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Let $U\subset \mathbb R^N$ be open and convex, and the function $f: U\to \mathbb R$ is differentiable in $U$. I've got to show that:

$f$ is Lipschitz continuous iff $\exists\: M>0$ such that $\|\nabla f(x)\|\leq M$ for all $x\in U$.

I understand that a function is Lipschitz continuous on a subset $E$ of $\mathbb R^N$ if for all $x, y\in E$,

$$|f(x)-f(y)|\le L|x-y|$$

for some $L>0$. But how do I work with this definition to get to my result? I sense that I would have to use the fact the norm is a convex function and show that

$$\|\nabla f(x)\| < \frac{|f(x)-f(y)|}{|x-y|}$$

but I'm not sure if this is the right approach.

creative
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user209396
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    For one direction use the mean value theorem. For the other prove that all directional derivatives of $f$ are bounded. – Jose27 Feb 09 '15 at 08:13

1 Answers1

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Assume $|\nabla f(x)|\leq M$ for all $x\in U$, and consider two points $p$, $q\in U$. The auxiliary function $$\phi(t):=f\bigl((1-t)p+tq\bigr)\qquad (0\leq t\leq1)$$ is differentiable with $\phi'(t)=\nabla f\bigl((1-t)p+tq\bigr)\cdot(q-p)$. The MVT therefore gives $$f(q)-f(p)=\phi(1)-\phi(0)=\phi'(\tau)=\nabla f(\xi)\cdot(q-p)\ ,$$ where $\xi:=(1-\tau)p+\tau q$, and this implies $|f(q)-f(p)|\leq M\>|q-p|$.

Conversely: Assume that $f:\>U\to{\Bbb R}$ is differentiable and $M$-Lipschitz on $U$. Consider a point $p\in U$ and let $\nabla f(p)=:a\ne0$. Then $$|a|^2=\nabla f(p)\cdot a=df(p).a=\lim_{t\to0+}{f(p+t a)-f(p)\over t}\ .$$ Since $$|f(p+ ta)-f(p)|\leq M | t a|=M \>t\> |a|\qquad(t>0)$$ we obtain $|a|^2\leq M|a|$, or $|a|\leq M$.

Christian Blatter
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