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Let $f\in C_{0}^{\infty}\left(\mathbb{R}^{n}\right)$. By Parseval's formula we know that $$ \left\Vert \widehat{f}\right\Vert _{L^{2}\left(\mathbb{R}^{n}\right)}=\left\Vert f\right\Vert _{L^{2}\left(\mathbb{R}^{n}\right)}, $$ where $\widehat{f}$ is the Fourier transform of $f$. Moreover, it is well-known that $\widehat{f}\in C\left(\mathbb{R}^{n}\right)$ for any $f\in C_{0}^{\infty}\left(\mathbb{R}^{n}\right)$. Thus $$ \intop_{\mathbb{S}^{n-1}}\left|\widehat{f}\left(x\right)\right|^{2}dx\;\;\text{ and }\intop_{\mathbb{S}^{n-1}}\left|f\left(x\right)\right|^{2}dx $$ make sense for any $f\in C_{0}^{\infty}\left(\mathbb{R}^{n}\right)$. Here $\mathbb{S}^{n-1}$ is the unit sphere, $\mathbb{S}^{n-1}:=\left\{ x\in\mathbb{R}^{n}:\left|x\right|=1\right\} $.

My question is: is it true that $$ \left\Vert \widehat{f}\right\Vert _{L^{2}\left(\mathbb{S}^{n-1}\right)}=\left\Vert f\right\Vert _{L^{2}\left(\mathbb{S}^{n-1}\right)} $$ for any $f\in C_{0}^{\infty}\left(\mathbb{R}^{n}\right)$?

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If you take any compactly supported function $f$ then $\|a^{n} f(a \cdot)\|_{\mathbb S^{n-1}} \to 0$ for $a \to \infty$. Moreover, we get by continuity of $\hat{f}$ that $\|\hat{f}(\cdot/a)\|_{\mathbb S^{n-1}} \to \hat{f}(0)$ (for simplicity, I assume that the measure on $\mathbb S^{n-1}$ is normalized). Since the Fourier transform of $a^{n} f(a \cdot)$ is $\hat{f}(\cdot/ a)$ this shows that your result does not hold if $\hat{f}(0) \neq 0$.

Your statement also does not hold if you additionally assume $\hat{f}(0) = 0$. If you analyze the argument above carefully, you see that your assumption implies that the Fourier transform of any compactly supported function needs to vanish on a neighborhood of $0$. By Paley-Wiener this will only work if $f = 0$.

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