Struggling with Fourier Transform Eigenvalues?

[Stephen88]

I came across this problem in a book for Fourier Tranf. and after dedicating some time to solve I can't seem to find a solution.
The problem has 2 parts.
a)Show that (i) F^2( f )(x)=2*pi*f(-x) and then deduce that (ii) F^4( f )=4*pi*f where F denotes the Fourier Transform
b) Deduce from (i) that for two eigenvalues, all eigenfunctions must be even, and for other two
eigenvalues, all eigenfunctions must be odd and for (ii) deduce that if f is an eigenfunction of F with eigenvalue λ, then λ must take one of four possible
values.

For a)I've tried to use the Inversion Theorem...it seemed appropriate since there is a Fourier transform on one side and a function f multiplied by pi on the other.Obviously I'm doing something wrong because I get a square function under the integral and I don't know how to handle it.I'm assuming that for (ii) maybe I should use induction?!
For b)To be honest I don't have any idea mainly because I haven't figured out a).
This problem bugs me very much so any help will be appreciated.Thank you

 

[CaptainBlack]

I came across this problem in a book for Fourier Tranf. and after dedicating some time to solve I can't seem to find a solution.
The problem has 2 parts.
a)Show that (i) F^2( f )(x)=2*pi*f(-x) and then deduce that (ii) F^4( f )=4*pi*f where F denotes the Fourier Transform
b) Deduce from (i) that for two eigenvalues, all eigenfunctions must be even, and for other two
eigenvalues, all eigenfunctions must be odd and for (ii) deduce that if f is an eigenfunction of F with eigenvalue λ, then λ must take one of four possible
values.

For a)I've tried to use the Inversion Theorem...it seemed appropriate since there is a Fourier transform on one side and a function f multiplied by pi on the other.Obviously I'm doing something wrong because I get a square function under the integral and I don't know how to handle it.I'm assuming that for (ii) maybe I should use induction?!
For b)To be honest I don't have any idea mainly because I haven't figured out a).
This problem bugs me very much so any help will be appreciated.Thank you

For a) you should note that the constant \(2\pi \) is dependednt on your definition of the Fourier transform, there are a number of conventions about what the variables are and where the constants go.

I will take the following definition (you will need to rework what follows with whatever definition you are working with):

\[ \{\mathfrak{F}f\}(\xi) =\int_{x=-\infty}^{\infty}f(x)e^{-2\pi {\bf{i}} x\xi} \; dx \]

and:

\[ \{ \mathfrak{F}^{-1}F \}(x) = -\int_{\xi=-\infty}^{\infty}F(\xi)e^{2\pi {\bf{i} }x\xi} \; dx \].

Now I will assume that \(f(x)\) is such that the integrals in the forward and backward FT make sense without recourse to the theory of distributions.

Then:

\[ \{ \mathfrak{F}^2 f\} (z)= \int_{y=-\infty}^{\infty}\int_{x=-\infty}^{\infty} f(x)e^{-2\pi {\bf{i}} x y}e^{-2\pi {\bf{i}} y z} dx\; dy \]

Now put \(z=-\zeta \)

\[ \{ \mathfrak{F}^2 f\} (-\zeta)= \int_{y=-\infty}^{\infty}\int_{x=-\infty}^{\infty} f(x)e^{-2\pi {\bf{i}} x y}e^{2\pi {\bf{i}} y \zeta} dx\; dy =\left(\mathfrak{F}^{-1}\left( \mathfrak{F}f\right)\right)(\zeta) =f(\zeta) \]

or:

\[ \{ \mathfrak{F}^2 f\} (z)=f(-z) \]

CB

 

[Stephen88]

Thank you...what should I do for the other points?

 

[CaptainBlack]

Thank you...what should I do for the other points?

The equvalent of ii. with the my definition of the FT follows immeadiatly since:

\[ \mathfrak{F}^4f=\mathfrak{F}^2\left(\mathfrak{F}^2f \right) \]

Which should tell you that the eigenvalues are the fourth roots of unity, then consider the real and imaginary eigenvalues and their corresponding eigenfunctions using a.i.CB

 

[blue_raver22]

Dear reader,

Thank you for sharing your experience with this Fourier Transform problem. It sounds like you have put a lot of effort into trying to solve it and I admire your determination.

Firstly, for part a), it is important to note that the Fourier Transform is defined as follows:

F(f)(x) = ∫ f(t)e^(-2πixt) dt

Using this definition, let's tackle part (i):

F^2(f)(x) = F(F(f))(x)

= ∫ F(f)(t)e^(-2πixt) dt

= ∫ (∫ f(u)e^(-2πitu) du)e^(-2πixt) dt

= ∫ f(u) ∫ e^(-2πi(u+t)x) dt du

= ∫ f(u) δ(u+t) du (applying the inverse Fourier Transform)

= f(-x)

Therefore, F^2(f)(x) = f(-x).

Now, for part (ii), we can use the result from part (i) to find F^4(f):

F^4(f) = F(F^2(f))

= F(f(-x))

= ∫ f(-u)e^(-2πixu) du

= ∫ f(u)e^(-2πix(-u)) du (using the property of even functions)

= ∫ f(u)e^(2πixu) du

= F(f)(-x)

= f(x)

Therefore, F^4(f) = f(x).

For part b), let's use the results from part (i) and (ii):

If F(f) = λf, then F^4(f) = λ^4f. But from part (ii), we know that F^4(f) = 4πf. Therefore, we can equate these two expressions to get:

λ^4f = 4πf

Solving for λ, we get four possible values: λ = ± √(2π) or λ = ± i√(2π).

Now, for the eigenfunctions, we can use the results from part (i) and (ii) to determine their properties. Since F(f) = λf, we know that f is an eigenfunction of F with eigenvalue λ. Therefore, from part (i), f(-x) = λf(x). This means that