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Convergence as measure vs in H^$H^{-2}$

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maria_c
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Convergence as measure vs in H^{-2}

Let the domain be the two dimensional torus $\mathbb{T}$, and let $f_n $ be a sequence bounded in $H^1$, such that $\sup_n |f_n|\le 1$, and $f_n \to f$ weakly in $H^1$. Let $u_n = f_n \frac{\partial f_n}{\partial x} \frac{\partial f_n}{\partial y}$.

My Question: do I have the lower semicontinuity result $$\liminf_n \|\nabla f_n\|_{L^2 }^2 +\epsilon \int_{\mathbb{T}} u_n dx\ge \|\nabla f\|_{L^2 }^2 +\epsilon \int_{\mathbb{T}} f \frac{\partial f}{\partial x} \frac{\partial f}{\partial y}dx $$ at least for sufficiently small $\epsilon$?

My Attempt: clearly the $L^2$ norm of the gradient is lower semicontinuous. To get the full semicontinuity, I try to show $$\int_{\mathbb{T}} u_n dx\to \int_{\mathbb{T}} f \frac{\partial f}{\partial x} \frac{\partial f}{\partial y}dx . $$ Since $f_n$ is uniformly bounded in $H^1$, we have $f_n \to f$ strongly in $L^2$, and $\frac{\partial f_n}{\partial x} \to \frac{\partial f}{\partial x} $ and $\frac{\partial f_n}{\partial y} \to \frac{\partial f}{\partial y} $ weakly in $L^2$. Thus $u_n\to f \frac{\partial f}{\partial x} \frac{\partial f}{\partial y}$ in some weaker topology, e.g. $H^{-2}$. Moreover, since $\sup_n|f_n|,|f|\le 1$, and $f_n\to f$ weakly in $H^1$, we have $u_n$ is uniformly bounded in $L^1$, hence it converges weak-* to some Radon measure $\mu$.

So (up to subsequence), $u_n$ converges to $\mu$ in the weak-* topology, and to $f \frac{\partial f}{\partial x} \frac{\partial f}{\partial y}$ in $H^{-2}$.

Main issue: can I say that $\mu=f \frac{\partial f}{\partial x} \frac{\partial f}{\partial y}$? My main worry is that $H^2$ functions are not dense in $L^\infty$...

Any help is greatly appreciated, thanks!