A function not in $L^2(\mathbb{R}^3)$

From this equation $$(p^2-\alpha)\hat{f}(p)=\frac{e^{-ip\cdot y}}{p^2+\lambda}$$ where $\hat{f}$ is the Fourier transform, $\alpha,\lambda>0$ e $y$ a fixed point in $\mathbb{R}^3$ can I conclude that $f$ is in $L^2(\mathbb{R}^3)$ if and only if it is the zero function?

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If I recall correctly, the Fourier transform is an isometric isomorphism on $L_2$, so you can probably just show that $\hat f$ is not in $L_2$. – Elmar Zander Mar 2 '13 at 9:52
You recall correctly – Mario Mar 2 '13 at 9:53
Pheew! However, I don't see anything like that in your conclusion. I don't even see how you come to your proposed conclusion at all. Maybe you can make that a bit clearer in your question. – Elmar Zander Mar 2 '13 at 9:56
@ElmarZander maybe the first part of my previous question was not very clear. The point is that I write now! – Mario Mar 2 '13 at 10:09
Sorry, I don't get it that way, either. Please see my answer. – Elmar Zander Mar 4 '13 at 19:49

What I would do is the following: If you move $p^2-\alpha$ to the right hand side, you have $$\hat{f}(p)=\frac{e^{-ip\cdot y}}{(p^2+\lambda)(p^2-\alpha)}$$ If $\hat f$ is to be in $L_2$, then we need $$\int_{\mathbb R^3}|\hat{f}(p)|^2=\int_{\mathbb R^3} \frac{1}{(p^2+\lambda)^2(p^2-\alpha)^2}<\infty$$ The problem has spherical symmetry and you get \begin{align} \int_{\mathbb R^3} \frac{1}{(p^2+\lambda)^2(p^2-\alpha)^2} dp &= S_2 \int_{\mathbb R} \frac{r^2}{(r^2+\lambda)^2(r^2-\alpha)^2} dr\\ &= S_2 \int_{\mathbb R} \frac{r^2}{(r^2+\lambda)^2(r-\sqrt\alpha)^2 (r+\sqrt\alpha)^2} dr\\ \end{align} where $S_2$ is the surface of the 2-sphere. Obviously, the integrand becomes singular at $r=\sqrt{\alpha}$, and since \begin{align} \int_{\sqrt\alpha-\delta}^{\sqrt\alpha+\delta} \frac{1}{(r-\sqrt\alpha)^2} dr=\infty \end{align} the whole integral cannot be finite.