Is $F(x)=\int_{(-\infty,x)}f\,d\lambda$ Lipschitz continuous

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ Lebesgue integrable. Is
$$F(x)=\int_{(-\infty,x)}f\,d\lambda$$
Lispchitz continuous? where $\lambda$ is the Lebesgue measure.

I think $F$ is not necessarily Lipschitz continuous. I tried with the functions $\chi_{[1,\infty)}x^{-2/3}$, $\chi_{(0,\infty)}x^{1/2}$ where $\chi_A$ is the characteristic function of $A$, and other combinations but non of them gives me a counterexample. Any suggestion?

Answer 1

Suppose $f\in L_1(\lambda)$ and set $F(x)=\int_{(-\infty,x]}f(t)\,dt$. If $F$ is Lipschitz over $\mathbb{R}$, then for some constant $L$ $$\frac{|F(y)-F(x)|}{|y-x|}\leq L$$ for all $x\neq y$. As $F$ is differentiable at every Lebesgue point of $f$, and $F'=f$ at such points, then
$|F'(x)|=|f(x)|\leq L$ almost surely; hence $\|f\|_\infty<\infty$

It follows that any $f\in L_1(\lambda)$ which is not in $L_\infty(\lambda)$, $F$ will not be Lipschitz over the whole real line. Here are two examples:

1. $f(x)=\frac{1}{\sqrt{|x|}}\mathbb{1}_{(0,1]}(|x|)$ is in $L_1$
2. $g(x) = \sum^\infty_{n=1}n\mathbb{1}_{(n-2^{-n},n+2^{-n}]}(x)$