Prove $f^T(x)(\nabla^T f(x)+\nabla f(x))f(x)+x\cdot f(x) = 0$ if and only if $x=0$

Question

Properties

Given a $f:\mathbb R^n\to \mathbb R^n$. The function $f$ has the following properties

• $f(x) = 0 $ if and only if $x=0$.
• $\nabla f(x)$ are negative semidefine for any $x$.
• $f(x)$ is continuously differentiable

Problem

Is the following iif statement hold? $$f^T(x)(\nabla^T f(x)+\nabla f(x))f(x)+x\cdot f(x) = 0 \Leftrightarrow x=0.$$ From right to left, it is straightforward. I am trying to prove the left to right.

Some tries

According to the negative semidefinite of $\nabla f(x)$, we have $\nabla^T f(x)+\nabla f(x)$ are also negative semidefinite, such that the first part $$f^T(x)(\nabla^T f(x)+\nabla f(x))f(x) \leq 0 .$$ Again using $\nabla f(x)$ are negative semidefine, by

lemma: If a mapping $f$ is continuously differentiable then $f$ is monotone if and only if its Jacobian matrix $\nabla f(x)$ is positive semidefinite

, we have the second part $$x\cdot f(x)\leq0.$$

So far, we have $$f^T(x)(\nabla^T f(x)+\nabla f(x))f(x)+x\cdot f(x) \leq 0.$$ This is the best I can do, I don't know where to go. Or maybe the iif statement is not true in the first place.

Answer 1

After an embarrassingly long time, I can now give my own proof of this problem. I will prove the result by contradiction.

Step 1: Let a point $x_3 \neq 0$, which can satisfy the statement $$f^T(x_3)(\nabla^T f(x_3)+\nabla f(x_3))f(x_3)+x_3\cdot f(x_3) = 0 .$$ And also since $x_3\neq0$, $f(x_3) \neq 0$ by property 1.

Step 2: Rewrite the statement as $$x_3\cdot f(x_3) = -f^T(x_3)(\nabla^T f(x_3)+\nabla f(x_3))f(x_3).$$ Since $\nabla^T f(x_3)+\nabla f(x_3)$ is a negative semidefinite matrix, then $-(\nabla^T f(x_3)+\nabla f(x_3))$ is a positive semidefinite matrix, we have $$x_3\cdot f(x_3) = -f^T(x_3)(\nabla^T f(x_3)+\nabla f(x_3))f(x_3)\geq 0$$ $$x_3\cdot f(x_3) \geq 0$$

Step 3: By the assumption $x_3$ and $f(x)$ are both not $0$, then $$x_3\cdot f(x_3) > 0$$

Step 4: By $\nabla f(x_3)$ being negative semidefinite, and using the lemma described above, we should have $$x_3\cdot f(x_3) \leq 0.$$ This is contradicting with the results of Step 3. That is, the point $x_3 \neq 0$ satisfying the statement would not exist.