# Thread: [SOLVED] Prove differentiability from Lipschitz condition

1. ## [SOLVED] Prove differentiability from Lipschitz condition

A function $fa,b)\to R" alt="fa,b)\to R" /> satisfies a Lipschitz condition at $x\in(a,b)$ iff there is $M>0$ and $\epsilon>0$ such that $|x-y|<\epsilon$ and $y\in(a,b)$ imply that $|f(x)-f(y)|\leq M|x-y|$. If $f$ is differentiable at $x$, prove that $f$ satisfies a Lipschitz condition at $x$.
The similarities between this condition and the definition of the derivative are obvious, but I can't seem to find an approach whereby I can actually prove a derivative exists in this case.

Any ideas?

Thanks!

2. Originally Posted by hatsoff
The similarities between this condition and the definition of the derivative are obvious, but I can't seem to find an approach whereby I can actually prove a derivative exists in this case.

Any ideas?

Thanks!
The question asks to show that "differentiable" implies "Lipschitz", not the converse (which is false) (however, a Lipschitz function is differentiable almost-everywhere, but this is a hard theorem).

To solve the question, you just have to write the $\varepsilon-\delta$ definition of the differentiability at $x$ for a fixed $\delta$, for instance $\delta=1$, and to use a triangle inequality: there is $\ell(=f'(x))$ and $\varepsilon>0$ such that for y such that $|x-y|<\varepsilon$, $|\frac{f(y)-f(x)}{y-x}-\ell|\leq 1$, hence $|\frac{f(y)-f(x)}{y-x}|\leq \ell + 1$.

3. Originally Posted by Laurent
The question asks to show that "differentiable" implies "Lipschitz", not the converse (which is false) (however, a Lipschitz function is differentiable almost-everywhere, but this is a hard theorem).

To solve the question, you just have to write the $\varepsilon-\delta$ definition of the differentiability at $x$ for a fixed $\delta$, for instance $\delta=1$, and to use a triangle inequality: there is $\ell(=f'(x))$ and $\varepsilon>0$ such that for y such that $|x-y|<\varepsilon$, $|\frac{f(y)-f(x)}{y-x}-\ell|\leq 1$, hence $|\frac{f(y)-f(x)}{y-x}|\leq \ell + 1$.
Oh, oops. Here I am worrying about my math abilities, and the problem is in my comprehension of English. No wonder I couldn't prove it!

Thanks!