Strongly Monotone and differentiable at $x_0$ implies $Df(x_0)$ is Strongly Monotone

Question

A function $f:\mathbb{R^{n}}\rightarrow\mathbb{R^{n}}$ is strongly monotone if there exists a constant $a>0$ such that : $\langle f(x)-f(y), x-y \rangle \geq a|x-y|^2$ for every $x,y$

I need to prove that if f is strongly monotone and differentiable at $x_0$ then $Df(x_0)$ is strongly monotone with same constant $a$.

What I did:

Write $E=Df(x_0)$, E is strongly monotone would mean that $\langle E(x-y), x-y \rangle \geq a|x-y|^2$, but $\langle E(x-y), x-y \rangle=(x-y)^TE^T(x-y)$

I also know that f is differentiable at $x_0$, then $\frac{f(x_0)-f(y)-E(x_0-y)}{|x_0-y|}\rightarrow0$ as $y \rightarrow x_0$.

But I couldn't connect both concepts.

Any suggestions ?

Answer 1

It's sufficient to prove that $\langle Eh,h\rangle \ge a |h|^2$ for all $h$.

Let $\epsilon > 0$ and choose $\delta > 0$ such that $|f(x_0 + h)-f(x_0) - Eh| < \epsilon |h|$ for $|h| < \delta$. We have:

$$\langle Eh,h \rangle = \langle Eh + f(x_0) - f(x_0+h) + f(x_0+h) - f(x_0), h\rangle \\ \ge \langle A(h), h\rangle + a|h|^2$$

Where $A(h) = Eh + f(x_0) - f(x_0+h)$. Recall that $-|h| |A(h)| \le \langle A(h), h\rangle$ (Cauchy-Schwarz). This gives:

$$\langle Eh, h\rangle \ge (a|h| - |A(h)|)|h| > (a|h|-\epsilon |h|)|h| = (a-\epsilon)|h|^2$$

Now let $z \neq 0$ and choose $h = \frac{\delta}{2|z|}z$. Then $|h| < \delta$. Substituting this $h$ in:

$$\langle Eh,h\rangle > (a-\epsilon)|h|^2$$

we get by linearity of $E$ that:

$$\langle Ez,z\rangle > (a-\epsilon)|z|^2$$

We deduce that for all $z$, $\langle Ez,z\rangle \ge (a-\epsilon)|z|^2$.

taking $\epsilon \to 0$, we get the desired inequality.