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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 $R \rightarrow \infty$.

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  • $\begingroup$ 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, 2019 at 7:02

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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 $C$.

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