Is this functional weakly lower semicontinuous? Define $F\colon L^2([0,1]) \to {\mathbb R}$ by
$$ F(R) = \int_0^1 \int_0^1 R(t) R(t') \exp\left(-|t-t'| - \left|\int_t^{t'} R(s)\,ds\right|\right)
\,dt\,dt'.$$
Is $F$ weakly lower semicontinous, that is, do we have $F(R) \leq \liminf_{n\to\infty}F(R_n)$ if $R_n$ converges weakly in $L^2$ to $R$?  
This is not a homework problem.
 A: This being last year's thread, I hope that the answer (which is affirmative) is not coming too late for cancer patients. The key point is the following 
Lemma. If $R_n\to R$ weakly in $L^2[0,1]$, then $\int_t^{t'} R_n\to \int_t^{t'} R$ uniformly in $t,t'\in [0,1]$.
Proof. We may assume $R=0$ (otherwise consider $R_n-R$). Since a weakly convergent sequence is bounded in $L^2$, we get the estimate $\left|\int_t^{t'} R_n\right|\le C\sqrt{t-t'}$ from the Cauchy-Schwarz inequality. Furthermore, for any positive integer $m$ and any integers $i,j\in\{0,\dots,m\}$ we have $\lim_{n\to\infty}\int_{i/m}^{j/m} R_n=0$ since the integral is the inner product of $R_n$ with a fixed $L^2$ function (characteristic function of the interval). 
Given $\epsilon>0$, pick $m$ such that $C\sqrt{1/m}<\epsilon/4$. Then pick $N$ such that $\left|\int_{i/m}^{j/m} R_n\right|<\epsilon/2$ for all $n\ge N$ and all $i,j\in\{0,\dots,m\}$. Given $t,t'\in [0,1]$, we can find $i,j$ such that $|t-i/m|<1/m$, $|t'-j/m|<1/m$, and therefore 
$$\left|\int_t^{t'} R_n\right|\le \left|\int_{i/m}^{j/m} R_n\right|+2C\sqrt{1/m}<\epsilon$$
whenever $n\ge N$. The lemma is proved.
Returning to the original problem, write $\Phi(t,t')=\exp\left(-|t-t'| -\left|\int_t^{t'}R(s)\,ds\right|\right)$ and note that 
$$
F(R_n)-\int_0^1\int_0^1 R_n(t)R_n(t')\Phi(t,t')\,dt\,dt' \to 0
$$
using the lemma and the fact that $R_n(t)R_n(t')$ is bounded in $L^1([0,1]^2)$. The integral $\int_0^1\int_0^1 R_n(t)R_n(t')\Phi(t,t')\,dt\,dt'$ is a positive quadratic form, hence convex, hence (sequentially) weakly lower semicontinuous.
