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I have a question from my tutorials and I don't know how to start...

Let $U$ be open in $\mathbb{R}^{n}$ and let $f:U\rightarrow \mathbb{R}$ a $C^{2}$ function. Let $p$ be a point in $U$ where $df_{p}$ of $f$ at $p$ does not vanish. Show that there exists a system of local coordinates $x^{i}$ defined in a neighbourhood of $p$ in which

\begin{eqnarray*} \frac{\partial^{2}f}{\partial x^{i}\partial x^{j}}=0 \end{eqnarray*}

for all $1\leq i,j\leq n$.

Thanks!

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1 Answer 1

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Let $y^1,\dots,y^n$ be the original coordinates of $\mathbb{R}^n$. Since $df_p\ne 0$, without loss of generality, we may assume that $\frac{\partial f}{\partial y^n}(p)\ne 0$. Now consider the map $$\varphi:U\to\mathbb{R}^n,\quad q\mapsto (y^1(q),\cdots,y^{n-1}(q),f(q)).$$ Then $\varphi$ is $C^2$, and its Jacobian determinant $\det( d\varphi)=\frac{\partial f}{\partial y^n}$. In particular, $\det( d\varphi)(p)\ne 0$. Therefore, by inverse function theorem, there exists a neighborhood $V$ of $p$, such that $\varphi:V\to \varphi(V)$ is invertible. As a result, $\varphi$ defines a system of local coordinates $x^1,\dots,x^n$ on $V$, i.e. for $q\in V$, $\varphi(q)=(x^1(q),\cdots,x^n(q))$. By definition, $f=x^n$ on $V$, and hence $\frac{\partial f}{\partial x^i\partial x^j}=0$ on $V$, $1\le i,j\le n$.

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