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I am trying to show that $\left\lbrace \widehat{\phi} : \phi \in C_c^{\infty}(\mathbb{R}) \right\rbrace$ is dense in $L^2(\mathbb{R})$, where $\widehat{\phi}$ is the Fourier transform of the function $\phi$.

I believe it may follow simply from Plancherel's Theorem and using the fact that $C_c^{\infty}(\mathbb{R})$ is dense in $L^2(\mathbb{R})$ but I can't quite seem to put the pieces together

Any pointers would be appreciated!

Joe Tilley
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  • I believe this is enough? https://math.stackexchange.com/questions/784949/is-continuous-image-of-dense-set-dense – Calvin Khor Jan 28 '19 at 11:59
  • The fourier transform induces a bijective isometry of L2 onto itself. Smooth compactly supported functions are dense in L2. Therefore, its image is also dense. – Ayoub Jan 28 '19 at 12:04

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If $f \in L^{2}$ then $\hat {f} \in L^{2}$ so there exists $\{\phi_n\} \subset C_c^{\infty}$ such that $\|\phi_n -\hat {f}\|_2 \to 0$. Since $f \to \hat {f}$ is an isometry and $\hat {\hat{f}}(x)=f(-x)$ we are done.

Kavi Rama Murthy
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