Let $f:O\subset \Bbb{R}^n\to\Bbb{R}^m$ be differentiable and O be convex. Prove that the following are equivalent:

$i$. $f'$ is bounded and $\sup\limits_{c\in O}\Vert f'(c) \Vert<\infty;$

$ii$. $f$ is globally Lipschitz.

My trial:

Proving $i\implies ii$. Let $x,y\in O$ such that $[x,y]\subset 0.$ Then, by MVT, $\exists \;c\in [x,y]$ such that \begin{align}\Vert f(x)-f(y) \Vert\leq \sup\limits_{c\in [x,y]}\Vert f'(c) \Vert\Vert x-y \Vert\end{align} \begin{align}\qquad\qquad\qquad\leq \sup\limits_{c\in O}\Vert f'(c) \Vert\Vert x-y \Vert\end{align} Since $f'$ is bounded and $\sup\limits_{c\in O}\Vert f'(c) \Vert<\infty;$ then, let $k=\sup\limits_{c\in O}\Vert f'(c) \Vert.$ So, we have \begin{align}\Vert f(x)-f(y) \Vert\leq k\Vert x-y \Vert\end{align} This implies that $f$ is globally bounded.

Proving $ii\implies i$. Let $f$ be globally globally Lipschitz, then $\exists\;k>0$ such that \begin{align}\Vert f(x)-f(y) \Vert\leq k\Vert x-y \Vert,\;\;\forall \;x,y\in\,O\end{align} \begin{align}\frac{\Vert f(x)-f(y) \Vert}{\Vert x-y \Vert}\leq k,\;\;\forall \;x,y\in\,O\end{align} Since $f$ is differentiable, we take the directional derivative as $x\to y$, we have \begin{align}\Vert f'(y) \Vert\leq k,\;\;\forall \;y\in\,O\end{align} Taking $\sup,$ we have that \begin{align}\sup\limits_{y\in O}\Vert f'(y) \Vert\leq k.\end{align} 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 $f(-1,\epsilon)-f(-1,-\epsilon)\to2\pi$ while $\epsilon\to0+$.

Given $x$, $y\in O$ you can argue as follows: If $z:=f(y)-f(x)\ne0$ consider the auxiliary real-valued function $$\phi(t)=z\cdot f\bigl((1-t)x+t y\bigr)\qquad(0\leq t\leq 1)$$ $\bigl($note that $(1-t)x+ty\in O$ when $0\leq t\leq 1\bigr)$. By the MVT there is a $\tau\in\>]0,1[\>$ such that $$z\cdot\bigl(f(y)-f(x)\bigr)=\phi(1)-\phi(0)= \phi'(\tau)=z\cdot\bigl( f'\bigl((1-\tau)x+\tau y\bigr).(y-x)\bigr)\ ,$$ and therefore $|z|^2\leq |z|\sup_{c\in O}\|f'(c)\|\>|y-x|$, 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 $f'(p)=:A\in{\cal L}({\mathbb R}^n,{\mathbb R}^m)$ at a given point $p\in O$ 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 $$f(p+t u)- f(p)= A.t u + o(t)\qquad(t\to0)\ .$$ This implies $$A.u={f(p+ tu)-f(p)\over t }+o(1)\qquad(t\to0)\ ,$$ hence $$|A.u|\leq{|f(p+ tu)-f(p)|\over |tu| }+o(1)\leq L+o(1)\qquad(t\to0)\ .\tag{1}$$ As the LHS of $(1)$ does not depend on $t$ it follows that $|A.u|\leq L$, and since $u\in S^{n-1}$ was arbitrary we may conclude that $\|A\|\leq L$.

  • $\begingroup$ But $\phi$ has to be $C^1$ for that to hold! $\endgroup$ Aug 22 '18 at 9:01
  • $\begingroup$ O.k. See my edit. (I still have to see a naturally occurring multivariate function that is differentiable, but not continuously differentiable$\ldots$) $\endgroup$ Aug 22 '18 at 10:13
  • $\begingroup$ Thanks for your feedback Sir! $\endgroup$ Aug 22 '18 at 13:37
  • $\begingroup$ Thanks for all but I think the first approach is right if I include the theorem I used. Mean Value Theorem for higher dimensions. $\endgroup$ Aug 22 '18 at 14:33

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