Stein 3.32

$\textbf{Exercise 32}$ : Let $f: \mathbb{R} \longrightarrow \mathbb{R}$. Prove that $f$ is Lipschitz continuous i.e. $\exists M>0$ s.t. $$\frac{|f(x)-f(z)|}{|x-z|} \leq M \quad \forall x,z\in{\mathbb{R}} \quad (\star)$$ iff $f$ is absolutely continuous $\textit{and}$ $|f'(x)| \leq M$ for a.e. $x$.

$\textit{Proof}$
"$(\Longrightarrow)$" Assume that $f$ has the above Lipschitz condition.
Remark: we should first establish that $f$ is absolutely continuous and then used the fact that an absolutely continuous function has a first derivative almost everywhere. This first derivative is then bounded by the Lipschitz constant $M$.
Recall that $f$ is $\textit{absolutely continuous}$ if $\forall \epsilon>0$, $\exists \delta(\epsilon)>0$ s.t. for any collection of disjoint intervals $\{(a_i,b_i)\}_{i=1}^n$ defined by a partition that $\sum_{i=1}^n(b_i-a_i) \leq \delta \Longrightarrow \sum_{i=1}^n|f(b_i)-f(a_i)| \leq \epsilon$. With the Lipschitz condition $(\star)$ let $\delta=\frac{\epsilon}{M}$. Then, $|f(b_i)-f(a_i)|\leq M |b_i-a_i|=M \frac{\epsilon}{M}=\epsilon$. Let $|b_i-a_i| \leq \frac{\delta}{n}$. Then, $|f(b_i)-f(a_i)|\leq \frac{\epsilon}{n}$ and so $$\sum_{i=1}^n(b_i-a_i) \leq n \cdot \frac{\delta}{n}=\delta \Longrightarrow \sum_{i=1}^n|f(b_i)-f(a_i)| \leq n \cdot \frac{\epsilon}{n} =\epsilon$$ So, $f$ is absolutely continuous.
Now, $f$ has a bounded first derivative, which is defined a.e. This can be seen as follows: $\forall x,h \in{\mathbb{R}}$, $$\left|\dfrac{f(x+h)-f(x)}{(x+h)-x}\right| \leq M$$ This statement is true for all $x$ and $h$, so it is therefore true in the limit $$\lim_{h \rightarrow 0}\left|\dfrac{f(x+h)-f(x)}{h}\right| =|f'(x)| \leq M$$ "$(\Longleftarrow)$" Assume that $|f'(x)| \leq M$ and $f$ is absolutely continuous. By Theorem 3.11, an absolutely continuous function $f$ has the properties that $f'$ exists a.e., $f'\in{L^1}$ and $f$ can be represented with the integral of its derivative: $$f(b)-f(a)=\int_a^bf'(t)dt \quad \forall a,b \in{\mathbb{R}}$$ Using the triangle inequality, the assumption $|f'(t)| \leq M$, and the linearity of the Lebesgue integral, $$|f(b)-f(a)| = \left|\int_a^bf'(t)dt\right| \leq \int_a^b|f'(t)|dt \leq \int_a^b M dt = M\int_a^b dt=M|b-a|$$ $\therefore$ $f$ is Lipschitz continuous. $\blacksquare$