Prove that $\int^1_0|f'(x)|dx \le 2\int^1_0 |f(x)|dx+\int^1_0|f''(x)|dx.$ $f(x),f'(x),f''(x)$ are  continuous functions, $f(0)f(1) \ge 0$, prove that $$\int^1_0|f'(x)|dx \le 2\int^1_0 |f(x)|dx+\int^1_0|f''(x)|dx.$$
This problem comes from here.
 A: Replacing $f$ with $-f$, we may assume that $f(0)\geq 0$.
If $f'$ has a root $a\in[0,1]$, then write $$\int_0^a|f'|=\int_0^a\left|\int_a^xf''(t)\,dt\right|dx\leq\int_0^a\int_x^a|f''(t)|\,dtdx=\int_0^a\int_0^t|f''(t)|\,dxdt\leq\int_0^a|f''|,$$
and similarly $\int_a^1|f'|\leq\int_a^1|f''|$, so the estimate holds.
If $f'$ does not have a root in $[0,1]$, then it has a fixed sign; replacing $f(x)$ with $f(1-x)$, we may assume that $f'>0$ in $[0,1]$. So, $f\geq 0$ in $(0,1)$.
We claim that there exists $\xi\in (0,1)$ such that $f'(\xi)\leq 2\int_0^1f$: if not, then $f'(x)>2\int_0^1f$ for all $x\in(0,1)$. Integrating in $(0,x)$ for $x\in(0,1)$, we then have $$f(x)-f(0)=\int_0^xf'>2x\int_0^1f,$$ therefore $$\int_0^1f(x)\,dx-f(0)=\int_0^1(f(x)-f(0))\,dx>\int_0^1f\int_0^12x\,dx=\int_0^1f(x)\,dx,$$ which implies that $f(0)<0$, a contradiction with $f\geq 0$ in $(0,1)$. So, for some $\xi\in (0,1)$, we have $f'(\xi)\leq 2\int_0^1f$. Then, \begin{align*}\int_0^1|f'(x)|\,dx&\leq\int_0^1|f'(x)-f'(\xi)|\,dx+|f'(\xi)|\\
&\leq\int_0^1\int_{\xi}^x|f''(t)|\,dtdx+2\int_0^1f,\end{align*} which implies the estimate.
