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Following problem in a mock exam:

Prove or falsify: (1) If $f_j \rightharpoonup f$, $g_j \rightarrow g$ in $L^4(\mathbb{R})$ then $f_j g_j \rightharpoonup fg$ in $L^2(\mathbb{R})$.

(2) If $f_j \rightharpoonup f$, $g_j \rightharpoonup g$ in $L^4(\mathbb{R})$ then $f_j g_j \rightharpoonup fg$ in $L^2(\mathbb{R})$.

My try:

We want to show that $\int_\mathbb{R} f_j g_j \psi dx \rightarrow \int_\mathbb{R} f g \psi dx$ for all $\psi \in L^2(\mathbb{R})$, so lets start:

$\vert \int_\mathbb{R} f_j g_j \psi dx - \int_\mathbb{R} f g \psi dx \vert \leq \vert \int_\mathbb{R} (f_j g_j - fg_j) \psi dx\vert + \vert \int_\mathbb{R} (f g_j - fg) \psi dx\vert$

The first summand equals $\int (f_j - f) g_j \psi dx$ and tends to zero because the assumption and $g_j \psi \in L^2$. The second summand equals $\int (g_j - g)f \psi dx$ and with $f\psi \in L^2$ it tends also to zero because $\int (g_j - g)f \psi dx \leq \vert\vert g_j - g\vert\vert_{L^4} \vert\vert f\psi\vert\vert_{L^{4/3}} \rightarrow 0$.

So that would be a proof for (1) but (2) would fail because this last estimate is no more true. Does anyone have a counterexample? Everything true in my steps above? :) Thanks for comments!

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It is not too hard to see for

$$ f_j (x) = e^{2\pi i j x} 1_{(0,1)}(x) $$ and $$ g_j (x) = e^{-2\pi i j x} 1_{(0,1)}(x) $$ that $f_j, g_j \rightharpoonup 0$. But $f_j g_j = 1_{(0,1)} $ does not converge weakly to $0$.

Also, in your proof, the integral $$\int(f_j - f) g_j \psi $$ needs some further thought, since $g_j \psi $ depends on $j $.

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  • $\begingroup$ thanks, but I do not see the point with the integral above, we only need ...$\in L^2$ isn't it? $\endgroup$ – tubmaster Mar 23 '17 at 22:01
  • $\begingroup$ @tubmaster: $f_j \rightharpoonup f$ weakly in $L^4$ means that for each fixed but arbitrary $h \in L^{4/3}$ (not $L^2$), we have $\int (f_j - f) \cdot h \, dx \to 0$. Since $h$ is fixed, it is not allowed to depend on $j$. One can easily find counterexample if $h$ is allowed to depend on $j$ (do it!). But what one can show is that if $h_j \to h$ strongly in $L^{4/3}$, then $\int (f_j - f) h_j \, dx \to 0$. $\endgroup$ – PhoemueX Mar 24 '17 at 8:32
  • $\begingroup$ but your last thing is true in our case, isn't it? $g_j \psi \rightarrow g\psi$, because $g_j \rightarrow g$ in $L^4$ and $\psi \in L^{4/3}$ $\endgroup$ – tubmaster Mar 24 '17 at 9:01
  • $\begingroup$ @tubmaster: Yes, it is. I am just saying that the general statement (the "last thing") is not completely trivial. $\endgroup$ – PhoemueX Mar 24 '17 at 10:47

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