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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.

  • 0
    What do you mean by "weakly lower semi-continuous"? Is it $R_n$ converges to $R$ weakly implies $\liminf F(R_n)\geq F(R)$ for all $R$?2011-12-30
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    Yes, that's what I mean.2012-01-01
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    You say it's not homework, but where is the problem from?2012-01-03
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    A medical physics application. Timing radiation treatments to minimize damage to healthy cells while killing cancerous cells.2012-01-05

1 Answers 1

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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.

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    excellent. I accepted your answer, but I am not 100% convinced that $\int_0^1 \int_0^1 R(t)R(t')\Phi(t,t')\,dt\,dt' > 0$ if $R \in L^2([0,1])$ and $R \neq 0$.2012-08-25
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    @bogus I'll think about it more.2012-08-25