Show that if $f$ is differentiable at $x$, then $f$ is Lipschitz of order $1$ at $x$

Question

Show that if $f$ is differentiable at $x$, then $f$ is Lipschitz of order $1$ at $x$. Is the converse true?

If $f$ is differentiable at $x$, the derivative must be defined there. So we must have $\displaystyle \lim_{y \to x} \dfrac{f(y)-f(x)}{y-x} = c$. Then we must show that this implies there is a constant $C > 0$ for which $|f(x) - f(y)| \leq C|x-y|$. We can't just multiply both sides by $y-x$ and get something useful and we don't know if $c$ is positive or not, so I don't know where to proceed.

Answer 1

It should be $f$ is at least of Lipschitz order $1$. If $c=0$, then the Lipschitz order can be larger than $1$.

Edit: Suppose $f\in C^{2}$ and $f'(0)=0$. By Taylor series with Lagrange remainder $$ f(x)=f(0)+f'(0)x+\frac{f''(\xi)x^2}{2} $$ where $|\xi|<|x|$. Thus $$ |f(x)-f(0)|=C|x|^2 $$ where $C=|\frac{f''(\xi)}{2}|$. So $f$ is of Lipschitz order $2$.

The converse is not true. An example is $$ f(x)=\begin{cases}x\sin {\dfrac1{x}}, \quad x\ne 0 \\0,\quad\quad\: \quad x=0 \end{cases} $$ Clearly $$ |f(x)-f(0)|\leqslant |x-0|=|x| $$ But $f$ is not differentiable at $0$.

Answer 2

By the existence of the limit defining the derivative, we know there exists $\;\delta>0\;$ such that for $\;|x-y|<\delta\;$ we have

$$\left|\frac{f(y)-f(x)}{y-x}-c\right|<1\implies |f(y)-f(x)-c(y-x)|<|x-y|$$

But $\;|f(y)-f(x)-c(y-x)|\ge|f(y)-f(x)|-|c(y-x)|\;$

So that we get that

$$|f(y)-f(x)|\le|c(y-x)|+|y-x|=|c+1||y-x|$$

Answer 3

If $f'(x) = c,$ then $ \lim_{y\to x}|(f(y)-f(x))/(y-x)| \to |c|.$ This implies $|(f(y)-f(x))/(y-x)|<(1+|c|)$ for $y$ close to $x,$ and that implies the local Lipschitz property.

For the converse consider $f(y) = |y|, x = 0.$