An Integral inequality involving vanishing boundary data Suppose that $f$ is twice differentiable on $[0,1]$ with $f(0)=f(1)=0$. Also $f$ is not identically zero. Show that $$|f(x)|\leq \frac{1}{4}\int_0^1 |f''(x)|dx,\ \forall\ x\in [0,1].$$
Thank @FisiaiLusia, I am sorry that on my computer "Mathematics Stack Exchange requires external JavaSrcipt from another domain, which is blocked or failed to load". So that I could not vote for your answer. 

We can assume that there exists point $c\in (0,1) $ such that $$ \sup_{v\in [0,1] } |f(v)|=f(c) .$$
By Lagrange Theorem we have that there exist points $u_1 \in (0,c) $ and $u_2 \in (c, 1)$ such that $$ f(c) =f(c) -f(0) =cf' (u_1 )$$ $$f(c) =f(c) -f(1) =(c-1)f' (u_2 )$$
so we have $$\int_0^1 |f''(s)|ds \geq \int_{u_1}^{u_2} |f''(s)|ds \geq \left|\int_{u_1}^{u_2} f''(s)ds\right| =|f'(u_2) -f'(u_1)|\geq f'(u_1) -f'(u_2) = \left(\frac{1}{c} +\frac{1}{1-c}\right) f(c) \geq 4f(c) .$$
