# is $u\in L_{loc}^1(\mathbb{R}^2)$ and 2. $u\in W_{loc}^{1,1}(\mathbb{R}^2)$?

Let $U\in C^1((0,\infty))$, $u:\mathbb{R}^2\to \mathbb{R}$ defined by $$u(x,y)=U(\sqrt{x^2+y^2}).$$ Under which conditions on $U$ is

1.$u\in L_{loc}^1(\mathbb{R}^2)$ and 2. $u\in W_{loc}^{1,1}(\mathbb{R}^2)$?

For 1. I have to check if $\int_K|u(x,y)|d(x,y)<\infty$ for all compact subsets $K\subseteq \mathbb{R}^2$ and I want to use polar coordinates $\varphi:\mathbb{R}_{\ge 0}\times (0,2\pi)\to \mathbb{R}^2,\; (r,\phi)\mapsto (rcos\phi,rsin\phi)$ and use the transformation theorem. Then we have $det(D\varphi(r,\phi))=r$ and $u(\varphi(r,\phi))=U(r)$. But I'm not sure how to apply the transformation theorem exactly and if I need further conditions on $U$ to apply this theorem.

For 2. If all weak partial derivatives $\partial_1u$, $\partial_2u$ of $u$ exist, then $u\in W_{loc}^{1,1}(\mathbb{R}^2)$. One requirement to check if the partial derivatives exist is that $u$ is in $L_{loc}^1(\mathbb{R}^2)$. Summarized, I'm stuck to find out, if there are functions $g_k\in L_{loc}^1(\mathbb{R}^2)$ such that $$\int_{\mathbb{R}^2}u\partial_k\psi dx=-\int_{\mathbb{R}^2}g_k \psi dx$$ $\psi\in C_c^{\infty}(\mathbb{R}^2)$ for $k=1,2$. CI appreciate your help.

The functions $U$, $U_x$ and $U_y$ are locally bounded, except possibly at $(x,y)=(0,0)$. If $B_1$ is the unit ball centred at $(0,0)$, We have $$\int_{B_1} \lvert u\rvert = \int_0^{2\pi}\int_0^1 r\lvert u(r\cos\vartheta,r\sin\vartheta)\rvert\,dr\,d\vartheta= \int_0^{2\pi}\int_0^1 r\lvert U(r)\rvert\,dr\,d\vartheta=2\pi\int_0^1 r\lvert U(r)\rvert\,dr.$$ Clearly, $u\in L_{\mathrm{loc}}^1(\mathbb R^2)$ iff $\int_0^1 r\lvert U(r)\rvert\,dr<\infty$.
Next $u_x(x,y)=U'(r)\frac{x}{r}$. Thus $$\int_{B_1} \lvert u_x\rvert =\int_0^{2\pi}\int_0^1 r\lvert u_x(r\cos\vartheta,r\sin\vartheta)\rvert\,dr\,d\vartheta= 2\pi\int_0^{2\pi}\int_0^1 r\lvert U'(r)\cos\vartheta\rvert\,dr\,d\vartheta$$ and $$\int_{B_1} \lvert u_y\rvert =\int_0^{2\pi}\int_0^1 r\lvert u_y(r\cos\vartheta,r\sin\vartheta)\rvert\,dr\,d\vartheta= 2\pi\int_0^{2\pi}\int_0^1 r\lvert U'(r)\sin\vartheta\rvert\,dr\,d\vartheta.$$ Clearly, $u_x,u_y\in L_{\mathrm{loc}}^1(\mathbb R^2)$ iff $\int_0^1 r\lvert U'(r)\rvert\,dr<\infty$.