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As I understand it, Fourier inversion theorem states that, for a Schwartz function $f: \mathbb{R} \to \mathbb{C}$, Fourier transform $$\mathcal{F}f(\omega) = \int_{\mathbb{R}} f(t) e^{i \omega t} dt $$

and its inverse $$\mathcal{F}^{-1}f(t) = \frac{1}{2\pi} \int_{\mathbb{R}} f(\omega) e^{-i \omega t} d\omega$$

the following holds: $$ \mathcal{F}^{-1}\mathcal{F}f(t) = f(t). $$

Yet Fourier transforms of generalized functions like Dirac delta are considered. Is that justified? Why?

Also, what about considering a function $f: \mathbb{C} \to \mathbb{C}$ and taking pointwise limit as $\Im[t] \to 0$? Would a Fourier inversion theorem be true in this limit?

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  • $\begingroup$ Your second question is unclear. What do you mean? If you do not specify, one can give you a trivial answer: if $f$ is defined on $\mathbb{C}$, one can identify $\mathbb{C}$ with $\mathbb{R}^2$ and define the Fourier transform of $f$ to be $$ \hat{f}(\xi+i\eta)=\int_{\mathbb_{R}^2} f(x+iy) e^{i x \xi+i y \eta}\, dxdy.$$ $\endgroup$ Jul 11, 2015 at 17:41
  • $\begingroup$ @GiuseppeNegro Well, my original concern was this: say I know the expression for some Fourier transform $\mathcal{F}g(\omega)$. Is it true that $\lim_{\eta \to 0} \mathcal{F}g(\omega + i\eta)$ = $\mathcal{F}g(\omega)$? $\endgroup$
    – Minethlos
    Jul 11, 2015 at 23:22
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    $\begingroup$ Ok. There are theorems asserting that, if $f$ decays sufficiently fast as infinity, then $\mathcal{F}(f)$ is analytic, at least in a strip containing the real axis. In particular, the property you seek holds. I guess that the condition $$\lvert f(x)\rvert\le C e^{-a\lvert x \rvert}$$ gives analyticity of $\mathcal{F}(f)$ in the strip $$\{z\in \mathbb{C}\ :\ \lvert\Im(z)\rvert < a\},$$ but you should check. $\endgroup$ Jul 11, 2015 at 23:26

2 Answers 2

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Addressing your first question:

Denote the space of Schwartz functions, $S(\mathbb{R})$, and consider the dual space, $S(\mathbb{R})^*$. If $T$ is in $S(\mathbb{R})^*$, then $T: S(\mathbb{R}) \mapsto \mathbb{R}$ is a linear functional. If $\phi \in S(\mathbb{R})$, you often you will see $T$ acting on $\phi$ written as $ \langle T, \phi \rangle$.

Define $\delta$ to be such that:

$$ \langle \delta, \phi \rangle = \phi(0) $$

Clearly, $\delta \in S(\mathbb{R})^*$. Also, note that if $f$ is a locally integrable function, you can define $T_f$ such that:

$$ \langle T_f, \phi \rangle = \int_{\mathbb{R}} f(x) \phi(x)dx $$

and $T_f \in S(\mathbb{R})^*$.


Now then, the Fourier transform on $S(\mathbb{R})^*$ is defined as

$$ \langle FT, \phi \rangle = \langle T, F\phi \rangle $$

Why? Consider when $f: \mathbb{R} \mapsto \mathbb{R}$ and $Ff$ "makes sense". Then:

\begin{align*} \langle FT_f, \phi \rangle &= \int_{\mathbb{R}} \left[ \int_{\mathbb{R}} f(y) \exp(-ixy)dy\right]\phi(x)dx \\ &= \int_{\mathbb{R}} \left[ \int_{\mathbb{R}} \phi(x) \exp(-ixy)dx \right] f(y)dy \\ &= \langle T_f, F\phi \rangle \end{align*}


Consider again the Dirac delta $\delta$ defined above.

\begin{align*} \langle F\delta, \phi \rangle &= \langle \delta , F\phi \rangle \\ &= (F\phi)(0) \\ &= \int_{\mathbb{R}} \phi(x) dx \\ &= \int_{\mathbb{R}} 1 \cdot \phi(x) dx \\ &= \langle T_1, \phi \rangle \end{align*}

So, the Fourier transform of $\delta$ is the linear functional acting on the space $S(\mathbb{R})$ induced by the constant, $1$. Now, here's the cool part that we can't do by treating things in the "normal" way. We can find the Fourier transform of 1:

\begin{align*} \langle FT_1, \phi \rangle &= \langle T_1, F\phi \rangle \\ &= \int_{\mathbb{R}} (F\phi)(x) dx \\ &= 2\pi \frac{1}{2\pi} \int_{\mathbb{R}} (F\phi)(x) \exp(ix\cdot 0) dx \\ &= 2\pi \phi(0) \\ &= \langle 2\pi\delta, \phi \rangle \\ \end{align*}

So $FT_1 = 2\pi \delta$.

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Concerning the second question, here's the (small)${}^{[1]}$ result mentioned in comments. Assume that the function $f\colon \mathbb{R}\to \mathbb{C}$ is integrable on the real line and satisfies the estimate \begin{equation} \left\lvert f(x)\right\rvert\le C e^{-a \lvert x \rvert} \end{equation} where $a>0$. (The value of $C>0$ is not important). Then the Fourier transform $$\tag{1} \mathcal{F}(f)(z)=\int_{-\infty}^\infty f(x) e^{-ix z}\, dx$$ is well-defined and analytic for $z\in S_a$, the latter denoting the strip $$ S_a=\left\{ z\in\mathbb{C}\ :\ \left\lvert \Im z\right\rvert < a\right\}. $$ To prove this one writes $z=\xi+i\eta$ and observes that $$ \left\lvert f(x) e^{-izx}\right\rvert\le Ce^{-\lvert x \rvert (a-\lvert \eta\rvert)}, $$ so the integral in $(1)$ is convergent for $\lvert \eta\rvert < a$. The convergence is also uniform on compact subsets of $S_a$, so it defines an analytic function because $f(x)e^{-ixz}$ is an analytic function of $z$. Indeed, one can decompose $f(x)e^{-ixz}$ into a Taylor series and the locally uniform convergence allows for termwise integration. $\square$

The following example${}^{[2]}$ shows that in general one cannot take a strip bigger than $S_a$: $$ \mathcal{F}\left( \frac{e^{-a\lvert x\rvert}}{2} \right) (z)= \frac1a\frac{1}{1+(z/a)^2}.$$ The right hand side has poles at $\pm ai$. Therefore the biggest strip of analyticity that contains the real axis is $S_a$, which is the one predicted by the result above.


Notes.

${}^{[1]}$ It is indeed the "baby" version of a family of famous results in analysis known as Paley-Wiener's theorems.

${}^{[2]}$ This computation arises when solving the Laplace equation on the half-plane. Look for the keyword "Poisson kernel" for more information. It is also easy enough to compute directly.

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  • $\begingroup$ I think you forgot the exponent in your example. The RHS form of the Poisson kernel is familiar to me, but I can't recognize what the exponent is supposed to be. Also thank you for this exposition - too bad I asked two questions that have different answers and I can't accept them both! $\endgroup$
    – Minethlos
    Jul 16, 2015 at 21:49
  • $\begingroup$ @Minethlos: What exponent? You mean, some factor of $\pi$? $\endgroup$ Jul 16, 2015 at 22:05
  • $\begingroup$ The LHS has brackets with '^' symbol in the exponent... $\endgroup$
    – Minethlos
    Jul 16, 2015 at 22:06
  • $\begingroup$ Ah, you probably meant the Fourier transform. My bad :) $\endgroup$
    – Minethlos
    Jul 16, 2015 at 22:11
  • $\begingroup$ @Minethlos: Yes, exactly. You are right it is not great notation! Editing $\endgroup$ Jul 16, 2015 at 22:31

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