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I need to prove that the Fourier transform of a $k$-homogeneous distribution $T \in \mathcal{S'}\mathbb{(R^n)}$ is $-(k+n)$-homogeneous.

I'm using Euler's formula $\mathrm{div}(x\,T)=(k+n)\,T$.

We get $\mathcal{F}(\mathrm{div}(x\,T)) =(k+n)\,\mathcal{F}(T)$

Can I use this formal computing without having to go through a test function ?

\begin{aligned} (k+n)\,\mathcal{F}(T)(\omega)&= \mathcal{F}(\mathrm{div}(x\,T))(\omega) \\ &=\mathcal{F}\Big(\sum_{j=1}^{n}\partial_j(x\,T)\Big)(\omega)\\ &=\sum_{j=1}^{n}\mathcal{F}(\partial_j(x\,T))(\omega)\\ &=\sum_{j=1}^{n}(i\omega_j\,\mathcal{F}(x\,T)(\omega) \end{aligned}

Now I know that from here that for a regular function $f$, the Fourier transform of $g(t) = t f(t)$ is $\mathcal{F}g(\omega)=i\,\partial_jf(\omega)$, can we use the same formula for $T$ ? i.e $\mathcal{F}(x\,T)(\omega) = i\frac{\partial T(\omega)}{\partial\omega_j}$

Using this would get me the desired result, but I don't know how to proceed. Thanks.

LL 3.14
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tareqath
  • 107

1 Answers1

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We define the Fourier transform of $f \in L^1$ as

$$\mathscr{F}\{f\}(\vec k)=\int_{\mathbb{R}^n}f(\vec x)e^{i\vec k\cdot \vec x}\,d^n\vec x$$

Now, suppose $T\in \mathbb{S}'$ and $\phi\in \mathbb{S}$. Then, we can write

$$\begin{align} \langle \mathscr{F}\{\vec xT\},\phi\rangle&=\langle T,\vec x\mathscr{F}\{\phi\}\rangle\\\\ &=i\langle T,\mathscr{F}\{\nabla \phi\}\rangle\\\\ &=i\langle \mathscr{F}\{T\},\nabla\phi\rangle\\\\ &\langle - i\nabla\left(\mathscr{F}\{T\}\right),\phi\rangle\\\\ \end{align}$$

So, in distribution we have

$$\mathscr{F}\{\vec xT\}(k)=-i\nabla_{\vec k} \mathscr{F}\{T\}(k)$$

And we are done!

NOTE:

If we had defined the Fourier transform of $f\in L^1$ as

$$\mathscr{F}\{f\}(\vec k)=\int_{\mathbb{R}^n}f(\vec x)e^{-i\vec k\cdot \vec x}\,d^n\vec x$$

then we would have found that

$$\mathscr{F}\{\vec xT\}(k)=i\nabla_{\vec k} \mathscr{F}\{T\}(k)$$

Mark Viola
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