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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!

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  • $\begingroup$ How on earth do you get that $f_n \partial_xf_n\partial_yf_n\to f\partial_xf\partial_y f$ "in a weak topology" (presumably $H^{-2}$, as you seem to believe)? you cannot simply take the product of weakly converging sequences and conclude that it is weakly converging to the product of the limits... I recommend cross-posting your question to Math.StackExchange, people there will certainly explain why your naive hope cannot hold $\endgroup$ Commented Jun 4, 2020 at 7:50
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    $\begingroup$ For what it's worth I think that the question may as well stay here, even though there is a problem with the reasoning as Leo Monsaingeon has pointed out. For instance, it is not immediately clear to me that the original question has a negative answer, merely that the OP's reasoning along the way is incorrect $\endgroup$
    – Yemon Choi
    Commented Jun 4, 2020 at 13:10
  • $\begingroup$ @leomonsaingeon The reason I "hope" this holds is gut feeling: what I hope is actually the lsc of the elastic part of the Landau de Gennes energy $\|\nabla Q\|_{L^2}^2 + \epsilon\sum_{i,j,k,l} Q^{kl} \partial_{x_k}Q^{ij} \partial_{x_l}Q^{ij}$, where $Q$ is a 2x2 Q-tensor. I just wrote one of the terms in the cubic term here, since I don't see lots of simplifications coming from the cubic term. With an added bulk term in $Q$, it was shown for the gradient flow to exist, under small initial data (Xu, Iyer, Zarnescu 2015). So for a gradient flow to exist, I'd guess its leading part to be lsc... $\endgroup$
    – maria_c
    Commented Jun 4, 2020 at 16:48
  • $\begingroup$ well I'm no expert in Ladau de Gennes theory, but I can tell you right ahead that something really intricate is going on here. In your simplified setting, and if all you know is that $f\in H^1$, the term $\int f \partial_x f\partial_y f$ may not even be integrable. Indeed $2^*=\frac{2d}{d-2}=+\infty$ at least formally, but this is a borderline case for the Sobolev embedding and $H^1$ functions fail to be $L^\infty$ in general in dimension $d=2$. Since the best one can hope for the crossed term is $\partial_x f\partial_yf\in L^2L^2=L^1$ clearly your $u$ term may very well fail to be integrable. $\endgroup$ Commented Jun 4, 2020 at 20:10
  • $\begingroup$ I would bet that the lower semi-continuity cannot really work, at least not for the weak $H^1$ convergence: because of my previous comment, and because you cannot take products of weak limits, there is no hope for a full "weak continuity" argument to work (as you were trying to use). So really the only hope is for true lower semicontinuity (by that I mean a real inequality between $\int\leq \liminf\int $, not an actual equality in general). But note that changing $f$ to $-f$ should give the reversed inequality, if any, so I really don't think that such a "true" lower sc can work here. $\endgroup$ Commented Jun 4, 2020 at 20:13

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