Proof that the derivative vanishes at a local extremum

Proof

We restrict $f$ to a line through $a$ and use the one-variable theorem.

1. Let $U$, $f$ and $a$ be as in the statement. Assume first that $f$ has a local maximum at $a$.
2. There is $\varepsilon > 0$ with $f(x) \le f(a)$ for all $x \in U$ with $\norm{x - a} < \varepsilon$. Since $U$ is open, we may shrink $\varepsilon$ so that all $x$ with $\norm{x - a} < \varepsilon$ lie in $U$.
3. Let $i$ be an index for which $\partial f / \partial x_i (a)$ exists. Let $e_i$ be the $i$-th standard basis vector, and define $\varphi(t) = f(a + t e_i)$ on the open interval $I = (-\varepsilon, \varepsilon)$.
4. For $t \in I$ we have $\norm{(a + t e_i) - a} = \abs{t} < \varepsilon$, so $\varphi(t) \le f(a) = \varphi(0)$. Thus $\varphi$ has a local maximum at $0$.
5. By the definition of the partial derivative, $\varphi$ is differentiable at $0$ with $\varphi'(0) = \partial f / \partial x_i (a)$.
6. By Fermat's theorem on stationary points, $\varphi'(0) = 0$. So $\partial f / \partial x_i (a) = 0$.
7. If $f$ has a local minimum at $a$, then $-f$ has a local maximum at $a$, and $\partial (-f)/\partial x_i (a) = -\partial f / \partial x_i (a)$. So step 6 applied to $-f$ gives again $\partial f / \partial x_i (a) = 0$.
8. Now let $f$ be differentiable at $a$. Then every partial derivative exists at $a$, and $Df(a) e_i = \partial f / \partial x_i (a)$, since $Df(a)$ has the Jacobian matrix of partial derivatives.
9. By steps 6 and 7, all these partial derivatives are $0$. So the gradient $\nabla f(a)$, the vector of partial derivatives, is $0$.
10. The linear map $Df(a)$ vanishes on each basis vector $e_i$, so it is the zero linear map.