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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?

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  • $\begingroup$ "I'm solving a very hard..." A random MSE user: And I have taken that personally $\endgroup$ Apr 10 at 10:22
  • $\begingroup$ 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. $\endgroup$ Apr 10 at 10:31
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    $\begingroup$ $|x-y|^a$ clearly blows up whenever $x=y$. Do you have any good reasons to believe that the bound you're looking for is true? $\endgroup$
    – Oscar
    Apr 10 at 11:16
  • $\begingroup$ 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. $\endgroup$ Apr 10 at 11:29
  • $\begingroup$ 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. $\endgroup$
    – whpowell96
    Apr 10 at 14:47

3 Answers 3

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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.

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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 $n=2$): $$ 1/p + a/4 = 1 + 1/2. $$

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    $\begingroup$ 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 $p>1$ $\endgroup$ Apr 10 at 12:44
  • $\begingroup$ Jacob Körner, +1, though. Thanks! $\endgroup$ Apr 10 at 13:10
  • $\begingroup$ thanks, that's true $\endgroup$ Apr 11 at 6:39
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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.

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