Fourier transform - Poisson Equation - Exercise

Question

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.

Answer 1

[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 $\int_{\mathbb R^n} \phi\ne 0$.