Show that the Fourier transform of a radial function $L^1 (\mathbb{R})$ is also radial

How do I prove that the Fourier transform of a radial function $f \in L^1 (\mathbb{R})$ is also radial function? I tried by polar coordinates but I dont got.

• what exactly is a radial function $f \in L^1(\mathbb{R})$? It seems all $f \in L^1(\mathbb{R})$ are "radial" to me. Perhaps you mean $f \in L^1(\mathbb{R}^n)$ $n > 1$? Aug 13 '13 at 22:57
• Well, not quite all $f \in L^1(\mathbb{R})$, maybe only the even $f$ can be construed as "radial". Aug 13 '13 at 23:00
• f is radial when $f(x)=f (y)$ whenever $|x|=|y|$ Aug 14 '13 at 0:03
• right, so every even $f \in L^1(\mathbb{R})$ is radial! Aug 14 '13 at 0:09

Just use the definition, and the change of variable $y=-x$: $$\widehat f(-\xi) = \int_{-\infty}^\infty f(x)e^{2\pi i \xi x}\,dx = -\int_{ \infty}^{-\infty} f(y)e^{-2\pi i \xi y}\,dy = \int_{ -\infty}^{ \infty} f(y)e^{-2\pi i \xi y}\,dy = \widehat f(\xi)$$
Let's also consider the general case $\mathbb R^n$, instead of just $\mathbb R$. If $A:\mathbb R^n\to\mathbb R^n$ is an invertible linear transformation, then $$\widehat{f\circ A}=\frac{1}{|\det A|} \widehat f\circ (A^{-1})^* \tag1$$ -- see this computation where the absolute value on $\det A$ was mistakenly omitted. How does (1) relate to your question? Saying that $f$ is radial is the same as saying that $f=f\circ A$ for every orthogonal transformation $A$. For such a transformation, $(A^{-1})^*=A$ and $|\det A|=1$. Therefore, $$\widehat f= \widehat {f\circ A}= \widehat f\circ A$$ which means that $\widehat f$ is radial.