sup-norm bound by second derivative If $f\in C[0,1]$, twice differentiable function, and $f(0)=0=f(1),$ then $$ \sup_{x\in[0,1]} |f(x)| \leq \frac{1}{8} \sup_{x\in[0,1]} |f''(x)|.$$
I am not able to get the constant ${1}/{8}$, here is my try:
For any $x\in[0,1]$, $$ |f(x)| = |f(x)-f(0)| \leq \int_0^x \int_0^t |f''(y)| \,dy\,dt \leq \sup_{x\in[0,1]} |f''(x)| \int_0^x t\,dt \leq \frac{1}{2} \sup_{x\in[0,1]} |f''(x)|. $$
Thanks for any help regarding this!
 A: Assume that $\sup_{x\in[0,1]}|f(x)|=|f(c)|$, we may assume that $c\in(0,1)$ or else the inequality holds immediately. It is not hard to see that $f'(c)=0$. And we have 
\begin{align*}
f(c)&=f(c)-f(0)\\
&=\int_{0}^{c}\int_{c}^{t}f''(y)dydt,
\end{align*}
so by letting $M:=\sup_{x\in[0,1]}|f''(x)|$,
\begin{align*}
|f(c)|\leq\dfrac{1}{2}c^{2}M.
\end{align*}
Assume at the moment that $c\in(0,1/2]$, then $|f(c)|\leq(1/8)M$. 
For $c\in[1/2,1)$, consider instead that
\begin{align*}
f(c)=f(c)-f(1)
\end{align*}
and use the same trick to get 
\begin{align*}
|f(c)|\leq\left(\dfrac{1}{2}c^{2}-c+\dfrac{1}{2}\right)\cdot M
\end{align*}
which has maximum $(1/8)M$ on $[1/2,1)$.
Explanation that $f'(c)=0$:
One may think of MVT gives you that $f'(d)=0$ due to $f(1)=f(0)=0$ but this does not entail that $c=d$. So we need to argue in a different way.
For $f(c)=0$, then $f$ is identically zero, there is nothing to prove. Suppose that $f(c)>0$, then a small neighborhood $(c-\delta,c+\delta)$ is such that $f(x)>0$ for $x\in(c-\delta,c+\delta)$, then $f(c)=|f(c)|\geq\sup_{x\in[0,1]}|f(x)|\geq\sup_{x\in(c-\delta,c+\delta)}|f(x)|=\sup_{x\in(c-\delta,c+\delta)}f(x)$, this shows that $f$ has a local maximum at $c$.
For $f(c)<0$, the same idea leads to some $(c-\delta,c+\delta)$ such that $f(x)<0$ for $x\in(c-\delta,c+\delta)$. Assume the moment that $f(x)<f(c)$ for some $x\in(c-\delta,c+\delta)$, then $|f(x)|=-f(x)>-f(c)=|f(c)|$, this violates that $|f(c)|=\sup_{x\in[0,1]}|f(x)|$, so it must be the case that $f(x)\geq f(c)$ for all $x\in(c-\delta,c+\delta)$, so $f$ has a local minimum at $c$.
A standard calculus theorem tells you that for local extremum of a differentiable function $f$ at $c$, then it follows that $f'(c)=0$. 
An Alternative Way:
For $f(c)\geq 0$, then $f(c)=|f(c)|=\sup_{x\in[0,1]}|f(x)|\geq|f(x)|\geq f(x)$, so $f$ has a global maximum at $c$.
For $f(c)<0$, then $-f(c)=|f(c)|=\sup_{x\in[0,1]}|f(x)|\geq|f(x)|\geq-f(x)$, so $f(c)\leq f(x)$, so $f$ has a global minimum at $c$.
In either case, $f$ has a local extremum at $c$.
A: Let $M = \sup\limits_{x \in [0,1]} |f''(x)|$. For any $t \in (0,1)$, consider the function
$$g(x) = f(x) - f(t) \frac{x(1-x)}{t(1-t)}$$
It is easy to see $g(0) = g(t) = g(1) = 0$. 
By Rolle's theorem, there is a $y_0 \in (0,t)$ and $y_1 \in (t,1)$ such that $g'(y_0) = g'(y_1) = 0$. 
Apply Rolle's theorem again, there is a $z \in (y_0, y_1) \subset (0,1)$ such that $g''(z) = 0$. Notice $g''(x) = f''(x) + \frac{2}{t(1-t)} f(t)$. This leads to
$$|f(t)| = \frac{t(1-t)}{2} |f''(z)| \le \frac{t(1-t)}{2} M 
= \frac{1 - (1-2t)^2}{8} M \le \frac18 M$$
Since this is true for all $t \in (0,1)$ and $f(0) = f(1) = 0$, we obtain
$$\sup_{x \in [0,1]} |f(x)| \le \frac18 M = \frac18 \sup_{x\in [0,1]} |f''(x)|$$
