Vtyjkyuk

Let $f:BbbR^ntoBbbR^n$ be a function of class $C^1$. We suppose that there exists $kin ]0,1[$ such that $$Vert g'(x)Vertleq k,forall;xinBbbR^n$$
where $f(x)=g(x)+x,forall;xinBbbR^n$

Prove that $(i)$ beginalignlangle f'(x)h,hranglegeq (1-k) Vert hVert^2endalign and $(ii)$ beginalignlimlimits_Vert xVertto inftyf(x)=inftyendalign

For the first:

I think we can use directional derivatives
beginalignlangle limlimits_tto 0 fracf(x+th)-f(x)t,hrangle=sum^n_i=1fracpartial f_i(x)partial x_ih^2_iendalign
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!

I do not know, if there is any way to use directional derivatives, but the following should work: We have $g = f - defidmathrmidid$ and hence for each $x in mathbb R^n$ that
$$ g'(x) = f'(x) - id iff f'(x) = g'(x) + id $$
Hence $$ def<#1>left<#1right> < f'(x)h,h> = <g'(x)h,h> + |h|^2 $$
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
beginalign*
<f(x),x> &= <f(0),x> + int_0^1 fracddt <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
endalign*
By Cauchy-Schwarz
beginalign*
|f(x)| &ge frac<f(x),x>\
&ge -|f(0)| + (1-k)|x|
endalign*
and the result follows.