I'm reading at the moment a Theorem about $\;W^{2,2}\;$ estimates and the following is an assumption that must hold in order for the Theorem to be valid:
Let $\;f:\mathbb R^{n-1} \to \mathbb R\;$ be a $\;\mathcal C^2\;$ function such that if $\;U \subset \mathbb R^n\;$ then $\;f\;$ decribes $\;\partial U\;$ in a neighborhood of $\;x_0 \in \partial U\;$: $x_n=f(x_1,\dots,x_{n-1})\;$. ASSUMPTION: $$\;\frac{{\partial}^2 f(x_0)}{\partial x_i \partial x_j}=0\;\forall i\neq j\;$$
However, our professor stated that this assumption is not really necessary since through a rotation of the co-ordinate system we can always deduce the above.
So, as I understand it, if $\;(x_1, \dots, x_{n-1})\;$ are the initial coordinates and $\;O\;$ stands for a rotation such that $\;O O^T=O^T O=I\;$ the new coordinates are given by: $\;y=Ox\;$ and $\;f(x)=\hat f(Ox)\;$.
Then after computation: $$\;\frac{{\partial}^2 f}{\partial x_i \partial x_j}=(O_{jm})^T\frac{{\partial}^2 \hat f}{\partial y_σ \partial y_m} O_{σi}\;, \text{ i.e. } \;{{\partial}^2}_x f=O^T {{\partial}^2}_y \hat f O\;(*)\;$$
Still I can't see why the above assumption could be ommited. Why is $\;(*)\;$ enough in order to deduce the given assumption?
I'm having a really hard time getting my head around it so any help would be valuable. Thanks in advance!
Since $f\in C^2$, the matrix $$\tag{Hessian} D^2f(x_0)=\left[ \frac{\partial f}{\partial x_i \partial x_j}(x_0)\right]_{1\le i, j\le n}$$ is symmetric, and so there exists an orthogonal matrix $O\in \mathbb R^{n\times n}$ such that $O\, D^2f(x_0)\, O^T$ is diagonal.
Define a new coordinate system by $$\tag{1} O\mathbf y= \mathbf x,$$ and set $g(y):=f(x)$, where, of course, $x$ and $y$ are related by (1). (That is, $g(y):=f(Oy)$). Using the chain rule, check that $$D^2g(y_0)=O\, D^2 f(x_0)\, O^T.$$ Conclude that the Hessian matrix of $g$ is diagonal at $y_0=O^Tx_0$.