# Thread: [SOLVED] Prove differentiability from Lipschitz condition

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

A function $\displaystyle fa,b)\to R$ satisfies a Lipschitz condition at $\displaystyle x\in(a,b)$ iff there is $\displaystyle M>0$ and $\displaystyle \epsilon>0$ such that $\displaystyle |x-y|<\epsilon$ and $\displaystyle y\in(a,b)$ imply that $\displaystyle |f(x)-f(y)|\leq M|x-y|$. If $\displaystyle f$ is differentiable at $\displaystyle x$, prove that $\displaystyle f$ satisfies a Lipschitz condition at $\displaystyle x$.
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 $\displaystyle \varepsilon-\delta$ definition of the differentiability at $\displaystyle x$ for a fixed $\displaystyle \delta$, for instance $\displaystyle \delta=1$, and to use a triangle inequality: there is $\displaystyle \ell(=f'(x))$ and $\displaystyle \varepsilon>0$ such that for y such that $\displaystyle |x-y|<\varepsilon$, $\displaystyle |\frac{f(y)-f(x)}{y-x}-\ell|\leq 1$, hence $\displaystyle |\frac{f(y)-f(x)}{y-x}|\leq \ell + 1$.

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