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In Stein's Fourier Analysis, there's an exercise:

The function $e^{-\pi x^2}$ is its own Fourier transform. Generate other functions [presumably in the Schwartz space $S(\mathbb{R})$] that, up to a constant multiple, are their own FTs. What must the constant multiples be? To decide this, prove that $F^4 = I$ where $F$ is the FT operator.

My problem is that I don't really understand what the question is asking. Is it asking us to find the class of all functions such that $F(f) = cf$ for constant c and prove the above identity, or is it asking us to just find other examples? Or, is the $F^4 = I$ identity for any $f \in S(\mathbb{R})$?

Finding every function would mean finding all $f$ such that $$f(\xi) = \int^{\infty}_{-\infty} f(x)e^{-2\pi i x \xi}dx$$ which seems rather difficult.

I can come up with a specific example: if I can find a function such that $\hat{f}(\xi) = 1$, I can just take $g = f \ast K_{\delta}(x)$ where $K_{\delta}$ is the family of Gaussian functions as approximations of the identity.

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  • $\begingroup$ You mean $f(\xi) = ...$ $\endgroup$ – Robert Israel Mar 27 '14 at 6:49
  • $\begingroup$ @RobertIsrael Yes, edited - thanks. $\endgroup$ – Lost Mar 27 '14 at 6:57
  • $\begingroup$ 1/cosh(x) works also :) $\endgroup$ – mathworker21 Dec 10 '16 at 23:05
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Because Fourier transform is a linear transformation in the Hilbert space of square-normalizable curves $L^2$, this is essentially an eigenvalue problem. You are looking for a function that doesn't change under Fourier transform. Beside the Gaussian function, which is very famously an eigenfunction of the Fourier transform, there is a whole class of them that also satisfy this eigenvalue condition of $Ff=cf$. They are essentially a Gaussian times a Hermite polynomial:

http://en.wikipedia.org/wiki/Hermite_functions#Hermite_functions_as_eigenfunctions_of_the_Fourier_transform

On the other hand, $F^4=I$ is universal. If you observe the Fourier transform integral, you can see that the Fourier transform and inverse Fourier transform have an opposite sign in the imaginary exponent. This is why $F$ is not exactly $F^{-1}$ but it's close. $F^2$ flips the function (reverses time) because of this mismatch in sign, but reversing time twice brings you back again, which is why $F^4=I$ works for any function, not just the eigenfunctions. This is related to $i^4=1$: $F^4=I$ tells you that the only eigenvalues of $F$ are $c\in\{\pm 1, \pm i\}$ (try $F^4 f=F^3 (cf)=c^4 f=f$ which gives you $c^4=1$). Of course, this means, that the Fourier space is highly degenerate. A linear combination of any number of eigenfunctions with the same eigenvalue is also an eigenfunction. You can use the Hermite functions (they sequentially cycle through eigenvalues $i^n$, so take every fourth) as a basis spanning each of the four subspaces: you get an infinite family of eigenfunctions for each eigenvalue.

Be warned that this depends on the normalization of the Fourier transform. You need the unitary version. If you take the regular definition (the one used in physics and engineering), you get additional factors of $\sqrt{2\pi}^n$, which the transform blows up or extinguishes the amplitude of the function, and is not cyclic anymore.


You will notice that you can split any function into 4 components with eigenvalues $\{1,i,-1,-i\}$ by doing this:

$$\frac{1}{4}(1+F+F^2+F^3)f=f_1$$ $$\frac{1}{4}(1-iF-F^2+iF^3)f=f_i$$ $$\frac{1}{4}(1-F+F^2-F^3)f=f_{-1}$$ $$\frac{1}{4}(1+iF-F^2-iF^3)f=f_{-i}$$

where $$Ff_1=f_1\quad Ff_i=if_i\quad Ff_{-1}=-f_{-1}\quad Ff_{-i}=-if_{-i}$$

Each of these components has a very specific symmetry. In particular, the $\pm i$ are odd and $\pm 1$ are even.

You can take any arbitrary function and use it to generate a function that is preserved under Fourier transform. This I think is the most general answer I can give you.

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    $\begingroup$ A lot more in-depth than the book - thanks for the great answer. $\endgroup$ – Lost Mar 27 '14 at 8:09
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The function $ f(\bf x) = \frac{1}{\sqrt{|x|}}$ is also its own Fourier Transform:

$$\mathcal{F}(f(x)) = \int^{\infty}_{-\infty} \frac{1}{\sqrt{|x|}}e^{-2\pi i x \xi}dx = \frac{1}{\sqrt{|\xi|}} = f(\xi) $$

Although $f(x)$:

  • is not continuous and defined at $ x = 0 $
  • is not integrable over $<-\infty , +\infty>$

Proof:

Let: $$g(\xi) = \mathcal{F}(f(x)) = \int^{\infty}_{-\infty} \frac{1}{\sqrt{|x|}}e^{-2\pi i x \xi}dx $$
Since $f(x) = \frac{1}{\sqrt{|x|}}$ is even (and real), $g(\xi)$ must be even and real. So to avoid problems with the absolute value, I'll work with $x > 0$ and $\xi > 0 $

$$g(\xi) = 2\int^{\infty}_{0} \frac{1}{\sqrt{x}}\cos(2\pi x \xi)dx \,\,\,\,\,\, \text{ for } \,\,\,\,\,\, \xi > 0 $$

Substitute: $$u =2\pi x \xi \implies \sqrt{x}=\sqrt{\frac{u}{2\pi \xi}} \implies dx = \frac{du}{2\pi \xi}$$

to get: $$g(\xi) = 2\int^{\infty}_{0} \frac{\sqrt{2\pi\xi}}{\sqrt{u}}\cos(u)\frac{du}{2\pi \xi} $$
$$g(\xi) = \frac{2}{\sqrt{2\pi \xi}}\int^{\infty}_{0} \frac{1}{\sqrt{u}}\cos(u)du $$
Substitute: $$ u = t^{2} \implies \sqrt{u}=t\implies du=2tdt $$ $$g(\xi) = \sqrt{\frac{2}{\pi}}\frac{1}{\sqrt{\xi}} \int^{\infty}_{0} \frac{1}{t}\cos(t^{2})2tdt $$
in order to get the $\color{blue}{\text{Fresnel integral}}$ : $$g(\xi) = 2\sqrt{\frac{2}{\pi}}\frac{1}{\sqrt{\xi}}\color{blue}{\int^{\infty}_{0} \cos(t^{2})dt} $$
$$g(\xi) = \sqrt{\frac{8}{\pi}}\frac{1}{\sqrt{\xi}}\color{blue}{\sqrt{\frac{\pi}{8}}} = \frac{1}{\sqrt{\xi}}$$

Now we use that $g(\xi)$ is even to expand its domain to $\xi \neq 0 $:

$$g(\xi) =\frac{1}{\sqrt{|\xi|}} = f(\xi)\,\,\,\,\,\,\,\,\,\, \blacksquare$$

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    $\begingroup$ I guess I'm not convinced that the Fourier transform of $f(x) = 1/\sqrt{|x|}$ can be legitimately calculated using the integral definition, when $f$ itself is not integrable. It seems that your change of variables $u = 2\pi i x \xi$ has introduced a decay $e^{-u}$ where there was none before. But has it really? Why are the limits of integration still $0$ to $\infty$ after the variable change? $\endgroup$ – Bungo Dec 9 '15 at 0:12
  • $\begingroup$ Hi Bungo, I have reworked it, instead of the complex FT, I use cosines only, and then make use of the Fresnel integral. There is still a problem with g(0), however, there is a symmetry with f(0), so maybe not perfect, but $\endgroup$ – Job Bouwman Dec 9 '15 at 0:55
  • $\begingroup$ Maybe it is better to derive it as a limiting case: $$g(\xi) = \lim_{A \downarrow 0} f(x) \int^{\infty}_{-\infty} \frac{1}{\sqrt{|x|+A}}e^{-2\pi i x \xi}dx $$ but I don't know at this moment if that's valid or how to proceed. Cheers! $\endgroup$ – Job Bouwman Dec 9 '15 at 1:06
  • $\begingroup$ I don't dispute the result - e.g. it appears as item 311 on the Wikipedia page for Fourier transforms. It's in the distributions section, so I'm guessing that some fiddly distribution-style calculation is required to do it rigorously (as opposed to a straightforward integration). It's certainly an interesting result in any case, one I haven't seen before. $\endgroup$ – Bungo Dec 9 '15 at 1:19
  • $\begingroup$ P.S. Your calculation using the Fresnel integral seems reasonably persuasive. I doubt there is any method which can assign a meaningful value for $g(0)$ (other than $+\infty$), so I don't view that as a flaw. $\endgroup$ – Bungo Dec 9 '15 at 1:35
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Hint: for any $f$, take a linear combination of $f$ and $F(f)$ and ...

Another hint for proving $F^4 = I$: Fourier inversion formula.

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  • $\begingroup$ To find a specific function, or for in general? $\endgroup$ – Lost Mar 27 '14 at 6:50
  • $\begingroup$ It works for any $f \in S({\mathbb R})$, (or in $L^2({\mathbb R})$ for that matter), once you've proven $F^4 = I$. $\endgroup$ – Robert Israel Mar 27 '14 at 6:51
  • $\begingroup$ My main issue is that I'm a bit lost with the flow of the problem. Am I supposed to prove that $F^4 = I$ first? $\endgroup$ – Lost Mar 27 '14 at 6:53
  • $\begingroup$ You could do that, or you could try to come up with examples first. $\endgroup$ – Robert Israel Mar 27 '14 at 6:54
  • $\begingroup$ Will that imply the existence of other functions that are invariant under the FT? How about determining the constant, as the question seems to ask? $\endgroup$ – Lost Mar 27 '14 at 6:54
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Not answer, only a recommendation:

Don't look for

$$f(x) = \int^{\infty}_{-\infty} f(x)e^{-2\pi i x \xi}dx$$

but (firstly) for

$$1 = \int^{\infty}_{-\infty} \int^{\infty}_{-\infty} f(x)e^{-2\pi i y \xi}e^{-2\pi i x \xi}dydx$$ which would be $F^2 = I$

And I guess there are also f(x) with

$$-1 = \int^{\infty}_{-\infty} \int^{\infty}_{-\infty} f(x)e^{-2\pi i y \xi}e^{-2\pi i x \xi}dydx$$ which would result in $F^4 = I$

At least I don't see another reason to use $F^4 = I$ instead of $F^2 = I$

Have a try

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    $\begingroup$ The reason to use $F^4$ rather than $F^2$ is that the statement $F^4 = I$ is true while $F^2 = I$ is false. That is, for every $f \in S({\mathbb R})$, $F^4(f) = f$, but there are some $f$ for which $F^2(f) \ne f$. $\endgroup$ – Robert Israel Mar 27 '14 at 6:48
  • $\begingroup$ But isn't the statement, a fouriertransformed of a function is the function itself not synonymous to $F^2 = I$. I understand your hint answer (really clever, by the way ;-)), but wouldn't the resulting answer be wrong as well, if $F^2$ wasn't $I$? $\endgroup$ – Lord_Gestalter Mar 27 '14 at 6:54
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    $\begingroup$ For some functions, $F(f) = f$. Then $F^2(f) = f$ also. But not all functions have this property. $\endgroup$ – Robert Israel Mar 27 '14 at 6:58
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Another example: the function $\frac{1}{\cosh(\pi x)}$ is its own Fourier transform, ie: $$\frac{1}{\cosh(\pi \xi)}=\int_{-\infty}^\infty e^{-2\pi i x \xi}\frac{1}{\cosh(\pi x)}dx$$

(Possibly off-topic, but since Job Bouwman posted an example I figured I'd do the same for future visitors!)

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Observe that, Fourier inversion formula gives us, $\widehat{(\hat{f})}(x) =f(-x)$; and temporally put, $g(x)= f(-x)$; again by Fourier inversion formula, $\widehat{(\hat{g})}(x) =g(-x)$ and clearly, $g(-x)= f(x).$

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