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Some time ago, I asked here:

How to use Hardy-Littlewood-Sobolev inequality to estimate an integral involving two fuctions and Riesz Potential.

about estimates involving the Hardy-Littlewood-Sobolev inequality. Hence, we know that if $u,v \in L^{2}(\mathbb{R}^{2})$, then $$\int_{\mathbb{R}^{2}}(I_{\beta} \ast |u|^{\frac{\beta}{2}+1})|u|^{\frac{\beta}{2}}|v|dx < +\infty. $$

Now, my question is: if $u_{n}$ or $v_{n}$ goes to infinity in $L^{2}(\mathbb{R}^{2})$, what I can say about $$\int_{\mathbb{R}^{2}}(I_{\beta} \ast |u_{n}|^{\frac{\beta}{2}+1})|u_{n}|^{\frac{\beta}{2}}|v_{n}|dx?$$

And if $u_{n}$ tends to infinity in $L^{2}$, what I can say about

$$\int_{\mathbb{R}^{2}}(I_{\beta} \ast |u_{n}|^{\frac{\beta}{2}+1})|u_{n}|^{\frac{\beta}{2}}u_{n}dx?$$

$\beta \in (0,2)$ .

BBVM
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  • When you say “goes to infinity” do you mean they are unbounded sequences in $L^2$? – JackT Aug 30 '21 at 04:25
  • Also, are you able to provide some more details on what specifically you are looking for? For example, are you hoping to conclude that that quantity will also be unbounded or something along those lines? – JackT Aug 30 '21 at 04:27
  • @JackT Hello, yes the sequences are unbounded and I hoping conclude that the quantity is unbounded too. – BBVM Aug 31 '21 at 19:19

1 Answers1

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No, there exists pairs of functions that are not in $L^2×L^2$ such that your integral is finite. For instance if $v$ is only in $L^1$ and $u$ is very nice (say $u∈ C^\infty_c$), then $$ \int_{\mathbb{R}^{2}} (I_{\beta} * |u|^{\frac{\beta}{2}+1})\,|u|^{\frac{\beta}{2}}\,|v|\,\mathrm dx ≤ \|(I_{\beta} * |u|^{\frac{\beta}{2}+1})\,|u|^{\frac{\beta}{2}}\|_{L^∞} \,\|v\|_{L^1} < \infty $$ If $u>0$ and $w= u^\frac{\beta+2}{2}$, notice that $$ N_β := ∫_{\mathbb R^2} (I_β * |u|^\frac{\beta+2}{2})\,|u|^\frac{\beta}{2}u = ∫_{\mathbb R^2} (-\Delta)^\frac{\beta-2}{2} (w)\,w \\ = ∫_{\mathbb R^2} |(-\Delta)^\frac{\beta-2}{4} w|^2 = \|w\|_{\dot{H}^{-(1-\beta/2)}}^2 $$ Hence your last question is equivalent to know if $\dot{H}^{-(1-\beta/2)} ⊆ L^\frac{4}{\beta+2}$ (since $∫|u|^2 = \int |w|^\frac{4}{\beta+2}$). However, the space $\dot{H}^{-(1-\beta/2)}$ contains distributions that are not even in $L^1_{\mathrm{loc}}$. Then any sequence of smooth compactly supported functions $u_n$ such that $u_n^{1+\beta/2}$ converges to such a distribution will in $\dot{H}^{-(1-\beta/2)}$ verify $\|u_n\|_{L^2}\to \infty$ and $N_\beta < C$.

As an example of function in $H^{-s}$ and not in $L^2$, see here Sobolev space with negative index

LL 3.14
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