If $f$ is differentiable in $x_0$ and $f$ reaches a local maximum in $x_0$ then $\nabla f(x_0)=0$

Question

If $f$ is differentiable in $\vec{x_0}$ and $f$ reaches a local maximum in $x_0$ then $\nabla f(\vec{x_0})=\vec{0}$

Proof:

Let $g:\mathbb{R}\rightarrow \mathbb{R}^n$ defined by $g(t)=\vec{x_o}+t\vec{v}$ and let $B=(f\circ g)$.

As $f$ reaches a local maximun in $\vec{x_0}$ then $B$ reaches the maximum in $t=0$.

This implies:

$$B(\vec{0})=(f\circ g)(\vec{0})=f(g(\vec{0}))\implies \vec{0}=B'(\vec{0})=f'(g(\vec{0}))g'(\vec{0})=D_\vec{v}f(g(\vec{0}))=D_\vec{v}f(\vec{x_0})= \langle\nabla f(\vec{x_0}),\vec{v}\rangle$$

I don't understand this line:

$$f'(g(\vec{0}))g'(\vec{0})=D_\vec{v}f(g(\vec{0}))=D_\vec{v}f(\vec{x_0})= \langle\nabla f(\vec{x_0}),\vec{v}\rangle$$

As $v$ is arbitrary, then $\nabla f(\vec{x_0})=\vec{0}$

Can someone explaint that step? Thanks!

Answer 1

This is because, given a scalar product $b$, $b(u,v)=0$ for every $v$ implies that $b(u,u)=0$ implies $u=0$, (positive definite).

Answer 2

Going back to the line $f'(g(0))\cdot g'(0)$. The derivative of $f$ is $\nabla f$, and it is evaluated at $g(0)=x_0$, and $g'(0)= v$. So, $f'(g(0))\cdot g'(0)=\nabla f(x_0)\cdot v=<\nabla f(x_0),v>$.

Now, the other expressions are different ways to rewrite the expression $f'(g(0))\cdot g'(0)$. The meaning $D_v(x_0)$, is the linear function that approximates $f$ at $x_0$, evaluated at $v$. The linear function that approximates $f$ is $\nabla f$, at $x_0$, that function is equal to $\nabla f(x_0)$. This function induces a linear transformation $D(x_0):\mathbb{R}^n\rightarrow \mathbb{R}^n$, given by $D(x_0)(v)= \nabla f(x_0) \cdot v$. A different notation for the previous expression is $D_v(x_0)= \nabla f(x_0) \cdot v$

If $<\nabla f(x_0),v>=0$ for all $v$, then $\nabla f(x_0)=0$.

Indeed, since it holds for all $v$, it holds for $v=\nabla f(x_0)$. In other words, $||\nabla f(x_0)||^2=<\nabla f(x_0),\nabla f(x_0)>=0$. And since $||w||=0\iff w=0$, we conclude.