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I have recently learnt about tempered distributions, and how one can define the Fourier transform of a tempered distribution $v$ to be $\hat v$ so that $$\langle\hat v,\varphi\rangle=\langle v,\hat \varphi\rangle.$$ for all Schwartz function $\varphi$. Here, let's say I take the convention that the Fourier transform is $$\hat\varphi(\xi)=\frac{1}{\sqrt{2\pi}}\int_\mathbb R e^{-i x\xi}\varphi(x) \, dx.$$ In particular, one can show that the Fourier transform of $1$ is the Dirac delta $\delta_0$. Here, $1$ is in the sense of tempered distribution, defined by $$\langle 1,\varphi\rangle=\int\varphi(x)\,dx.$$ No problem there.

However, I have come across a confusing usage of this fact: I was reading a proof of some estimate, the author claimed that ($g$ is just some function of $x$)

$$\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}\int_\mathbb{R} e^{-ixt}g(x)\,dt\,dx=\int_\mathbb{R}\delta(x)g(x)dx.$$ It looks like the author took the Fourier transform of $1$ "at the point $x$". But this is confusing to me because I don't understand what it means to take Fourier transform "at a point". I also don't understand what $\delta(x)$ means, and what it means to "integrate" it.

If the context is helpful, I am reading this set of seminar notes. https://www.math.ucla.edu/~visan/Oberwolfach2012.pdf Namely, page 11 Lemma 2.11 (Local Smoothing). I simplified the notation above the make the question less complicated.

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    $\begingroup$ $$\langle 1,\langle e^{-ixt},g(x) \rangle \rangle=\langle \langle 1,e^{-ixt} \rangle ,g(x) \rangle=\langle \delta(x),g(x)\rangle=g(0)$$ $\endgroup$
    – Mark Viola
    May 19, 2016 at 19:42
  • $\begingroup$ One way to understand the formula is by taking the Fourier-transform of $g$ and then taking the inverse transform (to get $g$ back) giving us: $$g(x) = \frac{1}{\sqrt{2\pi}}\iint g(x')e^{ik(x-x')}{\rm d}x'{\rm d}k$$ Now taking $x=0$ gives us $g(0) = $ your first integral. The defining relation of the $\delta$-function is that $\int g(x)\delta(x){\rm d}x = g(0)$ for all test-function $g$ so your last integral is also just $g(0)$. $\endgroup$
    – Winther
    May 19, 2016 at 19:44
  • $\begingroup$ You can think of it as notation: $\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{-ikx} d(x) dx$ is to be understood as the Fourier transform of the distribution $d$ while $\int_{-\infty}^\infty f(x) d(x) dx$ is to be understood as $d(f)$ if $d$ is a test function. Now what you have written can be understood as $\int_{-\infty}^\infty g(x) \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{-ixt} \cdot 1 dt dx$. So the inside is the distributional Fourier transform of $1$ and you go from there. $\endgroup$
    – Ian
    May 19, 2016 at 19:46

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Why so complicated? The inner integral is FT(g)(-t) and so we have to look at the FT(FT(g))(x) for x = 0, but FT(FT(g))(x) = g(-x) for test functions.

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Take g outside the inner integral on the left and you have the fourier transform of 1. This is the dirac delta, a distribution that integrates to 1 and is infinity at x. In this sense only the value of g where the dirac delta is non-zero becomes relevant. The point is then moving with x in your case.

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    $\begingroup$ It might be that the OP is looking for some justification for such manipulations with distributions (generalized functions) like the Dirac delta. $\endgroup$
    – hardmath
    May 19, 2016 at 21:11
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Actually it was (as you can check in the lecture notes you've referenced)

$$\frac 1 {2\pi} \int_{\mathbb R} \int_{\mathbb R} e^{-ixt} g(x)\ dt\ dx.$$ If you integrate by $t$ first then it can be rewritten as (I'll assume below that integration interval is always $\mathbb R$) $$ \int dx\ g(x) \left(\frac 1 {\sqrt{2\pi}} \int \frac{1}{\sqrt{2\pi}} e^{-ixt}\ dt \right) $$ Delta function is a generalized function that can be (roughly) defined as such function that for any $f(x)$ $$\int dx\ f(x)\ \delta(x) = f(0),$$ so (inverse in your convention) Fourier transform of it is $$\hat{\delta}(t) = \frac{1}{\sqrt {2\pi}}\int dx\ e^{ixt\ }\delta(x) = \frac{1}{\sqrt{2\pi}}$$ and from this follows that $$\frac 1 {\sqrt{2\pi}} \int \frac{1}{\sqrt{2\pi}} e^{-ixt}\ dt = \frac 1 {\sqrt{2\pi}} \int \hat{\delta}(t)\ e^{-ixt}\ dt = \delta(x).$$ So $$\int dx\ g(x) \left(\frac 1 {\sqrt{2\pi}} \int \frac{1}{\sqrt{2\pi}} e^{-ixt}\ dt \right) = \int dx\ g(x)\ \delta(x) = f(0).$$

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