prove that $\int_{0}^{1}|f(x)|dx \leq \int_{0}^{1}|f'(x)|dx$ I would appreciate if somebody could help me with the following problem
Q: prove that 
$$\int_{0}^{1}|f(x)|dx \leq \int_{0}^{1}|f'(x)|dx$$
where 
  $f'(x)$ is continuous on $(0,1)$, and $f(0)=0$.
 A: Using the fundamental theorem of  calculus and $f(0)=0$ we obtain $f(x)=\int_0^x f'(t)dt$ and hence $$\int_0^1 |f(x)|dx=\int_0^1 \left|\int_0^x f'(t)dt\right|dx \leq \int_0^1 \int_0^x |f'(t)|dt\ dx\leq \int_0^1 \int_0^1 |f'(t)|dt\ dx = \int_0^1 |f'(t)|dt$$ since $|f'(t)|\geq 0$.
A: Partial answer: If you suppose $f'\geq 0$, then $f$ is non decreasing, so $f \geq f(0) = 0$. Hence
$$\int_0^1 f(x) dx \leq \int_0^1 \max f dx = f(1) = f(1) - f(0) = \int_0^1 f'(x) dx $$
If you suppose $f' \leq 0$ you get the same result.
Now maybe one can work separately on the intervals where $f'$ is $\geq 0$ or $\leq 0$ alternately, using the fact that $f'$ is continuous.
A: By the fundamental theorem of Calculus,
$$
\left|f(x)\right| = \left|f(x) - f(0)\right| = \left|\int_0^x f'(t) dt\right| \le \int_0^x \left|f'(t)\right|dt \le \int_0^1 \left|f'(t)\right|dt.
$$
The result now follows by integrating with respect to $x$ from $0$ to $1$.
A: For any $f$, let us split the interval $[0,1]$ into subintervals such that $f'(x)$ has a constant sign. We can define a function $g(x)$ by "mirroring" all decreasing sections, i.e. $$g(x)=\int_0^x|f'(x)|dx.$$
We have $\int_0^x f'(x)dx\le\int_0^x|f'(x)|dx$, so that $f(x)\le g(x)$, and we have $|f'(x)|=g'(x)$. It is enough to consider $g(x)$.
By the mean value theorem,
$$\int_0^1g(x)dx\le g(1),$$ and $$\int_0^1g'(x)dx=g(1).$$

