Essentially Bounded Functions

Question

Show that if $f$ is Lipschitz then $f^{\prime}$ is essentially bounded.

Help me in this exercise.

Answer 1

This is mostly just using definitions. To show that $f’$ is essentially bounded, you must show that there exists a constant $C \in \mathbb{R}_{>0}$ so that $\{x:|\,f'(x)\,| > C\}$ has measure $0$. Since $f$ is Lipschitz, there exists $C \in \mathbb{R}_{>0}$ so that $$|f(x)-f(y)| \leq C\,|x-y|$$ for all $x$ and $y$. The same $C$ satisfies the definition of essentially bounded: at any $x$ where $f$ is differentiable, we have
$$|\,f’(x)\,| = \left| \,\lim_{y \to x} \frac{f(y)-f(x)}{y-x} \, \right|.$$ Since absolute value is a continuous function, the absolute value of a limit is the limit of the absolute value. Hence, this becomes $$|\,f’(x)\,| = \lim_{y \to x}\left|\,\frac{f(y)-f(x)}{y-x} \,\right| \leq \lim_{y \to x} \frac{C \, |y-x|}{|y-x|} = C.$$