# Fourier transform - Poisson Equation - Exercise

I have some troubles concerning the following question:

Let $\phi \in C_c^\infty(\mathbb{R^n})$. Prove using the Fourier transform that the Poisson equation $\Delta f=\phi$ has at most one solution $f$ that satisfies in addition $f\in S(\mathbb{R^n})$. Show that such a solution does not exist if $$\int_{\mathbb{R^n}}\phi(x)dx\neq0.$$

My problem is that I do not know how to start, I think $f$ is only twice continuously differentiable, so how to proceed. I have seen that any constant differential operator is injective from $C_c^\infty(\mathbb{R^n})$ to $C_c^\infty(\mathbb{R^n})$ (and from $S(\mathbb{R^n})$ to $S(\mathbb{R^n})$), but here I do not have sufficient information on $f$.

EDIT: Show that such a solution does not exist if $\int_{\mathbb{R^n}}\phi(x)dx\neq0.$, I was able to proof this. MY QUESTION, since I have some troubles with english..., do we assume that $f$ is a Schwartz function, or do I have to prove it? Thanks in advance.

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I think you need some boundary condition to get unicity. – Tomás Jan 29 '13 at 19:30
i added some more information, it should work without boundary conditions in this case – Mathoman Jan 29 '13 at 20:17
Can you please put your partial solution to this problem, so I can see what you know and maybe I can help you better. – Tomás Jan 29 '13 at 22:08
Suppose to have a solution $f$, then by taking the fourier on both sides you obtain $-|x|^2\hat{f}(x)=\hat{\phi}(x)$, evaluating this at $x=0$, you obtain the second part! My question is if in the problem, do I we assume f a Schwartz function or do we have to proof it? – Mathoman Jan 29 '13 at 22:22
You have to show that if $f$ exists then it is Schwartz. I dont know how to do this. I only know that the solution of you problem is given by $f=\phi\star E$, where $E$ is the fundamental solution of the laplacian. – Tomás Jan 30 '13 at 10:31

[Edited] We assume that $f$ is in the Schwartz class. Otherwise the problem statement would not false. For example, the equation $\Delta f=0$ has infinitely many solutions (harmonic functions) such as $f(x)=x_1$, $f(x)=2x_1$, etc... But if we assume that $f\in S(\mathbb R^n)$, then $f=0$ is the only solution.
The Fourier transform (not this one) converts the equation (under the assumption $f\in S(\mathbb R^n)$) into $|\xi|^2\hat f(\xi)=\hat \phi(\xi)$. This has no solutions when $\hat\phi(0)=0$, i.e., when $\int_{\mathbb R^n} \phi\ne 0$.
I have to use the Fourier transform, so I do not think that this helps, even if it is correct. Moreover, I think I have to show that $f$ is a Schwartz function, or is it an assumption) (I have some troubles with English) – Mathoman Jan 29 '13 at 17:39
Im confused now. If $f$ satisfies $\Delta f=\Phi$, is $f$ Schwartz? – Tomás Feb 2 '13 at 13:28
@Tomás No. I gave an example of linear $f$. Except for the zero function, harmonic functions are not in the Schwartz class. For other $\Phi$, you can take one solution and add a harmonic function to it. – user53153 Feb 2 '13 at 16:22