}(y_j)}\right) \\{} & {} {\mathop {=}\limits ^{(2.5)}} \lim _{t\rightarrow 0}\left( -\frac{1}{|\Sigma |_g}\int _{\Sigma } \nabla _y G_{Y_{j}^{-1}(y_j)}^{(1+t\theta )g}\theta dv_g\right. \\ {}{} & {} \left. \qquad +\int _{\Sigma } \frac{1}{t} \left( \frac{1}{|\Sigma |_g}-\frac{1+t\theta }{|\Sigma |_{(1+t\theta )g}}\right) \nabla _y G^g_{Y^{-1}_i(y_i)}dv_g \right) \\{} & {} \quad = - \frac{1}{|\Sigma |_g}\int _{\Sigma } \nabla _y \big (G^{g}(z,Y_i^{-1}(y_i) ) + G^g(z,Y_j^{-1}(y_j))\big )\cdot \theta (z) \,dv_g(z)\\{} & {} \quad =\nabla _y D_{\psi } {{\mathcal {H}}}^g(Y_i^{-1}(y_i), Y_j^{-1}(y_j), 1) [\theta ], \end{aligned}$$

where we used Proposition 1. Since $D_{\psi }{{\mathcal {F}}}_g(y,\psi )[\theta ] = D_{\psi }{{\mathcal {F}}}_{\psi g}(y, 1)\left[ \frac{\theta }{\psi }\right]$ we obtain

$$\begin{aligned} D_{\psi }{{\mathcal {F}}}_g(y,\psi )[\theta ] = - \frac{1}{|\Sigma |_{\psi g}}\int _{\Sigma } \nabla _y \big (G^{\psi g }(z,Y_i^{-1}(y_i) ) + G^{\psi g}(z,Y_j^{-1}(y_j))\big )\cdot \theta (z) \,dv_{g}(z). \end{aligned}$$

As in the proof of Proposition 1 we deduce that ${{\mathcal {F}}}_g(y,\psi )$ is ${{\mathcal {C}}}^1$ on U.

Now we need to check the assumptions (1)-(3) of Theorem 3.1. Obviously $D_y{{\mathcal {F}}}_g(y,\psi ):\mathbb {R}^{l+m}\rightarrow \mathbb {R}^{l+m}$ is a Fredholm operator of index $0<1$, hence (1) holds. Also, (3) is easy to prove: For $j\in \mathbb {N}$ the set

$$\begin{aligned} M_j:= \big \{Y(x)\in \mathbb {R}^{l+m}: x\in U,\ d_g\big (x,\Delta _X \cup \partial U\big ) \ge 2^{-j}\big \} \subset Y(U) \subset \mathbb {R}^{l+m} \end{aligned}$$

is compact as a continuous image of a compact set. Therefore the map

$$\begin{aligned} M_j\times {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+) \rightarrow {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+),\quad (y,\psi ) \mapsto \psi , \end{aligned}$$

is proper, hence its restriction to ${{\mathcal {F}}}_g^{-1}(0) \cap \big ( M_j\times {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)\big )$ is proper. Since $V = \bigcup _{j=1}^\infty M_j$, so ${{\mathcal {D}}}=\bigcup _{j=1}^\infty \big (M_j \times {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)\big )$ it follows that the map

$$\begin{aligned} {{\mathcal {F}}}_g^{-1}(0) = \bigcup _{j=1}^\infty \Big ({{\mathcal {F}}}_g^{-1}(0) \cap \big ( M_j\times {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+)\big )\Big ) \rightarrow {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}_+), \quad (y,\psi ) \mapsto \psi , \end{aligned}$$

is $\sigma$-proper.

Finally we prove the surjectivity of the derivative $D{{\mathcal {F}}}_g(y,\psi ): \mathbb {R}^{l+m}\times {{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}) \rightarrow \mathbb {R}^{l+m}$ at a point $(y,\psi )\in {{\mathcal {F}}}_g^{-1}(0)$. In fact, we shall prove that $D_\psi {{\mathcal {F}}}_g(y,\psi ):{{\mathcal {C}}}^{k+2,\alpha }(\Sigma ,\mathbb {R}) \rightarrow \mathbb {R}^{l+m}$ is onto. Since

$$\begin{aligned} D_{\psi } {{\mathcal {F}}}_g(y, \psi )[\theta ] = D_{\psi } {{\mathcal {F}}}_{\psi g}(y, 1)\left[ \theta / \psi \right] \qquad \text {for } \theta \in {{\mathcal {C}}}^{2+k,\alpha }(\Sigma , \mathbb {R}) \end{aligned}$$

it is sufficient to consider the case $\psi \equiv 1$. We observe for $\theta \in {{\mathcal {C}}}^{2+k,\alpha }(\Sigma , \mathbb {R})$:

$$\begin{aligned} D_{\psi }\big |_{\psi =1}\nabla _y G^{\psi g}(Y_i^{-1}(y_i), Y_j^{-1}(y_j) )[\theta ] = D_{\psi }\big |_{\psi =1}\nabla _y H^{\psi g}(Y_i^{-1}(y_i), Y_j^{-1}(y_j) )[\theta ]. \end{aligned}$$

Now Proposition 1 yields for $\theta \in {{\mathcal {C}}}^{2+k,\alpha }(\Sigma , \mathbb {R})$ with $\textrm{supp}(\theta )\subset \Sigma \setminus \big \{Y_1^{-1}(y_1),\dots ,Y_m^{-1}(y_m)\big \}$:

Consider an element $u=(u_1,\cdots ,u_m)\in \mathbb {R}^{l+m}$ with $u_1,\dots ,u_l\in \mathbb {R}^2$, $u_{l+1},\dots ,u_m\in \mathbb {R}$, that is orthogonal to the range of $D_\psi {{\mathcal {F}}}_g(y,1)$, i.e. it satisfies for every $\theta \in {{\mathcal {C}}}^{2+k,\alpha }(\Sigma , \mathbb {R})$ with $\textrm{supp}(\theta ) \subset \Sigma {\setminus } \big \{Y_1^{-1}(y_1),\dots ,Y_m^{-1}(y_m)\big \}$:

This implies, using $\sum _{j=1}^m\sigma _j\ne 0$:

$$\begin{aligned} \sum _{i=1}^m \sigma _i\big \langle u_i,\nabla _{y_i} G^g(z, Y_i^{-1}(y_i))\big \rangle = 0 \qquad \text {for } z\in \Sigma \setminus \big \{Y_1^{-1}(y_1),\dots ,Y_m^{-1}(y_m)\big \}.\nonumber \\ \end{aligned}$$

(3.1)

Setting $\kappa _i=2\pi$ for $i=1,\dots ,l$ and $\kappa _i=\pi$ for $i=l+1,\dots ,m$ we have

$$\begin{aligned} G^g(z,Y_i^{-1}(y_i)) = H^g\big (z,Y_i^{-1}(y_i)\big ) - \frac{1}{\kappa _i} \chi \big (4|Y_i(z)-y_i|/\delta _{\xi _i}\big )\log |Y_i(z)-y_i|. \end{aligned}$$

Now we define $z_i(t):= Y_i^{-1}(y_i+tu_i)$ for $i\in \{1,\dots ,m\}$ and observe that

$$\begin{aligned} \nabla _{y_j}G^g\big (z_i(t),Y_j^{-1}(y_j)\big ) = {{\mathcal {O}}}(1)\qquad \text {as } t\rightarrow 0 \text { for }j\ne i \end{aligned}$$

whereas

$$\begin{aligned} \big \langle u_i,\nabla _{y_i} G^g(z_i(t), Y_i^{-1}(y_i))\big \rangle \rightarrow \infty \qquad \text {as } t\rightarrow 0. \end{aligned}$$

Equation (3.1) implies $u_i=0$ because $\sigma _i \ne 0$. This holds for all i, hence $u=0$ and $D_\psi {{\mathcal {F}}}_g(y,\psi )$ must be onto.

4 Proof of Theorem 1.2

Theorem 1.2 follows easily from the following lemma.

Lemma 4.1

  If $\sigma _i>0$ for $i=1,\dots ,m$ then $\displaystyle \lim _{x\rightarrow \partial (X\setminus \Delta _X)}|\nabla ^g f_g(x)|_g=\infty$.

Proof of Theorem 1.2

Lemma 4.1 implies that the set of critical points of $f_g$ is compact. It follows that if $f_g$ is a Morse function, then it has only finitely many critical points, and any ${{\mathcal {C}}}^2$-perturbation of $f_g$ is also a Morse function. Therefore $\textrm{Riem}_{Morse}^{k+2,\alpha }(\Sigma )$ is an open subset of $\textrm{Riem}^{k+2,\alpha }(\Sigma )$. By Theorem 1.1 it is also a dense subset.

Proof of Lemma 4.1

Consider a sequence $x^n \in X\setminus \Delta _X$ such that

$$\begin{aligned} x^n \rightarrow x^\infty \in \partial (X\setminus \Delta _X) = \Delta _X\cup \big ((\partial \Sigma )^l\times (\partial \Sigma )^{m-l}\big ) \subset \Sigma ^l\times (\partial \Sigma )^{m-l}. \end{aligned}$$

Case 1. $x^\infty \in \Delta _X\subset {\mathring{\Sigma }}^l\times (\partial \Sigma )^{m-l}$.

Then, there exists $\xi \in \Sigma$ such that the set $I:=\big \{i\in \{1,\dots ,m\}: x^\infty _i=\xi \big \}$ contains at least two elements. Moreover, if $\xi \in {\mathring{\Sigma }}$, then $I\subset \{1,\cdots ,l\}$; if $\xi \in \partial \Sigma$, then $I\subset \{l+1,\cdots , m\}$. In either case, we have that

$$\begin{aligned} \big |\nabla ^g_{x_i}R^g(x^n_i)\big | = {{\mathcal {O}}}(1)\text { and } \big |\nabla ^g_{x_i}G^g(x^n_i,x^n_j)\big | = {{\mathcal {O}}}(1)\quad \text { as } n\rightarrow +\infty , \end{aligned}$$

for $i\in I$ and $j\notin I$, and $\big |\nabla ^g_{x_i}H^g(x^n_i,x^n_j)\big | = {{\mathcal {O}}}(1)$ for $i,j \in I$. Let $y_\xi :U(\xi )\rightarrow B^\xi \subset \mathbb {R}^2$ be the isothermal chart with $y_{\xi }(\xi )=0$ introduced in Sect. 2. Then for $i\in I$, setting $y^n_j:= y_\xi (x^n_j) \in \mathbb {R}^2$ for $j\in I$ and assuming $y^n_i \ne 0$, we obtain as $n\rightarrow \infty$:

$$\begin{aligned} \begin{aligned} \big |\nabla ^g f_g(x^n)\big |_g&\ge \big |\nabla ^g_{x_i} f_g(x^n)\big |_g = 2\left| \nabla ^g_{x_i}\sum _{j\in I\setminus \{i\}}\sigma _i \sigma _jG^g(x^n_i,x^n_j)\right| _g + {{\mathcal {O}}}(1)\\&= \frac{2}{\kappa (\xi )}\left| \nabla ^g_{x_i}\sum _{j\in I\setminus \{i\}}\sigma _i\sigma _j\log \big |y_\xi (x^n_i)-y_\xi (x^n_j)\big |\right| _g + {{\mathcal {O}}}(1)\\&\ge \frac{2c}{\kappa (\xi )}\left| \nabla _{y_i}\sum _{j\in I\setminus \{i\}}\sigma _i\sigma _j\log \big |y^n_i-y^n_j\big |\right| + {{\mathcal {O}}}(1)\\&= \frac{2c}{\kappa (\xi )}\left| \sum _{j\in I\setminus \{i\}}\sigma _i\sigma _j\frac{y^n_i-y^n_j}{|y^n_i-y^n_j|^2}\right| + {{\mathcal {O}}}(1)\\&\ge \frac{2c}{\kappa (\xi )}\left| \sum _{j\in I\setminus \{i\}}\sigma _i\sigma _j\left\langle \frac{y^n_i-y^n_j}{|y^n_i-y^n_j|^2},\frac{y^n_i}{|y^n_i|}\right\rangle \right| + {{\mathcal {O}}}(1)\ \end{aligned} \end{aligned}$$

The second inequality is a consequence of the fact that there exists $c>0$ such that for any function $\textrm{g}: U(\xi )\rightarrow \mathbb {R}$:

$$\begin{aligned} \big |\nabla ^g_x \textrm{g}(x)\big |_g \ge c\big |\nabla \big (\textrm{g}\circ y^{-1}_\xi \big )\big (y_{\xi }(x)\big )\big | \quad \text {for } x\in U(\xi ). \end{aligned}$$

Now for $n\in \mathbb {N}$ we choose $i(n)\in I$ with $|y^n_{i(n)}| \ge |y^n_j|$ for all $j\in I$. This implies $y^n_{i(n)} \ne 0$ and

$$\begin{aligned} \big \langle y^n_{i(n)}-y^n_j,y^n_{i(n)}\big \rangle \ge \frac{1}{2}\big |y^n_{i(n)}-y^n_j\big |^2 > 0\quad \text {for } j\in I\setminus \{i(n)\}, \end{aligned}$$

hence

$$\begin{aligned} \left\langle \frac{y^n_{i(n)}-y^n_j}{|y^n_{i(n)}-y^n_j|^2},\frac{y^n_{i(n)}}{|y^n_{i(n)}|}\right\rangle \ge \frac{1}{2|y^n_{i(n)}|} \quad \text {for }j\in I\setminus \{i(n)\}. \end{aligned}$$

As a consequence we obtain, using $\sigma _i\sigma _j>0$ for all $i,j\in I$:

$$\begin{aligned} \big |\nabla ^g f_g(x^n)\big |_g\ge & {} \big |\nabla ^g_{x_{i(n)}} f_g(x^n)\big |_g \ge \frac{c}{\kappa (\xi )}\pi \sum _{j\in I\setminus \{i(n)\}}\sigma _{i(n)}\sigma _j\frac{1}{|y^n_{i(n)}|}+ {{\mathcal {O}}}(1) \\\rightarrow & {} \infty \quad \text {as }n\rightarrow \infty . \end{aligned}$$

Case 2. $x^{\infty }\notin \Delta _X$. Then there exists $i\in \{1,...,l\}$ such that $x^n_i\rightarrow \xi$ for some $\xi \in \partial \Sigma$.

We fix an isothermal chart $(y_{\xi }, U({\xi }))$ around $\xi$ as introduced in Sect. 2. For any $\zeta \in U_{r_{\xi }}({\xi })$, we decompose $G^g(\cdot ,\zeta )$ as follows:

$$\begin{aligned} G^g(\cdot ,\zeta )= \tilde{H}^g(\cdot ,\zeta )-\frac{1}{\kappa (\zeta )}\chi (4|y_{\xi }(\cdot ))-y_{\xi }(\zeta )|/r_{\xi })\log {|y_{\xi }(\cdot )-y_{\xi }(\zeta ) |}. \end{aligned}$$

(4.1)

Equation (2.2) implies $R^g(\zeta )=\tilde{H}^g(\zeta ,\zeta )+ \frac{1}{2\kappa (\zeta )}\varphi _\xi \big (y_\xi (\zeta )\big )$. Let $\partial _1,\partial _2$ be the standard basis of $\mathbb {R}^2$ and define $\partial _{\zeta _1},\partial _{\zeta _2} \in T_{\zeta }\Sigma$ be the corresponding basis of the tangent space of $\Sigma$ at $\zeta \in U(\xi )$.

Fix $\zeta \in U_{r_{\xi }}(\xi )\cap {\mathring{\Sigma }}$ and set $\delta :=y_{\xi }(\zeta )_2>0$. The representation formula for $\tilde{H}^g$ yields:

$$\begin{aligned} \partial _{\zeta _1} \tilde{H}^g(\eta ,\zeta ) |_{\eta =\zeta }= & {} \frac{1}{|\Sigma |_g} \int _{\Sigma } \partial _{\zeta _1} \tilde{H}^g(\cdot ,\zeta )dv_g - \int _{\Sigma }G^g(\cdot ,\zeta ) \Delta _g \partial _{\zeta _1}\tilde{H}^g(\cdot ,\zeta ) dv_g\nonumber \\{} & {} + \int _{\partial \Sigma } G^g(\cdot ,\zeta )\partial _{\nu _g} \partial _{\zeta _1} \tilde{H}^g(\cdot ,\zeta ) ds_g\nonumber \\= & {} \frac{1}{2\pi } \int _{\partial \Sigma } G^g(\zeta ,x) \partial _{\nu _g}\partial _{\zeta _1}\nonumber \\ {}{} & {} \times \left( \chi \left( {4|y_{\xi }(x) -y_{\xi }(\zeta )|} /r_{\xi }\right) \log {|y_{\xi }(x)-y_{\xi }(\zeta ) |}\right) ds_g(x) + {{\mathcal {O}}}(1) \nonumber \\= & {} - \frac{1}{2\pi }\int _{B^{\xi }\cap \partial \mathbb {R}^2_{+}} G^g\left( \zeta , y_{\xi }^{-1}(y)\right) \chi (4|y-y_{\xi }(\zeta )|/ r_{\xi }) \nonumber \\ {}{} & {} \times \frac{ 2(y_1-y_{\xi }(\zeta )_1)(y_2-y_{\xi }(\zeta )_2)}{ |y-y_{\xi }(\zeta )|^4} dy + {{\mathcal {O}}}(1)\nonumber \\= & {} -\frac{1}{2\pi \delta }\int ^{\varepsilon /\delta }_{-\varepsilon /\delta } G^g\left( \zeta , y_{\xi }^{-1}\big ((\delta s,0)+ (y_{\xi }(\zeta )_1, 0)\big )\right) \nonumber \\{} & {} \times \frac{2s}{(s^2+1)^2} ds+{{\mathcal {O}}}\left( 1\right) . \end{aligned}$$

(4.2)

as $\delta \rightarrow 0$. Here $\varepsilon \in (0,\frac{r_{\xi }}{16})$ is chosen sufficiently small. Decomposing $G^g$ as in (4.1), we deduce for $\zeta \in U(\xi )$ with $y_{\xi }(\zeta )_2< \frac{r_{\xi }}{16}$:

$$\begin{aligned} G^g\left( \zeta , y_{\xi }^{-1}\big ((\delta s,0)+ (y_{\xi }(\zeta )_1, 0)\big )\right)= & {} \tilde{H}^g\left( \zeta , y_{\xi }^{-1}\big ((\delta s,0)+ (y_{\xi }(\zeta )_1, 0)\big )\right) \\ {}{} & {} -\frac{1}{\pi }\log |(\delta s, -\delta )|, \end{aligned}$$

Now we apply the mean value theorem for $\tilde{H}^g$ and obtain as $\delta s\rightarrow 0$:

$$\begin{aligned} \tilde{H}^g\big (\zeta , y_{\xi }^{-1}\big ((\delta s,0)+ (y_{\xi }(\zeta )_1, 0)\big )\big )= & {} \tilde{H}^g\left( \zeta ,y_{\xi }^{-1}\big (y_{\xi }(\zeta )_1,0\big )\right) \nonumber \\ {}{} & {} + {{\mathcal {O}}}\big (|\delta s|\sup _{x\in \partial \Sigma } \Vert \nabla _x\tilde{H}^g(\cdot ,x)\Vert _{{{\mathcal {C}}}(\Sigma )}\big ) \end{aligned}$$

(4.3)

This implies as $\delta \rightarrow 0$:

$$\begin{aligned}{} & {} \left| -\frac{1}{2\pi \delta }\int ^{\varepsilon /\delta }_{-\varepsilon /\delta } G^g(\zeta , y_{\xi }^{-1} \big ((\delta s,0)+ (y_{\xi }(\zeta )_1, 0)\big ) \frac{2s}{ (s^2+1)^2} ds\right| \\{} & {} \quad \le \left| -\frac{1}{2\pi \delta }\tilde{H}^g\big (\zeta ,y_{\xi }^{-1}\big (y_{\xi }(\zeta )_1,0\big )\big ) \int _{|s|\le \varepsilon /\delta }\frac{2s}{(s^2+1)^2} ds\right| \\{} & {} \qquad + \left| {{\mathcal {O}}}\left( \int _{|s|\le \varepsilon /\delta } \frac{2s^2}{(s^2+1)^2} ds\right) + \frac{1}{2\pi ^2\delta }\int _{|s|\le \varepsilon /\delta } \log (\delta \sqrt{s^2+1}) \frac{2s}{(s^2+1)^2} ds\right| \\{} & {} \quad \le {{\mathcal {O}}}(1). \end{aligned}$$

Now (4.2) yields $\partial _{\zeta _1}\tilde{H}^g(\zeta ,\zeta )={{\mathcal {O}}}(1)$ for $\zeta \in U_{r_{\xi }}(\xi )\cap \partial \Sigma$. Consequently, for $\zeta \in U_{r_{\xi }}(\xi )$ with $y_{\xi }(\zeta )_2< \frac{r_{\xi }}{16}$, we have proven that:

$$\begin{aligned} \partial _{\zeta _1}R^g(\zeta )= {{\mathcal {O}}}(1)\qquad \text {as } |y_{\xi }(\zeta )|\rightarrow 0. \end{aligned}$$

(4.4)

The representation formula of $\tilde{H}^g$ yields for $\zeta \in U_{r_{\xi }}(\xi )\cap {\mathring{\Sigma }}$ as $\delta \rightarrow 0$:

$$\begin{aligned} \partial _{\zeta _2} \tilde{H}^g(\eta ,\zeta ) |_{\eta =\zeta }{} & {} =\frac{1}{|\Sigma |_g} \int _{\Sigma } \partial _{\zeta _2} \tilde{H}^g(\cdot ,\zeta )dv_g - \int _{\Sigma }G^g(\cdot ,\zeta ) \Delta _g \partial _{\zeta _2}\tilde{H}^g(\cdot ,\zeta ) dv_g\\{} & {} \quad + \int _{\partial \Sigma } G^g(\cdot ,\zeta )\partial _{\nu _g} \partial _{\zeta _2} \tilde{H}^g(\cdot ,\zeta ) ds_g\\{} & {} = \frac{1}{2\pi } \int _{\partial \Sigma } G^g(\zeta ,x) \partial _{\nu _g} \partial _{\zeta _2}\\{} & {} \quad \times \left( \chi \left( {4|y_{\xi }(x)-y_{\xi }(\zeta )|} /r_{\xi }\right) \log {|y_{\xi }(x)-y_{\xi }(\zeta ) |}\right) ds_g(x)+ {{\mathcal {O}}}(1) \\{} & {} \le \frac{1}{2\pi }\int _{B^{\xi }\cap \partial \mathbb {R}^2_{+}} G^g(\zeta , y_{\xi }^{-1}(y)) \chi (4|y-y_{\xi }(\zeta )|/r_{\xi })\\{} & {} \quad \times \frac{(y_1- y_{\xi }(\zeta )_1)^2- (y_2- y_{\xi }(\zeta )_2)^2}{ |y-y_{\xi }(\zeta )|^4} dy+{{\mathcal {O}}}(1) \\{} & {} \le \frac{1}{2\pi \delta }\int ^{\varepsilon /\delta }_{-\varepsilon /\delta }G^g(\zeta , y_{\xi }^{-1}((\delta s,0)+(y_{\xi }(\zeta )_1, 0)) \frac{s^2-1}{ (s^2+1)^2}ds +{{\mathcal {O}}}(1) \\{} & {} {\mathop {\le }\limits ^{(4.3)}} \frac{1}{2\pi \delta } \tilde{H}^g(\zeta ,y_{\xi }^{-1}(y_{\xi }(\zeta )_1,0))\int _{|s|\le \varepsilon /\delta } \frac{s^2-1}{(s^2+1)^2} ds\\{} & {} \quad + {{\mathcal {O}}}\left( 1+\int _{|s|\le \varepsilon /\delta } \frac{|s(s^2-1)|}{(s^2+1)^2} ds\right) \\{} & {} \quad + \frac{\log (\delta ^{-1})}{2\pi ^2\delta }\int _{|s|\le \varepsilon /\delta } \frac{s^2-1}{(s^2+1)^2} ds\\{} & {} \quad + \frac{1}{2\pi ^2\delta }\int _{|s|\le \varepsilon /\delta }\log \left( \frac{1}{\sqrt{s^2+1}}\right) \frac{s^2-1}{(s^2+1)^2} ds +{{\mathcal {O}}}(1)\\{} & {} \le \frac{1}{2\pi ^2\delta }\int _{|s|\ge 1} \log \left( \frac{\sqrt{s^{-2}+1}}{\sqrt{s^2+1}}\right) \frac{s^2-1}{(s^2+1)^2} ds\\{} & {} \quad - \frac{1}{2\pi ^2\delta }\int _{|s|\ge \varepsilon /\delta }\log \left( \frac{1}{\sqrt{s^2+1}}\right) \frac{s^2-1}{(s^2+1)^2} ds\\{} & {} \quad - \frac{\log (\delta ^{-1})\varepsilon }{\pi ^2(\delta ^2+\varepsilon ^2)}+{{\mathcal {O}}}(\delta ^{-1/2}) +{{\mathcal {O}}}(\log (\delta ^{-1})+1) \\{} & {} \le - \frac{1}{2\pi ^2\delta }\int _{|s|\ge 1} \log \left( s^{2}\right) \frac{s^2-1}{(s^2+1)^2} ds + {{\mathcal {O}}}(\delta ^{-\frac{1}{2}})\le -\frac{1}{4\pi \delta }. \end{aligned}$$

The last inequality used the identity

$$\begin{aligned} \int _{|s|\ge 1} \log (s^2) \frac{s^2-1}{(s^2+1)^2} ds =\pi . \end{aligned}$$

From the above estimate we deduce for $\zeta \in {\mathring{\Sigma }}\cap U_{r_{\xi }}(\xi )$:

$$\begin{aligned} \partial _{\zeta _2} R^g(\zeta )\le -\frac{1}{2\pi |y_{\xi }(\zeta )_2|}\qquad \text {as }|y_{\xi }(\zeta )_2|\rightarrow 0. \end{aligned}$$

(4.5)

Now if $I:=\big \{i\in \{1,\dots ,m\}: x^n_i\rightarrow \xi \big \}$ contains only a single element i then (4.5) yields

$$\begin{aligned} \begin{aligned} \big |\nabla ^g f_g(x^n)\big |_g&\ge \big |\nabla ^g_{x_{i}} f_g(x^n)\big |_g \ge \big |\sigma _i^2\nabla ^g_{x_{i}} R^g(x_i^n) + 2\sigma _i\sum _{j\ne i} \nabla ^g_{x_{i}} G^g(x^n_i,x^n_j)+ \nabla ^g_{x_{i}}h(x^n)\big |_g\\&\ge \sigma _i^2 |\nabla ^g_{x_i}R^g(x^n_i)|+{{\mathcal {O}}}(1) \ge \frac{ c\sigma _i^2 }{2\pi |y_{\xi }(x^n_i)_2|}+{{\mathcal {O}}}(1)\\&\rightarrow \infty \qquad \text {as }n\rightarrow \infty . \end{aligned} \end{aligned}$$

Here $c>0$ is a constant such that for any function $F:U(\xi )\rightarrow \mathbb {R}$:

$$\begin{aligned} \big |\nabla ^g_x F(x)\big |_g \ge c\big |\nabla \big (F\circ y^{-1}_\xi \big )\big (y_{\xi }(x)\big )\big | \quad \text {for } x\in U(\xi ). \end{aligned}$$

(4.6)

Next we consider the case that I contains at least two elements. Then $\xi ^n_i\in U(\xi )$ for n large and $i\in I$. By a direct calculation we obtain for $j\in I\setminus \{i\}$, $\iota =1,2$ and $n\rightarrow \infty$:

$$\begin{aligned} \partial _{\zeta _\iota } G^g(x^n_j,\zeta ) |_{\zeta =x^n_{i}}= & {} \partial _{\zeta _\iota } \tilde{H}^g(x^n_j,\zeta ) |_{\zeta =x^n_{i}}\nonumber \\{} & {} + \frac{1}{\kappa (x^n_i)} \log \frac{1}{|y_{\xi }(x^n_j)- y_{\xi }(x_{i}^n)|} \partial _{\zeta _\iota }\chi (4|y_{\xi }(x^n_j)- y_{\xi }(\zeta )|/r_{\xi })|_{\zeta =x_{i}^n}\nonumber \\{} & {} + \left. \frac{1}{\kappa (x^n_i)} \chi (4|y_{\xi }(x^n_j)- y_{\xi }(x_{i}^n)|/r_{\xi })\partial _{\zeta _\iota } \log \frac{1}{|y_{\xi }(x^n_j)- y_{\xi }(\zeta )|}\right| _{\zeta =x_{i}^n}\nonumber \\= & {} \partial _{\zeta _\iota } \tilde{H}^g(x^n_j,\zeta ) |_{\zeta =x^n_{i}} -\frac{1}{\kappa (x^n_i)}\frac{(y_{\xi }(x_{i}^n)- y_{\xi }(x^n_j) )_{\iota }}{|y_{\xi }(x^n_j) - y_{\xi }(x_{i}^n)|^2} +{{\mathcal {O}}}(1). \end{aligned}$$

(4.7)

For $n\in \mathbb {N}$ we set

$$\begin{aligned} \varrho _n:=\max \Big (\big \{ y_{\xi }(x^{n}_i)_2: i\in I\big \}\cup \big \{ |y_{\xi }(x_{i}^n)_1-y_{\xi }(x^{n}_j)_1|: i,j\in I\text { with } i\ne j \big \}\Big ). \end{aligned}$$

If there exists $i(n)\in I$ such that $\varrho _n=y_{\xi }\big (x^n_{i(n)}\big )_2$, then $i(n)\in \{1,\dots ,l\}$ satisfies

$$\begin{aligned} y_{\xi }(x^n_{i(n)})_2-y_{\xi }(x^n_{j})_2\ge 0 \text { and } y_{\xi }(x^n_{i(n)})_2\ge |(y_{\xi }(x_{i}^n)_1-y_{\xi }(x^{n}_j)_1| \quad \text {for every }i,j\in I. \end{aligned}$$

Given $\zeta =x^n_{i(n)}$ and $\eta = x^n_j$ with $j\in I\cap \{1,\ldots ,l\}{\setminus }\{i(n)\})$, we will calculate the upper bound of $\partial _{\zeta _2}\tilde{H}^g(\eta ,\zeta )$ as $n\rightarrow 0$. Setting $a=y_{\xi }(\eta )_2>0$ and $b=y_{\xi }(\zeta )_1-y_{\xi }(\eta )_1$ we have for $|b|\le \varrho _n$ as $n\rightarrow \infty$:

$$\begin{aligned} \partial _{\zeta _2} \tilde{H}^g(\eta ,\zeta ){} & {} =\frac{1}{|\Sigma |_g} \int _{\Sigma } \partial _{\zeta _2} \tilde{H}^g(\cdot ,\zeta )dv_g -\int _{\Sigma }G^g(\cdot ,\eta ) \Delta _g \partial _{\zeta _2}\tilde{H}^g(\cdot ,\zeta ) dv_g\\{} & {} \quad + \int _{\partial \Sigma } G^g(\cdot ,\eta )\partial _{\nu _g} \partial _{\zeta _2} \tilde{H}^g(\cdot ,\zeta ) ds_g\\{} & {} = {{\mathcal {O}}}(1) +\frac{1}{2\pi } \int _{\partial \Sigma } G^g(\eta ,x) \partial _{\nu _g}\partial _{\zeta _2} \big (\chi \big ({4|y_{\xi }(x)-y_{\xi }(\zeta )|} /r_{\xi }\big )\\{} & {} \quad \times \log {|y_{\xi }(x)-y_{\xi }(\zeta ) |}\big ) ds_g(x)\\{} & {} \le {{\mathcal {O}}}(1) + \frac{1}{2\pi }\int _{B^{\xi }\cap \partial \mathbb {R}^2_{+}} G^g(\eta , y_{\xi }^{-1}(y)) \chi (4|y-y_{\xi }(\zeta )|/r_{\xi })\\{} & {} \quad \times \frac{(y_1- y_{\xi }(\zeta )_1)^2-(y_2- y_{\xi }(\zeta )_2)^2}{ |y-y_{\xi }(\zeta )|^4} dy\\{} & {} \le {{\mathcal {O}}}(1) +\frac{1}{2\pi \varrho _n }\int ^{\varepsilon /\varrho _n}_{-\varepsilon /\varrho _n} G^g(\eta , y_{\xi }^{-1}((\varrho _n s,0)+ (y_{\xi }(\zeta )_1, 0)) \frac{ s^2-1}{ (s^2+1)^2} ds \\{} & {} {\mathop {\le }\limits ^{(4.3)}}\frac{1}{2\pi \varrho _n} \tilde{H}^g(\eta ,y_{\xi }^{-1}(y_{\xi }(\zeta )_1,0)) \int _{|s|\le \varepsilon /\varrho _n} \frac{s^2-1}{(s^2+1)^2} ds\\{} & {} \quad + {{\mathcal {O}}}\left( 1+\int _{|s|\le \varepsilon /\varrho _n} \frac{|s(s^2-1)|}{(s^2+1)^2} ds\right) \\{} & {} \quad -\frac{\log (a)}{2\pi ^2\varrho _n}\int _{|s|\le \varepsilon /\varrho _n} \frac{s^2-1}{(s^2+1)^2} ds\\{} & {} \quad + \frac{1}{2\pi ^2\varrho _n} \int _{|s|\le \varepsilon /\varrho _n}\log \left( \frac{1}{\sqrt{1+(\varrho _n s+b )^2/a^2}}\right) \frac{s^2-1}{(s^2+1)^2} ds\\{} & {} \le -\frac{1}{2\pi ^2\varrho _n}\int _{|s|\ge \varepsilon /\varrho _n} \left( \log (a^{-1})+ \log \left( \frac{1}{\sqrt{1+(\varrho _n s+b )^2/a^2}}\right) \right) \\{} & {} \quad \times \frac{s^2-1}{(s^2+1)^2} ds\\{} & {} \quad +\frac{1}{2\pi ^2\varrho _n}\int _{|s|\ge 1} \log \left( \frac{\sqrt{1+(\varrho _n s^{-1}+b )^2/a^2}}{\sqrt{1+(\varrho _n s+b )^2/a^2}}\right) \frac{s^2-1}{(s^2+1)^2} ds\\{} & {} \quad +{{\mathcal {O}}}\big (1+\log (\varrho _n^{-1})\big )\\{} & {} \le {{\mathcal {O}}}\big (\log (\varrho _n^{-1})\big ) \end{aligned}$$

Here we used the inequalities:

$$\begin{aligned} \log \left( \frac{\sqrt{1+(\varrho _n s^{-1}+b )^2/a^2}}{\sqrt{1+(\varrho _n s+b )^2/a^2}}\right) \le 0, \text { for }|b|\le \varrho _n, |s|\ge 1 \end{aligned}$$

and

$$\begin{aligned} 0\le \log \sqrt{a^2+(\varrho _n s+b)^2}\frac{s^2-1}{(s^2+1)^2} \le C \frac{\varrho _n ^{\frac{1}{2} }}{s^{\frac{3}{2}}} \end{aligned}$$

for some constant $C>0$, any $s\in \{ s\in \mathbb {R}:|s|\ge \varepsilon /\varrho _n,\ a^2+(\varrho _ns+b)^2\ge 1\}$. We notice that we have for $j\in I\cap \{l+1,\ldots ,m\}$:

$$\begin{aligned} |\partial _{\zeta _2} \tilde{H}^g(x^n_j,\zeta ) |_{\zeta =x^n_{i(n)}}|= |\partial _{\zeta _2} \tilde{H}^g(\zeta ,x^n_j) |_{\zeta =x^n_{i(n)}}|\le {{\mathcal {O}}}(1) \qquad \text {as }n\rightarrow \infty ; \end{aligned}$$

(4.8)

and for $j\in I\cap \{1,\dots ,l \}\setminus \{ i(n)\}$, $\varrho _n=y_{\xi }(x^n_{i(n)})$:

$$\begin{aligned} \big |\partial _{\zeta _2} \tilde{H}^g(x^n_j,\zeta ) |_{\zeta =x^n_{i(n)}}\big |\le {{\mathcal {O}}}\big (\log (\varrho _n^{-1})\big ) \qquad \text {as }n\rightarrow \infty . \end{aligned}$$

(4.9)

Now (4.5)-(4.9) imply for $n\rightarrow \infty$:

$$\begin{aligned} \begin{aligned} |\nabla ^g f_g(x^n)|_g&\ge \left| \nabla ^g_{x_{i(n)}}\left( \sigma _i^2 R^g(x^n_{i(n)})+2 \sigma _{i(n)} \sum _{j\ne i(n)} \sigma _jG^g(x^n_{i(n)}, x^n_j)+ h(x^n)\right) \right| _g\\&\ge c \left| \partial _{(x_{i(n)})_2} \left( \sigma ^2_{i(n)}R^g(\xi ^n_{i(n)})+ 2\sigma _{i(n)}\sum _{j\in I\setminus \{i(n)\} }\sigma _j G^g(\xi _{i(n)}, \xi _j)) \right) \right| +{{\mathcal {O}}}(1)\\&\ge {{\mathcal {O}}}\big (\log (\varrho _n^{-1})\big ) + \sigma ^2_{i(n)} \frac{ c }{2\pi \varrho _n}\\&\rightarrow \infty . \end{aligned} \end{aligned}$$

Here $c>0$ is as in (4.6). If $\varrho _n>\max \{y_{\xi }(x^n_i): i\in I\}$, then we take $i(n)\in I$ such that $y_{\xi }(x^n_{i(n)})_1=\max \{ y_{\xi }(x^n_i): i\in I\}$. The proof of Lemma 6 in [16] implies for $x\in U(\xi )$ and $y^*_{\xi }(x):= (y_{\xi }(x)_1, -y_{\xi }(x)_2)$:

$$\begin{aligned} \left\| G^g(\cdot ,x)+\frac{1}{2\pi }\log \big |y_{\xi }(\cdot )-y_{\xi }(x)\big | +\frac{1}{2\pi }\log \big |y_{\xi }(\cdot )-y^{*}_{\xi }(x)\big |\right\| _{{{\mathcal {C}}}^1(U_{r_{\xi }}(\xi ))}\le C \end{aligned}$$

where $C>0$ depends only on $(\Sigma ,g)$ and on $\xi$. Therefore we have for $j\in I\setminus \{i(n)\}$ as $n\rightarrow \infty$:

$$\begin{aligned} \partial _{\zeta _1} G(x^n_j, \zeta )\big |_{\zeta =x^n_{i(n)}}\le & {} -\frac{1}{2\pi } \frac{y_{\xi }(x^n_{i(n)})_1- y_{\xi }(x^n_j)_1}{|y_{\xi }(x^n_{i(n)})- y_{\xi }(x^n_j)|^2}\nonumber \\ {}{} & {} -\frac{1}{2\pi } \frac{y_{\xi }(x^n_{i(n)})_1- y_{\xi }(x^n_j)_1}{|y_{\xi }(x^n_{i(n)})- y^*_{\xi }(x^n_j)|^2}+ {{\mathcal {O}}}(1). \end{aligned}$$

(4.10)

We can take $i'(n)\in I{\setminus }\{ i(n)\}$ such that $\varrho _n= y_{\xi }(x^n_{i(n)})_1-y_{\xi }(x^n_{i'(n)})_1$ by the assumption. The inequality (4.10) yields

$$\begin{aligned} \partial _{\zeta _1} G(x^n_{i'(n)}, \zeta )\big |_{\zeta =x^n_{i(n)}}\le -\frac{1}{4\pi \varrho _n}\qquad \text {as }n\rightarrow \infty . \end{aligned}$$

(4.11)

From (4.4) together with (4.10) and (4.11) we derive the following estimate for the gradient of $f_g$ as $n\rightarrow \infty$:

$$\begin{aligned} \begin{aligned} |\nabla ^g f_g(x^n)|_g&\ge \left| \nabla ^g_{x_{i(n)}}\left( \sigma _i^2 R^g(x^n_{i(n)})+2 \sigma _{i(n)} \sum _{j\ne i(n)} \sigma _jG^g(x^n_{i(n)}, x^n_j)+ h(x^n)\right) \right| _g\\&\ge c \left| \partial _{(x_{i(n)})_1} \left( \sigma ^2_{i(n)}R^g(\xi ^n_{i(n)})+ 2\sigma _{i(n)}\sum _{j\in I\setminus \{i(n)\} }\sigma _j G^g(\xi _{i(n)}, \xi _j)) \right) \right| +{{\mathcal {O}}}(1)\\&\ge {{\mathcal {O}}}(1)+ \sigma _{i(n)}\sigma \big (i'(n)\big ) \frac{ c }{2 \pi \varrho _n}\\&\rightarrow \infty \end{aligned} \end{aligned}$$

Again $c>0$ is as in (4.6). This completes the proof of Lemma 4.1.