# The behavior of a potential-like integral at infinity

• I
• wrobel
Thank you for this conversation. In summary, the conversation discusses a problem involving a function in certain spaces and a number, and attempts to prove a limit involving an integral. One approach involves using the Uniform Boundedness Principle, but the speaker is unsure if this is a valid method. The other speaker suggests using the Banach-Steinhaus theorem and a dense set of compactly supported functions. There is also a discussion about a potential counterexample involving an alternating function.

#### wrobel

I need a help in the following problem. I feel that the question is stupid.
Take a function ##f\in C(\mathbb{R}^3)\cap L^1(\mathbb{R}^3)## and a number ##\alpha\in(0,3)##.
Prove that
$$\lim_{|x|\to\infty}\int_{\mathbb{R}^3}\frac{f(y)dy}{|x-y|^\alpha}=0.$$
I can prove this fact by the Uniform Boundedness Principle only. This frustrates me much.

Maybe I do not take the problem properly, the integral is rewritten as
$$\int |t|^{-\alpha}f(t+x)dt$$
So for your expectation we should require in the limit of large |x|
$$f(x) \rightarrow 0$$
and the effect of divergence of ##|t|^{-\alpha}## at t=0 should be investigated.

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wrobel
It seems I do not know how to prove it. Perhaps it is not true.

wrobel said:
I feel that the question is stupid.
Why say this?

Klystron
anuttarasammyak said:
So for your expectation we should require in the limit of large |x|
f(x)→0
yes that is one of possible ways to save the matter

I have no time at the moment to search myself, but the keyword to look for is Newton potantial.

the correct assertion is as follows.
Take a function $$g\in L^p(\mathbb{R}^3) \cap L^1(\mathbb{R}^3)$$ and a number $$\alpha\in(0,3),\quad p>\frac{3}{3-\alpha}.$$
Then one has
$$\lim_{|x|\to\infty}\int_{\mathbb{R}^3}\frac{g(y)dy}{|x-y|^\alpha}=0.$$
The question is the same: is there a direct proof? Besides the Uniform Boundedness Principle

Would you mind showing us the proof? How do you get rid of the singularity at ##y=x##?

fresh_42 said:
Would you mind showing us the proof? How do you get rid of the singularity at ##y=x##?

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Why is
$$\int_{B_1}\dfrac{d\xi}{|\xi|^{q\alpha }} < \infty$$
and wouldn't we need ##|f(y)|## or ##|g(y)|##, too, which makes the ##u_j## non-linear?

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fresh_42 said:
Why is
$$\int_{B_1}\dfrac{d\xi}{|\xi|^{q\alpha }} < \infty$$
and wouldn't we need ##|f(y)|## or ##|g(y)|##, too, which makes the ##u_j## non-linear?
1) The integral in the spherical coordinates takes the form
$$\int F(\psi,\theta)d\psi d\theta \int_{0}\frac{r^2}{r^{q\alpha}}dr,$$
and employ the inequalities from the beginning of the text.

2) for what?

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wrobel said:
1) The integral in the spherical coordinates takes the form
$$\int F(\psi,\theta)d\psi d\theta \int_{0}\frac{r^2}{r^{q\alpha}}dr,$$
and employ the inequalities from the beginning of the text.

2) for what?
Sorry, but I do not see this. What if ##\alpha =1\, , \,p=q=2##?
$$\int_{B_1}\dfrac{d\xi}{|\xi|^2}=\int_0^1\int_0^{\pi}\int_0^{2\pi}\dfrac{1}{r^2}\,d\theta\,d\psi\,dr=2\pi^2\int_0^1\dfrac{dr}{r^2}$$
I think I could construct a counterexample for a function ##g(x\cdot y)## but we have ##g(x+y),## so I'm not sure whether I can use it.

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$$dx\wedge dy\wedge dz=r^2\sin\theta dr\wedge d\theta\wedge d\psi$$
do not forget ##r^2##

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fresh_42
The parameters are chosen really tight. Your handling of the singularity - catching it in a compact ball - still looks a bit like cheating. But before I get too deep into it, how do you finally conclude? I see that you get from Banach-Steinhaus that ##\|u_j\| <\infty ##. How does this prove the theorem, esp. where does the norm go?

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I use the following version of Banach-Steinhaus (due to Laurent Schwartz Analyse Mathematique. vol 2 Hermann 1967.

Let ##X## be a Banach space and let ##X_0\subset X## be a dense set. Assume that ##u_k:X\to\mathbb{R}## is a uniformly bounded sequence of linear functions such that ##u_j(x)\to 0## for any ##x\in X_0##. Then ##u_j(x)\to 0## for any ##x\in X##.

The assertion is valid for the given function ##g\in X## (see the manuscript) as long as it is valid for the dense set ##X_0## of compacly supported functions.

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Sorry, if it annoys you that I try to understand the details since your question was a different one, so what I'm asking here is strictly off-topic.

Just an unimportant question: Where did you use ##p>3/(3-\alpha )##. I get that ##p>0 ## is sufficient for ##2-q\alpha >-1##. (Do you understand papers in German. I have nice slices on that subject. It's what I use to construct a counterexample. I mean if ##g(y)=1## we still have that singularity somewhere.)

See
https://www.wolframalpha.com/input?...+(1/((x-10)^2+(y-10)^2+(z-10)^2))dz)dy)dx++==

The limit is outside of the integral, not inside, so we don't have a vanishing mass at infinity.

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From the conditions of the theorem there must be ##g\in L^1(\mathbb{R}^3)##. For ##g=1## it is not so. Thus it is not a counterexample.
Unfortunately I do not speak German.

Post #2 in polar coordintes of ##\mathbf{t}(r,\theta,\phi),##
$$\int\int\int |t|^{-\alpha}f(t+x)d^3t=\int_0^\infty r^{2-\alpha} dr \int_0^\pi sin\theta d\theta \int_0^{2\pi} d\phi \ \ f(\mathbf{t}+\mathbf{x})$$
When f(x) has a finite support, the integral is zero for large |x|.

anuttarasammyak said:
When f(x) has a finite support, the integral is zero for large |x|.
This is exactly what I do not see. The radius covers the entire space, so the integral matches the singularity at some point, regardless of how far away it is. The limit is always taken afterward! So I imagine an "alternating" function (to bypass ##L^1##), something in the realms of ##f(x,y)=x |x| \exp(-(xy)^2).## I know this isn't a counterexample, it is only my template to find one.

wrobel said:
From the conditions of the theorem there must be ##g\in L^1(\mathbb{R}^3)##. For ##g=1## it is not so. Thus it is not a counterexample.
Unfortunately I do not speak German.
Yes, but increase the power in the denominator and make ##g(y) \sim 1/|y|^\beta ## and we're back in business. I am not saying it is wrong. I simply do not see it. But I'm more an algebraic person than a topological.

There is still the magic trick with your compact balls to lock out the singularity, and the fact that you have ##\lim \int## and not ##\int \lim.##

fresh_42 said:
This is exactly what I do not see. The radius covers the entire space, so the integral matches the singularity at some point, regardless of how far away it is. The limit is always taken afterward!
Say the support does not exceed r=R,
$$|\int\int\int |t|^{-\alpha}f(t+x)d^3t| =|\int_0^\infty r^{2-\alpha} dr \int_0^\pi sin\theta d\theta \int_0^{2\pi} d\phi \ \ f(\mathbf{t}+\mathbf{x})|$$$$\leq |\int_0^R r^{2-\alpha} dr \int_0^\pi sin\theta d\theta \int_0^{2\pi} d\phi |\ \ max(| f(\mathbf{t}+\mathbf{x})|)$$ where maximum is taken within the region of sphere of radius R centered at vector ##\mathbf{x}##.
For large enough |x| of ##|\mathbf{t}+\mathbf{x}|>>R,|\mathbf{t}|<R##, ##max(| f(\mathbf{t}+\mathbf{x})|)=0##. So
$$\int\int\int |t|^{-\alpha}f(t+x)d^3t=0$$
for large enough |x|. But I am not sure whether the original question allows f(x) to have a finite support.

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anuttarasammyak said:
But I am not sure whether the original question allows f(x) to have a finite support.
##f \in L^1## is a strong statement. During my attempts to find a counterexample, I observed that ##\alpha \in (0,3)## and ##f\in L^1## are the crucial conditions here. So whatever "easier" proof there is, it will have to make use of them.