Fourier transform in $L^2$

Question

I have a function $\phi\in L^2(\mathbb{R}^3)$ and I know that its Fourier transform satisfies the following equation: $$(p^2-A)\hat{\phi}(p)=Q\frac{A+\lambda}{p^2+\lambda}$$ where $Q$ is a constant, $A\geq 0$ and $\lambda>0$. Can I conclude that $\hat{\phi}$ cannot be in $L^2$ unless it is identically zero?

Answer 1

Note that $\hat{\phi}(p)=Q\frac{A+\lambda}{p^4-\lambda}$ almost everywhere, hence

\begin{eqnarray} \int_{\mathbb{R}^3}|\hat{\phi}(p)|^2dp &=& \int_{\mathbb{R}^3}\Big|Q\frac{A+\lambda}{p^4-\lambda}\Big|^2 dp \nonumber \\ &=& Q^2(A+\lambda)^2\int_0^{\infty}\int_{S^1}\frac{r^2}{|r^4-\lambda|^2}drd\sigma \nonumber \end{eqnarray}

where $S^1$ is the sphere with center in the origin and radius 1, and $d\sigma$ is the volume element of the sphere (note that I did the transformation $p=r\sigma$ with $\sigma\in S^1$). The last integral is finite if and only if your function is identically zero. This case occurs when $Q=0$.