# $L^2$ functions with compactly supported Fourier transforms form a Hilbert space

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.
• 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. – Kabo Murphy May 9 at 7:20