Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.