How to compute $\partial^2_{n,n} u(x_0)$ for $x_0\in\partial\Omega$? Here is the setting: Let $\Omega\subset\mathbb{R}^2$ is a bounded smooth strictly convex domain, $\mu_1(\Omega)$ is the first positive Neumann eigenvalue and $u$ is the corresponding eigenfunction. Assume that $u$ attains its maximum on $\bar{\Omega}$ at $x_0\in\partial\Omega$. I want to prove that $\partial^2_{n,n}u(x_0)\neq 0$, where $n$ is the outer normal at $x_0$. Here the notation means
$$
\frac{\partial}{\partial n}\left(\frac{\partial u}{\partial n}\right)\bigg|_{x=x_0}.
$$
Here is my point: Since $\Omega$ is strictly convex, I can write $\partial\Omega$ as the image of a strictly concave function $f$. I want to deduce a formula like $\partial^2_{n,n} u(x_0)=f''\cdot b$, where $b$ is positive, and then I can get the desired answer.
However, by the assumption, at point $x_0$, it's easy to find out that $\partial u_n=\partial u_t=\partial^2 u_{n,t}=0$, and so if we use the parameter representation of $\partial\Omega$, we can not obtain anything useful.
Any help will be appreciated a lot! Thanks!
 A: More a long comment than an answer, and also a comment oriented toward showing that the calculations involved in an hypothetical answer are not trivial at all. The reason for which is so difficult to answer to this question is the fact that the evaluation of the iterated normal derivative imply the knowledge of how to multiply distributions in a meaningful and unique way.
To see this, let's first consider the standard classical definition of normal derivative $\partial_n$:
$$\DeclareMathOperator{\Grad}{\nabla\!}\DeclareMathOperator{\dist}{dist}
\frac{\partial}{\partial {n}} u(x)\triangleq n(x) \cdot 
\Grad u(x)\big|_{x\in\partial\Omega},
$$
where the dot $\cdot$ stands for the scalar product in $\Bbb R^N$ (and implicitly  $\Omega\Subset\Bbb R^N$). It seems to be a simple definition of a trace, if we do not pay attention to the fact that $n(x)$ is a vector function supported only on the boundary of $\Omega$. Digging a little deeper, we see that the reality is very different. A formal, mathematically rigorous ([1], pp. 143-144 and [2], p. 6-7) definition tells us that
$$
n(x)=\Grad\chi_\Omega.\label{1}\tag{1}
$$
and thus
$$
\frac{\partial}{\partial {n}} u(x)\triangleq \Grad\chi_\Omega \cdot \Grad u(x),
$$
where $\chi_\Omega$ is the characteristic function of the domain $\Omega$.
Precisely, if $\Omega$ is at least a Caccioppoli set (as all smooth domains are), the second member of \eqref{1} is a finite vector Radon measure supported on $\partial\Omega$. Intuitively it has the following structure
$$
n(x)\simeq\nu(x)\delta\big(\dist(x,\partial\Omega)\big)
$$
where

*

*$\nu:\Bbb R^N\to\Bbb R^N$ is a vector function such that $\|\nu(x)\|=1$ and

*$\delta\big(\dist(x,\partial\Omega)\big)$ is a Dirac measure supported on $\partial\Omega$ with $\dist(\cdot,\partial\Omega):\Bbb R^N\to\Bbb R_{\ge 0}$ defined as the distance from the set $\partial\Omega$.

Said that, let's proceed formally to the calculation of the iterated normal derivative. Since $\Grad u =0$ in $x_0$ as in the OP $u$ is assumed to have a maximum there, then
$$\begin{split}
\left.\frac{\partial^2 }{\partial {n}^2} u(x)\right|_{x=x_0} &=\Grad\chi_\Omega \cdot \Grad \big(\Grad\chi_\Omega \cdot \Grad u\big) =
\Grad\chi_\Omega \cdot \big(\Grad\Grad\chi_\Omega \cdot \Grad u+ \Grad\chi_\Omega \cdot \Grad \Grad u\big)\\
&=\Grad\chi_\Omega \cdot \big(\Grad\chi_\Omega \cdot \Grad \Grad u\big) \\
& = \Grad\chi_\Omega \cdot \left[\Grad\chi_\Omega \cdot
\begin{pmatrix}
\dfrac{\partial^2 u}{\partial {x^2_1}} & \dfrac{\partial^2 u}{\partial {x_1} \partial{x_2}} & \cdots & \dfrac{\partial^2 u}{\partial {x_1} \partial{x_N}}\\
\dfrac{\partial^2 u}{\partial {x_2x_1}} & \dfrac{\partial^2 u}{\partial {x^2_2}} & \cdots &  \dfrac{\partial^2 u}{\partial {x_1} \partial{x_N}} \\
\vdots & \vdots & \ddots &\vdots\\
\dfrac{\partial^2 u}{\partial {x_Nx_1}} & \dfrac{\partial^2 u}{\partial {x_Nx_2}}& \cdots & \dfrac{\partial^2 u}{\partial {x^2_N}}
\end{pmatrix}
\right]
\end{split}\label{2}\tag{2}
$$
Expression \eqref{2} shows the problem: going on calculating the scalar products, you'll obtain an expression containing products of finite vector Radon measures which are de facto Dirac type distributions. Now there are several theories trying to associate a meaning to this kind of product, but none of them is entirely satisfactory in all aspects, so here lies the core difficulty of the problem.
References
[1]  Luigi Ambrosio, Nicola Fusco, Diego Pallara, Functions of bounded variation and free discontinuity problems, Oxford Mathematical Monographs, New York and Oxford: The Clarendon Press/Oxford University Press, New York, pp. xviii+434 (2000), ISBN 0-19-850245-1, MR1857292, Zbl 0957.49001.
[2] Enrico Giusti (1984), Minimal surfaces and functions of bounded variations, Monographs in Mathematics, 80, Basel–Boston–Stuttgart: Birkhäuser Verlag, pp. XII+240, ISBN 978-0-8176-3153-6, MR 0775682, Zbl 0545.49018
