Suppose $f: \mathbb{R}^n \to \mathbb{R}$, show that $Df(c)=\vec0$ if $f$ has a local minimum and is differentiable at $c$. Suppose $f: \mathbb{R}^n \to \mathbb{R}$, show that $Df(c)=\vec0$ if $f$ has a local minimum and is differentiable at $c$.
My attempt: The problem is straightforward when $n=1$; but I struggle when $n\geq 2$. Clearly, I want to show that $$\lim_{h\to0}\frac{\lVert f(c+h)-f(c)\rVert}{\lVert h\rVert}=0$$, because then if we let $Df(c)(h)=\vec0$, it follows that $$\lim_{h\to0}\frac{\lVert f(c+h)-f(c)-Df(c)(h)\rVert}{\lVert h\rVert}=0$$. Hence by the uniqueness of $Df(c)$, we have $Df(c)(h)=\vec0$. But how exactly do I utilize $\exists\varepsilon>0$ such that $f(c)\leq f(c+h) \; \forall \lVert h \rVert<\varepsilon$? Any "hint" would be greatly appreciated.
 A: Since $\displaystyle D_{j}f(\mathbf{c})=\lim_{t\searrow 0}\frac{f({\bf c}+t{\bf e}_{j})-f({\bf c})}{t}$ and $f({\bf c})\leqslant f({\bf c}+t{\bf e}_{j})$ when $|t|<\varepsilon$ for some sufficiently small $\varepsilon>0$ by definition of local minimum. Hence $\displaystyle \frac{f({\bf c}+t{\bf e}_{j})-f({\bf c})}{t}\geqslant 0$ for some $t\in \left]0,\varepsilon\right[$ and $\displaystyle \frac{f({\bf c}+t{\bf e}_{j})-f({\bf c})}{t}\leqslant 0$ for some $t\in \left]-\varepsilon ,0 \right[$ and then $D_{j}f({\bf c})\geqslant 0$ and $D_{j}f({\bf c})\leqslant 0$, then $D_{j}f({\bf c})=0$ for $1\leqslant j\leqslant n$.  Therefore $f'({\bf c})=0$ due to $(*)$.
$(*):$ If $f$ maps an open set $U\subseteq \mathbb{R}^{n}$ into $\mathbb{R}^{m}$ and $f$ is a differentiable function at a point ${\bf c}\in U$. Then $\displaystyle f'({\bf c}){\bf e}_{j}=\sum_{1\leqslant i\leqslant m}(D_{j}f_{i})({\bf c}){\bf u}_{i}$ with $1\leqslant j\leqslant n$ and $\{{\bf e}_{1},{\bf e}_{2},\ldots,{\bf e}_{n}\}$ is a basis for $\mathbb{R}^{n}$ and $\{{\bf u}_{1},{\bf u}_{2},\ldots,{\bf u}_{n}\}$ is a basis for $\mathbb{R}^{m}$.
Of course, a similar argument allows to change the hypothesis of local minimum to local maximum and the proof works too.
