Given $D\subset \mathbb{R}^n$. $f:D\longrightarrow\mathbb{R}^n$ is said to be a monotone operator if $$\left\langle f(x)-f(y),x-y\right\rangle\ge 0,\text{ for all }x,y\in D.$$ I wonder if this hypothesis is true about monotone operators.
Given $D\subset \mathbb{R}^n$ and a differentiable operator $f:D\longrightarrow\mathbb{R}^n$. Then, two following statements are equivalent
(i) $f$ is a monotone operator.
(ii) $\left\langle u,f'(x)(u)\right\rangle\ge 0$ for all $x\in D$ and $u\in\mathbb{R}^n$.
Can you show me the proof (or brief proof) if this is true or show me the counterexample if this is false?
Thank you!
I will assume that $D$ is convex. You can use the mean value theorem, applied to the auxiliary function $$ h(t)=f(y+t(x-y)). $$ In terms of $h$ condition (i) reads $\langle h(1)-h(0),x-y\rangle\ge 0$ If $f$ is differentiable, so is $h$ and, by the mean value theorem $$ h(1)-h(0)=h'(\xi)=f'(z)(x-y) $$ where $0<\xi<1$ and, equivalently, $z$ is in the segment joining $x$ and $y$. Then, your inequality certainly holds if $$ \langle f'(z)(x-y),x-y\rangle\ge 0 $$ for all $x,y\in D$ and all $z\in D$, which is implied by (ii). For the necessity, you can argue by contradiction, showing that if condition $(ii)$ fails, you can choose a point $y$ close to $x$ in the direction of $u$ such that inequality $(i)$ is violated (just by linear approximation).