# Is the integration of Lipschitz function still Lipschitz?

Let $f:(0,1)\to\mathbb R$

1) Is the integration of a Lipschitz function still Lipschitz?

2) Is the integration of a bounded function still Lipschitz?

3) (Claim) If $f$ is Lipschitz, then the intersection between $f$ and x-axis is finite, almost surely. (In other word, if $f$ intersect with x-axis at countable points, then the number of those points is finite)

For Q1, I have no idea.

For Q2, a counterexample is the integration of $f(x)=sin(1/x)$?

For Q3, I think it is true. The weak derivative of L-Lipschitz $f$ is bounded. Loosely speaking, I could denote the set of the y-coordinates of all $f$'s local minima above x-axis as $A$, and the y-coordinates of all local maxima below x-axis as $B$. All elements of $A$ and $B$ are real numbers. Find the min among $A$; denote it as $a$. Find the max among $B$, denote it as $b$.

For example, if $(x_1,y_1),...,(x_m,y_m)$ are the local min and all $y>0$, then we denote $A=\{y_1,...,y_m\}$ and $a=y_\text{min}$.

We have: $$L/(a-b)\geq \frac{|A|+|B|}2$$

QED?

(Not homework)

• Generally, integration smooths things out. That is, if you integrate a function, the result is simpler. For example, the integral of $\sin(1/x)$ must certainly be Lipschitz since it has a bounded derivative - namely, $\sin(1/x)$. Oct 12, 2017 at 23:01
• For (1), write $F(x) = \int_{1/2}^xf(t)\,dt$ and use that $f$ is bounded. Oct 12, 2017 at 23:01
• (2) is true. (1) is true because a Lispchitz function extends to a continuous function on the whole $[0,1]$ and, thus, it is bounded. (3) is false.
– user228113
Oct 12, 2017 at 23:06

Let $f$ be bounded, i.e., $|f(x)|\le C$ for all $x\in (0,1)$. Then $$\left|\int_{1/2}^xf(t)\,dt - \int_{1/2}^y f(t)\,dt\right|\le\int_x^y|f(t)|\,dt\le C|y-x|.$$ Hence, the integral is Lipschitz. Since a Lipschitz function on $(0,1)$ is bounded (to show that, use the opposite triangle inequality), (1) and (2) can be answered in the affirmative. However, (3) is false. An example is $f(x) = x^2\sin(1/x)$.
• Thanks! What is the sufficient and necessary condition of $f$ if $f'$ is Lipschitz? (Great username BTW) Oct 13, 2017 at 3:24
• I mean, what is the mildest conditions $f$ need to satisfy, if we want $f'$ be Lipschitz? Oct 13, 2017 at 18:10
• I don't know. Consider the functions $f : (0,1)\to\mathbb R$ having the property that$$F(x,y) := \frac{f(x)-f(y)}{(x-y)^2},\qquad x,y\in (0,1)$$is bounded. Maybe they do. Oct 13, 2017 at 22:36