Vtyjkyuk

Let $f:Osubset BbbR^ntoBbbR^m$ be differentiable and O be convex. Prove that the following are equivalent:

$i$. $f'$ is bounded and $suplimits_cin OVert f'(c) Vert<infty;$

$ii$. $f$ is globally Lipschitz.

My trial:

Proving $iimplies ii$. Let $x,yin O$ such that $[x,y]subset 0.$ Then, by MVT, $exists ;cin [x,y]$ such that beginalignVert f(x)-f(y) Vertleq suplimits_cin [x,y]Vert f'(c) VertVert x-y Vertendalign
beginalignqquadqquadqquadleq suplimits_cin OVert f'(c) VertVert x-y Vertendalign
Since $f'$ is bounded and $suplimits_cin OVert f'(c) Vert<infty;$ then, let $k=suplimits_cin OVert f'(c) Vert.$ So, we have beginalignVert f(x)-f(y) Vertleq kVert x-y Vertendalign This implies that $f$ is globally bounded.

Proving $iiimplies i$. Let $f$ be globally globally Lipschitz, then $exists;k>0$ such that beginalignVert f(x)-f(y) Vertleq kVert x-y Vert,;;forall ;x,yin,Oendalign
beginalignfracVert f(x)-f(y) VertVert x-y Vertleq k,;;forall ;x,yin,Oendalign
Since $f$ is differentiable, we take the directional derivative as $xto y$, we have
beginalignVert f'(y) Vertleq k,;;forall ;yin,Oendalign
Taking $sup,$ we have that
beginalignsuplimits_yin OVert f'(y) Vertleq k.endalign
Please, I'm I correct? If not, alternative proofs will be highly regarded.

(i) There is something fishy here: You cannot say "$exists c$ such that $Phi$ holds" and then take the sup over all $c$ in $Phi$. Furthermore the statement is wrong if $O$ is not convex. As an example let $f$ be the principal value of the polar angle in the plane slit along the negative $x$-axis. Here $f(-1,epsilon)-f(-1,-epsilon)to2pi$ while $epsilonto0+$.

Given $x$, $yin O$ you can argue as follows: If $z:=f(y)-f(x)ne0$ consider the auxiliary real-valued function
$$phi(t)=zcdot fbigl((1-t)x+t ybigr)qquad(0leq tleq 1)$$
$bigl($note that $(1-t)x+tyin O$ when $0leq tleq 1bigr)$. By the MVT there is a $tauin>]0,1[>$ such that
$$zcdotbigl(f(y)-f(x)bigr)=phi(1)-phi(0)= phi'(tau)=zcdotbigl( f'bigl((1-tau)x+tau ybigr).(y-x)bigr) ,$$
and therefore $|z|^2leq |z|sup_cin O|f'(c)|>|y-x|$, or
$$bigr|f(y)-f(x)bigl|=|z|leq k>|y-x| .$$

(ii) It's unclear what the directional derivative is for a vector valued function. We need a clear cut proof that the derivative $f'(p)=:Aincal L(mathbb R^n,mathbb R^m)$ at a given point $pin O$ has norm $leq L$, where $L$ is a global Lipschitz constant for $f$. To this end we go back to the definition of $f'(p)$. Let $u$ be an arbitrary unit vector. Then
$$f(p+t u)- f(p)= A.t u + o(t)qquad(tto0) .$$
This implies
$$A.u=f(p+ tu)-f(p)over t +o(1)qquad(tto0) ,$$
hence
$$|A.u|leq+o(1)leq L+o(1)qquad(tto0) .tag1$$
As the LHS of $(1)$ does not depend on $t$ it follows that $|A.u|leq L$, and since $uin S^n-1$ was arbitrary we may conclude that $|A|leq L$.