# Does the inequality $\iint_{\mathbb R^2} \frac{|f(x)|^2}{|x-y|^a} \ dy\,dx\le k ||f||_{L^2(\mathbb R)}^2$ hold?

I am solving an exercise about integration in dimension 2 and I am stuck at this point.

Suppose that $$f\in L^2(\mathbb R)$$ and let $$a\in (1, 2)$$. Consider the integral $$\iint_{\mathbb R^2} \frac{|f(x)|^2}{|x-y|^a} \ dy\,dx.$$

I would like to prove that $$\iint_{\mathbb R^2} \frac{|f(x)|^2}{|x-y|^a} \ dy\,dx\le k ||f||_{L^2(\mathbb R)}^2,$$ for a positive constant $$k$$. There is any trick which can allow me to state that the previous inequality hold?

I have tried the integration by parts, cauchy-schwarz inequality, divide the integration domain. Nonetheless, I am still not able to find a way to get the desired inequality.

Do you have any hint?

• "I'm solving a very hard..." A random MSE user: And I have taken that personally Apr 10 at 10:22
• Just to say that I find the exercise to be very hard for my current level of math. I can remove that part if it bothers you so much or if it offends someone's sensibilities. Apr 10 at 10:31
• $|x-y|^a$ clearly blows up whenever $x=y$. Do you have any good reasons to believe that the bound you're looking for is true? Apr 10 at 11:16
• To embark on @Oscar's thoughts: This bound would imply that $\int_{[-\varepsilon, \varepsilon]^2} \tfrac{1}{\lvert x - y \rvert^a}~\mathrm{d}(x, y) < \infty$ for any $\varepsilon > 0$. My quick Fubini computations (that could be off) tell me that this does not hold. Apr 10 at 11:29
• This looks very similar to some sort of fractional Sobolev seminorm, in which case the numerator would be $f(x) - f(y)$. Can you give a bit more context on where you encountered this integral? If it comes from that setting, it may be just a typo. Apr 10 at 14:47

For $$f\neq 0$$ the left hand side is equal $$\infty,$$ as $$\int\limits_{-\infty}^\infty|f(x)|^2\left [\int\limits_{-\infty}^\infty{dy\over |x-y|^\alpha}\right ]\,dx\\ =\int\limits_{-\infty}^\infty|f(x)|^2\left [\int\limits_{-\infty}^\infty {dy\over |y|^\alpha}\,dy\right ]\,dx=\infty$$ The function $$f$$ does not matter at all as long as it is nonzero. So no inequalities may help.

Your estimate is a special case of the weak Young-inequality (used in Potential theory) which follows from the Hardy-Littlewood-Sobolev inequality, see the first answer of

https://mathoverflow.net/questions/181669/generalized-hardy-littlewood-sobolev-inequality

It holds that $$\iint \frac{\vert f(x) \vert^2 }{ \vert x-y\vert ^a} \, dy \, dx = \Vert \, \vert \cdot \vert^{-a/2} * f \Vert_2 \leq C \Vert f \Vert_p.$$ The exponent $$p$$ depends on $$a$$ via (we use $$n=2$$): $$1/p + a/4 = 1 + 1/2.$$

• The inequality does not help. In order to apply HL you need to estimate something of the form $$\int\left (\int {|f(x)|^2\over |x-y|^a}\,dx \right )^pdy$$ with $p>1$ Apr 10 at 12:44
• Jacob Körner, +1, though. Thanks! Apr 10 at 13:10
• thanks, that's true Apr 11 at 6:39

The claim as stated is not true. Suppose that $$f$$ can be reduced to a single variable function, i.e $$\exists \phi:\mathbb R_{\geq 0}\to\mathbb R$$ such that $$f(x)=\phi(|x|).$$ Now see that $$\Vert f\Vert^2_{L^2(\mathbb R)}=\int_{\mathbb R^2}|f(x)|^2\mathrm d^2 x=\int_0^\infty \int_{\partial\mathbb B(0,r)}|f(x)|^2\mathrm d^1x~\mathrm dr \\ =\int_0^\infty {\phi(r)}^2~2\pi r~\mathrm dr<\infty$$ This means that the function $$\psi(r)={\phi(r)}^22\pi r$$ satisfies $$\psi\in L^1(\mathbb R_{\geq 0})$$.

Now suppose for instance $$y=0$$ and consider your integral in question:

$$\int_{\mathbb R^2}|f(x)|^2 |x-y|^{-a}\mathrm d^2 x=\int_{\mathbb R^2}|f(x)|^2 |x|^{-a}\mathrm d^2 x=\int_0^\infty \int_{\partial\mathbb B(0,r)}|f(x)|^2|x|^{-a}~\mathrm d^1 x~\mathrm dr \\ =\int_0^\infty 2\pi r~{\phi(r)}^2~r^{-a}\mathrm dr \\ =\int_0^\infty \psi(r)~r^{-a}\mathrm dr$$

So the question is now, can we find a function $$\psi:\mathbb R_{\geq 0}\to\mathbb R$$ such that $$\psi\in L^1(\mathbb R_{\geq 0})$$ but the function $$r\mapsto \psi(r)r^{-a}$$ is not $$L^1$$, for some $$a\in(1,2)$$? The answer is of course, yes. Consider $$a=3/2$$ and

## $$\psi(r)=\mathrm e^{-r}$$

You can construct $$\phi(r)=\sqrt{\frac{\mathrm e^{-r}}{2\pi r}}$$ and see that the associated 2D function $$f(x)=\sqrt{\frac{\mathrm e^{-|x|}}{2\pi|x|}}$$ Is indeed $$L^2(\mathbb R^2)$$, but however that $$\int_{\mathbb R^2}|f(x)|^2|x|^{-a}\mathrm d^2 x$$ does not converge.