Strong Convergence in $L^2(I)$ Let $I=[0,1]$, and let $(u_n)$ be a sequence in $L^\infty(I)$ converging to $u\in L^\infty(I)$ in the weak$\vphantom{}{^*}$ topology of $L^\infty(I)$. Let $f:\mathbb R\to\mathbb R$ be a $C^2$ function with $f''(t)>0$ for any real $t$. Assume that $$\lim_{n\to\infty}\int_I f(u_n(x))\mathrm dx=\int_If(u(x))\mathrm dx.$$ Prove then that $u_n$ converges strongly in $L^2(I)$ to $u$.
EDIT since the question has been raised in the comments I want to remark that the $f$ in the problem is not arbitrary but it is a specific strictly convex function.. 
EDIT 2. I've thought quite a long time about this problem but yet I didn't came up with anything interesting. But a reliable way to follow it seems to me Taylor Expansion in the sense that one can write $$f(u(x))=f(u_n(x))+f'(u(x))(u_n(x)-u(x))+\frac 12 f''(u(x))(u_n(x)-u(x))^2+h_2(u(x))(u_n(x)-u(x))^2$$ and then try to integrate and pass to the limit and see what happens. Is this useful?
Please let me know because i feel lost in front of this problem.
 A: Since $u_n$ converges to $u$ weak* in $L^\infty(I)$, we know that $\|u_n\|_{L^\infty}$ is uniformly bounded. Indeed, the weak* convergence implies
$$
|\langle u_n,\phi\rangle| \leq |\langle u,\phi\rangle| + 1 \leq \|u\|_{L^\infty} + 1,
$$
for $\phi\in L^1$ such that $\|\phi\|_{L^1}=1$, and for $n$ large.
This gives a uniform bound on $|\langle u_n,\phi\rangle|$
for any fixed $\phi\in L^1$, and an application of the uniform boundedness principle, followed by the duality between $L^\infty$ and $L^1$, shows that $\|u_n\|_{L^\infty}$ is uniformly bounded.
Then as Siminore observed in the comments,
$$
f(u_n(x))-f(u(x))-f'(u(x))(u_n(x)−u(x))=\frac12f''(\xi_n(x))(u_n(x)−u(x))‌^2\geq c|u_n(x)-u(x)|^2,
$$
for almost every $x\in I$, with some constant $c>0$.
Integrating over $I$, we get
$$
\int \left(f(u_n)-f(u)\right) - \int f'(u) (u_n-u) \geq c\|u_n-u\|_{L^2}^2.
$$
In the limit $n\to\infty$, the first integral on left hand side goes to $0$ by assumption, and so does the integral term by weak* convergence. Hence $u_n\to u$ in $L^2$.
