Sufficient condition for bounded Fourier transform

[mnb96]

Hello,

Let's suppose we are given a function f:\mathbb{R}\rightarrow \mathbb{R}, and we assume its Fourier transform F=\mathcal{F}(f) exists and has compact support.

What sufficient condition could we impose on f, in order to be sure that F is also bounded?

 

[mathman]

f is L1 is certainly sufficient, even if F does not have compact support.

 

[mnb96]

Thanks a lot!
I think I figured out a way to prove it, but it relies on the Fourier shift theorem and the compact support assumption.
The proof I sketched out is more or less like this:

- Let's suppose F is not bounded and has compact support
- Then there must be at least one point in the support of F where the Fourier transform goes to infinity.
- Let u be that point, and translate the Fourier transform so that u coincides with the origin.
- This means that the inverse FT of the translated F must have integral=infinity.
- But from the Fourier shift-theorem we know that such a function is simply f itself multplied by the complex sinusoid e^-iux
- Note that |\int_{-\infty}^{+\infty}f(x)e^{-iux}dx| \leq \int_{-\infty}^{+\infty}|f(x)|dx which means that f is not absolute integrable.
- Now if we assumed f was in L1 we would reach a contradiction

However I don't see how to generalize this argument to the case when F does not have compact support, e.g. F(u) might go to infinity only for u-->infinity which would render the "translation trick" useless.

 

[mathman]

I don't understand your argument. If f is in L1 then, as you noted, the Fourier transform is bounded at every point by the integral of |f(x)|.

 

[mnb96]

Yes. I have just realized that I in my proof I don't need the statement "there must be at least one point in the support of F where the Fourier transform goes to infinity". The inequality that I wrote simply says that the magnitude of the F at each point is bounded by the absolute integral of f.
Thanks again for clarifying this.

 

[WWGD]

Thanks a lot!
I think I figured out a way to prove it, but it relies on the Fourier shift theorem and the compact support assumption.
The proof I sketched out is more or less like this:

- Let's suppose F is not bounded and has compact support
- Then there must be at least one point in the support of F where the Fourier transform goes to infinity.
- Let u be that point, and translate the Fourier transform so that u coincides with the origin.
- This means that the inverse FT of the translated F must have integral=infinity.
- But from the Fourier shift-theorem we know that such a function is simply f itself multplied by the complex sinusoid e^-iux
- Note that |\int_{-\infty}^{+\infty}f(x)e^{-iux}dx| \leq \int_{-\infty}^{+\infty}|f(x)|dx which means that f is not absolute integrable.
- Now if we assumed f was in L1 we would reach a contradiction

However I don't see how to generalize this argument to the case when F does not have compact support, e.g. F(u) might go to infinity only for u-->infinity which would render the "translation trick" useless.

Cant we use a continuity argument together with compactness to argue boundedness of the transform?

 

[mathman]

Yes. I have just realized that I in my proof I don't need the statement "there must be at least one point in the support of F where the Fourier transform goes to infinity". The inequality that I wrote simply says that the magnitude of the F at each point is bounded by the absolute integral of f.
Thanks again for clarifying this.

If you look at the reverse question. F bounded with compact support implies f is L1.