We know that as tempered distributions $S'$ on $\mathbb{R}^3$, the Fourier transform of $1/4\pi|x|$ is $1/k^2$. There are many ways to argue that this is true. One particular way I have seen in physics is that you compute

$$ \int_{|x|<R} \frac{e^{-ikx}}{4\pi|x|} dx= \frac{1}{k^2}[1-\cos{(|k|R)}] $$ Then by taking $R \rightarrow \infty$, we see that the second term $\rightarrow 0$ as a tempered distribution, which does kind of make sense since you expect the cosine term to oscillate so rapidly that it annihilates any Schwartz function. However, could any one provide a rigorous argument of how the oscillations annihilates Schwartz functions?

EDIT: I just realized that $\cos{|k|R}$ acting on a Schwartz function $\phi (k)$ is basically the Fourier transform of the Schwartz function $\phi (k)$ at $R$ (maybe some linear combination or you may need a bound on $|\phi|$). Since the Fourier transform maps Schwartz functions to Schwartz functions, we see that it must $\rightarrow 0$ as $R \rightarrow \infty$.

  • $\begingroup$ One common way you could make the oscillations canceling out rigorous is through "integration by parts" and taking the distributional limit that way. For example, in 1D, $\lim_{k\to \infty} \int e^{ikx}\phi(x)dx = \lim_{k\to \infty} \frac{i}{k}\int e^{ikx} \phi'(x) dx = 0$ by dominated convergence since $\phi$ is a smooth, compactly supported function. But there is another way to compute this Fourier transform which generalizes to higher dimensions which I will type up as an answer. $\endgroup$ Sep 18 '19 at 7:02

Instead of doing that trick, we will use a clever property of the Fourier Transform involving homogeneity.

$\mathbf{\text{Def.}}$ A function $f\in\mathcal{S}(\Bbb R^n)$ is homogeneous of degree $s$ if $\forall a \neq 0$

$$f(ax) = a^sf(x)$$

$\mathbf{\text{Lemma.}}$ Let $f$ be homogeneous degree $s$. Then $\hat{f}$ is homogeneous degree $-n-s$.

Proof: $$ \begin{split} \hat{f}(ak) &= \frac{1}{(2\pi)^{\frac{3}{2}}}\int_{\Bbb R^n}f(x)e^{-i(ak) \cdot x}dx \\ &= \frac{1}{(2\pi)^{\frac{3}{2}}}\int_{\Bbb R^n}f\left(\frac{u}{a}\right)e^{-ik \cdot u}\frac{du}{a^n}\\ &=\frac{1}{(2\pi)^{\frac{3}{2}}}\int_{\Bbb R^n}\left(\frac{1}{a}\right)^sf(u)e^{-ik \cdot u}\frac{du}{a^n} = a^{-n-s}\hat{f}(k) \end{split} $$

Even though we proved this in the Schwartz function case, the same property applies to tempered distributions, albeit with a lot more work.

Notice that $|x|^{-1}$ is radial and homogeneous degree $-1$. Using the above lemma, we have that its Fourier transform should be homogeneous with degree $-3+1 = -2$. The only possible radial homogeneous distribution with that degree is $C|k|^{-2}$ for some constant $C$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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