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)
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 $f(x) = x^2\sin(1/x)$.