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!


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 $.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.