Maximum value of a twice continuously differentiable function on an interval Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be an everywhere twice continuously differentiable function, and suppose $f(1)=f'(1)=f(-1)=f'(-1)=0$. What is the maximum possible value for $f(0)$ given $|f''(x)|\leq 1$ for all $x\in\mathbb{R}$?
Thanks for the help!
 A: By continuity, $f$ assumes its maximum on $[-1,1]$ at some point $a$.
Wlog. $a\ge 0$.
By the condition, $f(1-h)\le \frac12h^2$.
From $f'(a)=0$, also by the condition $f(a+h)\ge f(a)-\frac12h^2$.
This implies $$ f(a)-\frac{(1-a)^2}8\le f\left(a+\frac{1-a}2\right)= f\left(1-\frac{1-a}2\right)\le \frac{(1-a)^2}8$$
hence
$$f(0)\le f(a)\le \frac{(1-a)^2}{4}\le \frac14.$$
This is sharp as we see from
$$ f(x)=\begin{cases}\frac12(1+x)^2&\text{if }x\le-\frac12\\
\frac14-\frac12x^2&\text{if }-\frac12\le x\le \frac12\\
\frac12(1-x)^2&\text{if }x\ge \frac12\end{cases}$$
which is twice dfferentiable except at $\pm\frac12$, but can be approximated arbitrarily well with a smoother function:
For $0<\epsilon<\frac12$ let 
$$ g(x)=\begin{cases}1&\text{if }x<-\frac12-\epsilon\text{ or } x>\frac12+\epsilon\\
\frac1\epsilon(x+\frac12)&\text{if }-\frac12-\epsilon\le x\le-\frac12+\epsilon\\
-1&\text{if }-\frac12+\epsilon<x<\frac12+\epsilon\\
\frac1\epsilon(\frac12-x)&\text{if }+\frac12-\epsilon\le x\le+\frac12+\epsilon,\end{cases}$$
then $G(x)=\int_{-1}^x g(t)\,\mathrm dt$ and $F(x)=\int_{-1}^x G(t)\,\mathrm dt$. Then $F(x)$ fulfills the conditions and almost coincides with $f(x)$. In fact $|G(x)-f'(x)|\le \int_{-\epsilon}^\epsilon \frac t\epsilon\,\mathrm dt =\epsilon$ and therefore $|F(0)-f(0)|\le \epsilon$ (because $F(-1)=f(-1)=0$).
