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Let $f\in L^{1}_{\text{loc}}(\mathbb{R}^{2})$. If $f$ is radial then $f\circ R_{\theta}=f$, where $R_{\theta}=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$ is the rotation in the plane through angle $\theta\in[0,2\pi]$.

I want to show that $f$ is radial as a distribution. I know that $f$ is radial as a distribution if and only if $\langle f,\varphi\rangle=\frac{1}{2\pi}\int_{0}^{2\pi}\langle f\circ R_{\theta},\varphi\rangle\,d\theta$; so this is what I need to show.

For all $\varphi\in\mathcal{D}(\mathbb{R}^2)$, we have:

$$\begin{aligned}\langle f,\varphi\rangle&=\langle f\circ R_{\theta},\varphi\rangle \\ &=\langle f,\varphi\circ R_{-\theta}\rangle \\ &=\langle f\circ R_{\theta},\varphi\circ R_{-\theta}\rangle \\ &=\int_{\mathbb{R}^{2}}f(R_{\theta}(x))\varphi(R_{-\theta}(x))\,dx,\end{aligned}$$ but I don't really see how to derive the solution from here...

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THe following chain of equalities should do the trick: $$\langle f\circ R_\theta,\phi \rangle = \int_{\Bbb R^2} (f\circ R_\theta )(x)\phi(x)dx = \int_{\Bbb R^2} f (x)\phi(x)dx =\langle f ,\phi \rangle $$

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    I think I see it. If $f\circ R_{\theta}=f$, then: $$\frac{1}{2\pi}\int_{0}^{2\pi}\langle f\circ R_{\theta},\phi\rangle\,d\theta=\langle f,\phi\rangle\cdot\frac{1}{2\pi}\int_{0}^{2\pi}d\theta=\langle f,\phi\rangle.$$ – 2017-01-21