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I am wondering if it is possible to prove the properties of the Fourier transform (such as scaling or differentiation) formally using their action on test functions?

We know that if $f(t)$ is a generalized function, its Fourier transform is also a generalized function whose action on a test function $\varphi$ is

$$\langle F(\nu), \Phi(\nu)\rangle = \langle f(t), \varphi (-t)\rangle.$$

And inverse Fourier transformation is:

$$\langle f(t), \varphi (t)\rangle=\langle F(\nu), \Phi(-\nu)\rangle.$$

My attempt so far:

(i) $f(at) \leftrightarrow \frac{1}{|a|}F(\frac{\nu}{a})$

I started from the definition of distributions

$$\langle f(at), \varphi(t)\rangle = \intop^\infty_{-\infty} f(at)\varphi(t) dt $$

I need to arrive at $\langle \frac{1}{|a|}F(\frac{\nu}{a}), \Phi(-\nu)\rangle$. How can we do this? I played with some manipulations such as substitution $u=at$, but they didn't work.

(ii) $\frac{df(t)}{dt} \leftrightarrow j2\pi \nu F(\nu)$

Just like the previous case, I don't see how I can arrive at the required result:

$$\langle \frac{df(t)}{dt}, \varphi(t)\rangle =^? \langle j2\pi \nu F(\nu), \Phi(-\nu)\rangle$$

I am not sure if this would be useful, but for differentiation we have a property that:

$$\langle f'(t), \varphi(t)\rangle = -\langle f(t), \varphi '(t)\rangle.$$

I have never seen a proof of this in any textbook. Any suggestion would be greatly appreciated.

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  • $\begingroup$ I'm not sure about your definition, the Fourier transform of a distribution in $S'$ is itself a distribution in $S'$ defined by the action $<\hat{u},\phi> = <u,\hat{\phi}>$ $\endgroup$ – plebmatician Jun 7 '18 at 9:47
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For (i), I managed to find:

$$f(at)\Leftrightarrow\frac{1}{|a|}F\left(\frac{\nu}{a}\right)$$

$$\langle F(\nu),\Phi(\nu)\rangle=\langle f(t),\phi(-t)\rangle$$

$$\langle f(at),\phi(-t)\rangle=\frac{1}{|a|}\langle f(t),\phi(\frac{-t}{a})\rangle$$

$$=\frac{1}{|a|}\langle F(\nu),|a|\Phi(a\nu)\rangle$$

$$=\langle F(\frac{\nu}{a}),\phi(\nu)\rangle$$

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  • $\begingroup$ Think to a distribution $T$ as the limit (in the sense of distributions) of a sequence of Schwartz functions $\phi_n$, then the properties of the Fourier transform of Schwartz functions extend naturally to $T$. $\endgroup$ – reuns May 13 at 2:00

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