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.
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 $u$) .
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.
