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?


$f_n \to f$ 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.

  • $\begingroup$ 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

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