Some time ago, I asked here:
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
$$\int_{\mathbb{R}^{2}}(I_{\beta} \ast |u_{n}|^{\frac{\beta}{2}+1})|u_{n}|^{\frac{\beta}{2}}u_{n}dx?$$
$\beta \in (0,2)$ .
Best Answer
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