Let $I=[0,1]$ and $u\in L^\infty(I)$. Suppose $u_n$ is a bounded sequence in $L^\infty(I)$ with the property that $\lim_{n\to\infty}\int_If(x,u_n(x))\mathrm dx=\int_If(x,u(x))\mathrm dx,$ for all continuous functions $f\;\colon\;I\times\mathbb R\to\mathbb R$. Prove then that $u_n$ converges strongly in $L^1(I)$ to $u$. Does anybody have an hint to solve the problem? I think I can start from supposing that $u\in C(I)$ and then maybe proceed by approximation in the general case, however I'm stuck even in this simpler case, so I welcome also every hint for the case of $u\in C(I)$.
Unusual sufficient condition to have strong convergence in $L^1([0,1])$
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real-analysis
functional-analysis
1 Answers
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Taking functions $f$ of the form $f(s,t)=P(s)\cdot t$, we can see that $\lim_{n\to +\infty}\int_IP(x)u_n(x)dx=\int_IP(x)u(x)dx$ for all polynomial $P$. Using the fact that $I$ is compact and the density of $ C(I)$ in $L^2(I)$, we can see that $u_n$ converges weakly to $u$ in $L^2(I)$. Considering $f(s,t)=t^2$, we get that the convergence is strong in $L^2$ hence in $L^1$.