A question about the integral of convex function 
Let $F:\mathbb{R}\to\mathbb{R}$ be a strictly convex function. Let $u:[0,1]\to\mathbb{R} $ be a continuous function, with
  $$\int_{0}^{1}u(x)\,dx=0$$
  Show that
  $$\int_{0}^{1}F(u(x))\,dx\leqslant\frac{F(\| u\|_\infty)+F(-\| u\|_\infty)}{2}$$
  where
  $$\| u\|_\infty:=\sup_{x\in [0,1]}|u(x)|$$
  Also determine where the equality occurs.

I have tried to use Jensen inequality but I failed.
 A: Note that if $||u||_{\infty} = 0$ then your question is trivially true with equality. Now assume that $||u||_{\infty} > 0$. 
Observe 
$F(u(x)) \leq (\frac{-u(x) + ||u||_{\infty}}{2||u||_{\infty}})F(-||u||_{\infty}) + (\frac{+u(x) + ||u||_{\infty}}{2||u||_{\infty}})F(||u||_{\infty})$ by convexity of $F$ thus
$\int_{0}^1F(u(x)) dx \leq \int_{0}^1(\frac{-u(x) + ||u||_{\infty}}{2||u||_{\infty}})F(-||u||_{\infty}) + (\frac{+u(x) + ||u||_{\infty}}{2||u||_{\infty}})F(||u||_{\infty})$ $dx$
$= \frac{1}{2}(F(-||u||_{\infty})+F(||u||_{\infty}))$
For equality we must have that for $y$ in the range of $u$ 
$F(y) = (\frac{-y + ||u||_{\infty}}{2||u||_{\infty}})F(-||u||_{\infty}) + (\frac{+y + ||u||_{\infty}}{2||u||_{\infty}})F(||u||_{\infty})$ almost everywhere (in the range of $u$) .
A: All that is required is that $F$ is convex (the strict convexity is not needed). If $\|u\|_{\infty}=0$ the statement is vacuous. Otherwise, for each $x$ we may write $u(x)$ as a convex combination of $\|u\|_{\infty}$ and $-\|u\|_{\infty}$ as follows:
$$
u(x)=\frac{\|u\|_{\infty}+u(x)}{2\|u\|_{\infty}}\|u\|_{\infty}+\frac{\|u\|_{\infty}-u(x)}{2\|u\|_{\infty}}(-\|u\|_{\infty})
$$
and apply the definition of convexity to obtain that
$$
F(u(x))\leq \frac{\|u\|_{\infty}+u(x)}{2\|u\|_{\infty}}F(\|u\|_{\infty})+\frac{\|u\|_{\infty}-u(x)}{2\|u\|_{\infty}}F(-\|u\|_{\infty}),
$$
which after integrating leads to
$$
\int_0^1F(u(x))\ dx\leq \frac{F(\|u\|_{\infty})+F(-\|u\|_{\infty})}{2}.
$$
The strict convexity helps simplify the equality case, since it implies that equality holds in the pointwise bound only when $u(x)$ is equal to either $\|u\|_{\infty}$ or $-\|u\|_{\infty}$. Since this must hold for all $x\in[0,1]$ by continuity, it follows that $u(x)$ vanishes identically.
Note, another (more concrete) way to argue in the equality case is to suppose that $\|u\|_{\infty}\not=0$ and show that there is a strict inequality, by restricting to an interval where $|u|$ is bounded away from $0$ and its maximum, and reversing the steps sketched in the previous paragraph.
