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Q : I need to prove that if $K$ is tempered distribution on $\mathbb{R}$ satisfying: \begin{equation} K = K*e^{-\pi |x|^2} \end{equation} then $K$ is first degree polynomial. mean $K(x) = Ax + b$

Remark: The question was changed. The original was to prove that if $K = K * e^{-\pi |x|^2}$ then $K$ is constant, which is false.

The first thing I did is to apply fourier transform on both sides to work with multiplication instead of convolution. and I got $\hat{K} = e^{-\pi |x|^2} \hat{K}$.

I succeeded to prove $\hat{K}$ is supported at the origin and by theorem 1.7 at page 110, from Stein and Shakarchi functional analysis(Can't find the pdf online) or theorem 6.25 at page 165 from Rudin Functional analysis: \begin{equation} \hat{K} =\sum_{|\alpha| \leq N} a_{\alpha} \partial^{\alpha}\delta \end{equation}.

Now, if I apply the inverse fourier transform I get that $K$ is a polynomial.

The solution will arise if I will prove that if $p$ is a polyomial in $\mathbb{R}^{d}$ satisfying $p*e^{-\pi |x|^2} = p$, then $p$ is constant.

It sounds true(which is not, please see the comments) but I think it is kind of "ugly" to prove and I am pretty sure that there is another way for me to continue.

Hot to continue?

Thanks :)

shestak
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  • So $(e^{-\pi|x|^2}-1)\hat K = 0.$ What does this imply for $\hat K$? – md2perpe Aug 06 '20 at 17:51
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    Is it actually true? If $d=1$ and $K$ is the multiplication operator corresponding to the polynomial $p(x)=x$, then $$(p*e^{-\pi|\cdot|^2})(x)=\int_{\mathbb{R}}(x-y)e^{-\pi y^2},\mathrm{d}y=x=p(x).$$ This also seems to be suggesting that some lengthy or dirty argument is unavoidable. – Sangchul Lee Aug 06 '20 at 17:54
  • @md2perpe I was thinking to continue to work with the fourier transform but, I already proved that it is supported at the origin, and any distribution, $F$ which supported at the origin satisfying that $e^{-\pi |x|^2}F = F$ so I dont think I have something else to do. Also $$e^{-\pi |x|^2} - 1$ is not schwartz function, so I dont think that I it is defined. – shestak Aug 06 '20 at 17:59
  • @SangchulLee It is question from my homework. so I hope it is true :) But thanks to you I know that the proposition I wanted to prove is false. – shestak Aug 06 '20 at 17:59
  • In fact, if $p$ is any harmonic polynomial, then it is invariant under the evolution by heat equation, and so, $p=p*e^{-\pi|\cdot|^2}$ holds as well. I suspect that the converse is also true, but I have no good idea to prove it right now. – Sangchul Lee Aug 06 '20 at 18:28
  • @SangchulLee maybe the main clain is false? – shestak Aug 06 '20 at 18:50
  • I think so. (Most of all, we have counter-examples!) – Sangchul Lee Aug 06 '20 at 18:52
  • In $d=1$ it is possible to show that the most general solution admitted is $K(x)=Ax+B$, which is a trivial version of the polynomials suggested by @SangchulLee . The proof I did is pure brute force and consists of looking at the image of arbitrary degree mononomials under the transformation and showing that ${1,x}$ are eigenvectors of the convolution, while all the others are not and that there exist no linear transformations that can make them eigenvectors of the operator (the eigenspace of eigenvalue 1 is 2-dimensional) which suggests the conclusion. – K. Grammatikos Aug 06 '20 at 20:18
  • @shestak. $e^{-\pi |x|^2} - 1$ is not a Schwartz function, but it is a tempered distribution. – md2perpe Aug 06 '20 at 20:50
  • @DinosaurEgg Last change. Indeed $d = 1$. Thanks. – shestak Aug 07 '20 at 17:52

1 Answers1

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As I mentioned in the comment, the problem is false. In fact, let $p$ be any harmonic polynomial. Then by noting that

$$ \Phi(x,t) = \frac{1}{(4\pi t)^{d/2}}e^{-\frac{|x|^2}{4t}} $$

is the fundamental solution of the heat equation $\partial_t\Phi = \Delta\Phi$, we have

$$ \partial_t(p*\Phi) = p*(\partial_t\Phi)=p*(\Delta\Phi)=(\Delta p)*\Phi=0. $$

Together with $ p(x) = \lim_{t\to 0} (p*\Phi)(x,t) $, this implies that $p = p*\Phi$. Then plugging $t = \frac{1}{4\pi}$ proves

$$ p = p * e^{-\pi|\cdot|^2}. \tag{*} $$

I suspect that the converse is also true, since the condition $\text{(*)}$ implies that $p(x) = (p*\Phi)(x,\frac{n}{4\pi})$ for any integer $n \geq 1$. This surely sounds like another line of interesting question, although I have no good idea to prove this right now.

Sangchul Lee
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  • Thank you for your help! The question was changed to proved that K is liner polynomial(affine) – shestak Aug 07 '20 at 11:18
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    @shestak, it is true that the harmonic polynomials for $d=1$ are polynomials of degree at most one. Even for $d=2$, however, the harmonic polynomials are precisely the real/imaginary parts of complex polynomials in $z=x+iy$, and so, they can assume arbitrarily high degree. For instance, $p(x,y)=x^2-y^2$ works! – Sangchul Lee Aug 07 '20 at 15:23
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    Thanks again. About the inverse, correct me if I am wrong, but if condition $()$ is satisfied, then since p is a polynomial at $x$(does not depend on $t$), then $\Delta_x(p\Phi) = \dotsb=\partial_t(p\Phi) = (\partial_t p) \Phi = 0$. since this is true for all $t$ we can plugging $ t = \frac{1}{4\pi} $ to recieve that $\Delta p = 0$ – shestak Aug 07 '20 at 16:01
  • Last change in the exercise. $ d = 1 $. Thanks – shestak Aug 07 '20 at 17:53
  • @shestak, We cannot conclude that $\partial_t(p*\Phi)$ because $t$ is not the variable involved in convolution. Indeed, if $p$ is not a harmonic polynomial then this expression can change in time. – Sangchul Lee Aug 08 '20 at 08:18