0
$\begingroup$

Given a fixed compact subset of $\mathbb{R}$, I want to show that square integrable functions on the real line whose fourier transforms are supported in the given compact set form a Hilbert space in the $L^2$ inner product. It was easy to show that the functions form a subspace. However I am stuck at the closedness of the subspace. Could anyone please help me?

$\endgroup$
2
$\begingroup$

$f_n \to f$ in $L^{2}$ implies $\hat {f_n} \to \hat {f}$ in $L^{2}$ and this implies that some subequence of $(\hat {f_n})$ converges almost everywhere to $\hat {f}$. Hence $\hat {f}$ is also supported by the compact set.

$\endgroup$
  • $\begingroup$ How can one show that the L^2 convergence of the original series implies the L^2 convergence of the fourier transform? I think I saw some similar contents in the past but my memories are faded away now. Could you please explain? $\endgroup$ – Keith May 9 at 6:56
  • $\begingroup$ The map which takes a a function $f \in L^{2}$ to its Fourier transform is an isometry. The very construction of FT for $L^{2}$ goes through this isometry property. $\endgroup$ – Kabo Murphy May 9 at 7:20
  • $\begingroup$ Oh is this construction called the Plancherel theorem according to Folland Real Analysis? $\endgroup$ – Keith May 9 at 7:32
  • 1
    $\begingroup$ Yes, it is called the Plancherel Theorem. $\endgroup$ – Kabo Murphy May 9 at 7:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.