# Differentiability and Lipschitz Continuity

I've seen some questions on this site that are similar to the following, but not precisely the same.

Let $f:[a,b]\to\mathbb{R}$ be a differentiable function and assume $f'$ is continuous in $[a,b]$. Prove that $f$ is Lipschitz continuous. What is the best possible Lipschitz constant?

My work so far:

Assume $f$ differentiable on $[a,b]$, $f'$ continuous on $[a,b]$. Since $f$ differentiable on $[a,b]$, then $f$ is continuous on $[a,b]$. Consider any $x_1,x_2\in[a,b]$ with $x_1<x_2$. Since $[x_1,x_2]\subseteq[a,b]$, then $f$ differentiable and continuous on each such $[x_1,x_2]$, and $f'$ continuous on $[x_1,x_2]$. Then, by the Mean Value Theorem, exists $c\in(x_1,x_2)$ such that $f'(c)=(f(x_2)-f(x_1))/(x_2-x_1).$ Since $f'$ continuous on compact set $[a,b]$, by the min-max value theorem, $f'$ achieves its minimum and maximum values at some $x_m,x_M\in[a,b]$. Let $A=max\{|f(x_m)|,|f(x_M)|\}.$ Then$$|f'(c)|=|(f(x_2)-f(x_1))/(x_2-x_1)|\leq A\to|f(x_2)-f(x_1)|\leq A|x_2-x_1|$$

$\forall x_1,x_2\in[a,b]$

Is this correct? I appreciate the help...

• – philmcole Jan 16 '18 at 17:15

More or less... More directly, $|f(x_2)-f(x_1)| = |f'(c)| |x_2-x_1| \leq \max_{c \in [a,b]} |f'(c)| |x_2-x_1|$. Hence $A=\|f'\|_\infty = \max_{c \in [a,b]} |f'(c)|$ is one possible Lipschitz constant.

On the other hand, if $f(x)=mx+q$, then $\|f'\|_\infty=m$, and clearly $m$ is also the best Lipschitz constant.