Let $f$ be a $C^2$ function defined on a neighborhood of $0$ in $\mathbb R^n$ such that $f(0)=0, df(0)\not=0$. By the Implicit Function Theorem, it is easy to get (after a rotation) that near 0 $$ f(x)=\bigl(x_n-\alpha(x')\bigr) e(x),\quad\text{$\alpha\in C^2$ near the origin in $\mathbb R^{n-1}$, $e\in C^1$.} $$ Moreover I claim that for $T=(t',t_n)\in \mathbb R^{n-1}\times \mathbb R$ $$ f''(x',\alpha(x'))T^2=-e(x', \alpha(x'))\alpha''(x')(t')^2 +2 (e'(x', \alpha(x'))\cdot T)(t_n-\alpha'(x')\cdot t'), $$ where $g''$ stands for the symmetric matrix of second derivatives. It would be a simple consequence of Leibniz formula if $e$ were $C^2$, but $e$ is no better than $C^1$. I guess a "direct" proof would do.
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$\begingroup$ Sorry, your notation is a bit confusing. What does $f''$ mean? It's not $\partial^2 f / \partial x_n^2$ is it? $\endgroup$– Igor KhavkineCommented Jan 30, 2017 at 12:54
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$\begingroup$ @Igor Khavkine I wrote in the text above that $f''$ stands for the matrix of second derivatives. $\endgroup$– BazinCommented Jan 30, 2017 at 13:35
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$\begingroup$ Sorry for another silly question, but isn't the cross-term $(x_n-\alpha(x'))' e'(x',\alpha(x'))$ missing then? $\endgroup$– Igor KhavkineCommented Jan 30, 2017 at 14:24
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$\begingroup$ Can't you just approximate $f''$ by smooth functions? $\endgroup$– Christian RemlingCommented Jan 30, 2017 at 16:04
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$\begingroup$ @Igor Khavkine Sorry, I wrote now something which hopefully makes sense. $\endgroup$– BazinCommented Jan 31, 2017 at 8:18
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