I understand that one way to define the radial Lebesgue space $L_\text{rad}^{p}(\mathbb{R}^n)$ is by the completation of the space of radial smooth function with compact support, i.e, $L_\text{rad}^p(\mathbb{R}^n)=\overline{C_\text{rad,0}^{\infty}(\mathbb{R}^n)}^{\left\|\cdot\right\|_{p}}$ but, it can also be defined as $L_\text{rad}^p(\mathbb{R}^n)=\left\{u\in L^p(\mathbb{R}^n)\mid \text{$u$ is radial function }\right\}$.
Question. Do these two definitions coincide? Thanks.
Yes, for $1\le p<\infty$. The space $L_{\text {rad}}^p(\mathbb R^n)$ of $L^p$ radial functions $f$ ( that is rotation-invariant: $f\circ R=f$ for all $R\in SO(n)$) is a closed subspace of $L^p(\mathbb R^n)$ . So it contains the closure of the radial functions with compact support. On the other hand, the latter space is dense. In fact, you can approximate every $f\in L_{\text {rad}}^p(\mathbb R^n)$ in $L^p$ norm by a convolution $f_\epsilon:=(f\chi_{B(0,r)})*\phi_\epsilon$, which is smooth, compactly supported and radial, provided $\phi$ is a smooth radial convolution kernel with compact support (recall that in general for $ f\in L^p$ and $g\in L^1$ and $R\in SO(n)$ one has $(f\circ R)*(g\circ R)=(f*g)\circ R$, so if $f$ and $\phi$ are radial, so is $f*g$).
Also note that you can isometrically represent $ L_{\text {rad}}^p(\mathbb R^n)$ by $ L^p\big([0,\infty),\mu\big)$ for $d\mu:=\frac{2\pi^{n/2}}{\Gamma({n/2})}r^{n-1}dr$ via the correspondence $f(x)=u(\|x\|)$.
