Relation to ∇f(x)⋅x and laplacian

Question

Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ Proof that:

If $f\in C^2(\mathbb{R}^n)$ and $\forall x \in \mathbb{R}^n$,

$$(\nabla f(x),x)\leq 0 $$

where (,) is the scalar product

$\Rightarrow \Delta f(0)\leq 0$ (where $\Delta$ is the Laplacian).

I've been thinking about the definition of $ \Delta f = \nabla \cdot \nabla f $. But the $(\nabla f(x),x)$ is different from $\nabla f$ , so if I try to $\nabla \cdot (\nabla f(x),x)$ I have different results... I think I miss some properties of Laplacian or of $\nabla f$.

Answer 1

Herein I use $v \cdot w$ for $(v, w)$ with $v, w \in \Bbb R^n$.

If

$\nabla^2 f(0) > 0, \tag 1$

then since $f(x) \in C^2(\Bbb R^n)$, there is a closed ball $B(\epsilon, 0) = \{z \in \Bbb R^n \mid \Vert z \Vert \le \epsilon \}$ of radius $\epsilon > 0$, centered at $0$, such that for $y \in B(\epsilon, 0)$,

$\nabla^2f(y) > 0; \tag 2$

we donote the spherical surface, or boundary, of this ball by $S(\epsilon, 0)$. Then by the divergence theorem,

$\displaystyle \int_{S(\epsilon, 0)} \nabla f \cdot \vec n \; dS = \int_{B(\epsilon,0)}\nabla^2 f(y) \; dV > 0, \tag 2$

by (1), where $\vec n$ is the outward-pointing unit normal vector field on $S(\epsilon, 0)$. (2) clearly forces

$\nabla f(y) \cdot \vec n(y) > 0 \tag 3$

somewhere on $S(\epsilon, 0)$. But then

$\nabla f(y) \cdot y = \nabla f(y) \cdot \Vert y \Vert \vec n(y) = \Vert y \Vert \nabla(y) \cdot \vec n(y) > 0, \tag 4$

contradicting the hypothesis

$\nabla f(x) \cdot x \le 0 \tag 5$

we have placed upon $f(x)$. Thus (1) cannot hold; we have instead

$\nabla^2 f(0) \le 0. \tag 6$.

Answer 2

By showing that $0$ ∈ $\mathbb{R}^n$ is a global maximum for f. Sia x ∈ $\mathbb{R}^n$. By the mean value Theorem, $$f(x) − f(0) = (∇f(tx), x)$$ $t\in (0,1)$. Hence $$ f(x) − f(0) = \frac{1}{t}(∇f(tx), tx) \leq 0 $$ Then $$(∇f(tx), tx) ≤0$$ so $0 ∈ \mathbb{R}^n$ it is a global maximum. So, as a global maximum for f, all the eigenvalues of Hessian matrix in $0$ are necessarially not positive so $λi ≤ 0 i = 1, 2 . . . n.$ $$\sum f_{xi,xi}(0) = ∆f(0) ≤ 0,$$