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Let $f:\Bbb{R}^n\to\Bbb{R}^n$ be a function of class $C^{1}$. We suppose that there exists $k\in ]0,1[$ such that $$\Vert g'(x)\Vert\leq k,\forall\;x\in\Bbb{R}^n$$ where $f(x)=g(x)+x,\forall\;x\in\Bbb{R}^n$

Prove that $(i)$ \begin{align}\langle f'(x)h,h\rangle\geq (1-k)\ \Vert h\Vert^2\end{align} and $(ii)$ \begin{align}\lim\limits_{\Vert x\Vert\to \infty}f(x)=\infty\end{align}

For the first:

I think we can use directional derivatives \begin{align}\langle \lim\limits_{t\to 0} \frac{f(x+th)-f(x)}t,h\rangle=\sum^{n}_{i=1}\frac{\partial f_{i}(x)}{\partial x_i}h^2_i\end{align} To me, directional derivatives may not be the only way to show this but I don't even seem to be finding my way through. Please, could someone help? Thanks for your time and efforts!

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I do not know, if there is any way to use directional derivatives, but the following should work: We have $g = f - \def\id{\mathrm{id}}\id$ and hence for each $x \in \mathbb R^n$ that $$ g'(x) = f'(x) - \id \iff f'(x) = g'(x) + \id $$ Hence $$ \def\<#1>{\left<#1\right>} \< f'(x)h,h> = \<g'(x)h,h> + \|h\|^2 $$ By Cauchy-Schwarz $$ \<g'(x)h,h> \ge -\|g'(x)h\| \|h\| \ge -k\|h\|^2 $$ and hence $$ \<f'(x)h,h> = \<g'(x)h,h> +\|h\|^2 \ge (1-k)\|h\|^2 $$ For (ii), use (i). Note that for $x \in \mathbb R^n$ we have \begin{align*} \<f(x),x> &= \<f(0),x> + \int_0^1 \frac{d}{dt} \<f(tx),x>\, dt\\ &= \<f(0),x> + \int_0^1 \<f'(tx)x,x> \, dt\\ &\ge \<f(0),x> + \int_0^1 (1-k)\|x\|^2\, dt\\ &\ge -\|f(0)\|\|x\| + (1-k)\|x\|^2 \end{align*} By Cauchy-Schwarz \begin{align*} \|f(x)\| &\ge \frac{\<f(x),x>}{\|x\|}\\ &\ge -\|f(0)\| + (1-k)\|x\| \end{align*} and the result follows.

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  • $\begingroup$ Thanks a lot! I'll go through the proof, ask questions before thicking! $\endgroup$ Aug 23, 2018 at 6:02
  • $\begingroup$ Please, can you explain how you got \begin{align*} \<f(x),x> &= \<f(0),x> + \int_0^1 \frac{d}{dt} \<f(tx),x>\, dt\\ \end{align*} and how did you arrive at this point? \begin{align*} &= \<f(0),x> + \int_0^1 \<f'(tx)x,x> \, dt \end{align*} $\endgroup$ Aug 23, 2018 at 6:51
  • $\begingroup$ The first is just the fundamental theorem of calculus. Consider the function $h(t) = \left<f(tx), x\right>$, then we have $$ h(1) - h(0) = \int_0^1 h'(t) \, dt $$ which is just the first statement. The second statement is derived from the chain rule, note that $$ h'(t) = \left< \frac{d}{dt} f(tx), x\right> = \left< f'(tx) x, x\right> $$ by the chain rule, as $f'(tx)$ is the outer and $x$ the inner derivative (note that $\frac{d}{dt}(tx) = x$). $\endgroup$
    – martini
    Aug 23, 2018 at 10:05
  • $\begingroup$ Thanks for the explanation! Everything is clear! $\endgroup$ Aug 23, 2018 at 11:13

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