2
$\begingroup$

Recall that if $f$ is a distribution, then its Fourier transform is defined as follows:

$$ \def\R{\mathbb{R}} \langle \hat f ,\phi\rangle:=\langle f ,\hat \phi\rangle, $$ where $\phi$ is a Schwartz function.

I would like to use this definition in order to calculate directly (without using the inverse Fourier transform formula) the Fourier transform of $f\equiv1$. We have $$ \def\R{\mathbb{R}} \langle \hat 1 ,\phi\rangle=\langle 1 ,\hat \phi\rangle=\int_\R\int_\R \exp(i\lambda x)\phi(x)\, dx d\lambda. $$

I know that $\hat 1=\delta(x)$. Thus we need to check that the double integal equals $\phi(0)$. How do we formally check that?

$\endgroup$
3
  • $\begingroup$ In what context will you assign meaning to this double integral? $\endgroup$ Jun 20, 2017 at 17:20
  • $\begingroup$ @uniquesolution The double integral is well defined as a standard Riemann integral. You integrate first with respect to the $x$ variable and then with respect to the $\lambda$ variable. $\endgroup$
    – Oleg
    Jun 20, 2017 at 17:22
  • $\begingroup$ Good question, I should keep this around to ask on an exam some day. $\endgroup$
    – icurays1
    Jun 20, 2017 at 18:18

2 Answers 2

1
$\begingroup$

Proving this essentially amounts to proving the Fourier inversion theorem. To fix conventions, let's define

$$ \hat{f}(\lambda) = \int_{-\infty}^\infty f(x)e^{i\lambda x}dx $$

Now, suppose you can prove "directly" that for any $\phi\in\mathcal{S}$,

$$ \langle \hat{1},\phi\rangle = \langle 1,\hat{\phi}\rangle = 2\pi\phi(0) $$ (You seem to be missing a $2\pi$, but this depends on the transform convention). Then, simply shift $\phi(\cdot)$ to $\phi_x(\cdot) = \phi(\cdot+x)$; then $\phi_x\in\mathcal{S}$, and $\phi_x(0) =\phi(x)$. By the shift theorem for the Fourier transform,

$$ \widehat{\phi_x}(\lambda) = \mathcal{F}[\phi(\cdot+x)](\lambda) = \exp(-i\lambda x)\hat{\phi}(\lambda) $$ and so (by what we've already shown)

$$ 2\pi\phi(x) = 2\pi\phi_x(0)=\langle \hat{1},\phi_x\rangle = \langle 1,\widehat{\phi_x}\rangle = \int \hat{\phi}(\lambda) e^{-i\lambda x}d\lambda $$and thus

$$ \phi(x) = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\phi}(\lambda)e^{-i\lambda x}d\lambda $$

So, long story short, if you want to prove $\langle \hat{1},\phi\rangle \propto\delta$, you'll probably need to apply the same tricks used to prove Fourier inversion, namely convolution with an approximate identity and dominated convergence theorem (see the Wikipedia page on Fourier inversion theorem)

$\endgroup$
0
$\begingroup$

$$ \langle \hat{1},\phi\rangle = \langle 1,\hat{\phi}\rangle = \sqrt{2\pi}(\hat{\phi})^{\vee}|_{x=0} = \sqrt{2\pi}\phi(0) = \sqrt{2\pi}\langle \delta_0,\phi\rangle \\ \hat{1}=\sqrt{2\pi}\delta_0 $$ Normalization constants vary depending on how you define the Fourier transform and its inverse.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .