Consider $\mathbb{R}^N = \mathbb{R}^{N_1} \times \mathbb{R}^{N_2}$ and $B$ the unit ball in $\mathbb{R}^N$. Define $f_\alpha(x,y) = |x|^{\alpha}$ for $\alpha \geq 0$ and $(x,y) \in B$. Given $u \in L^2(B)$, I am wondering if the function $\frac{u(x,y)}{f_\alpha(x,y)}$ is in $L^1_{loc}(B)$ for all $\alpha > 0$. I know that $\frac{1}{f_\alpha} \in L^1_{loc}(B)$ if $\alpha < N_1$. In fact, denoting $B^{N_i}$ the unit ball in $\mathbb{R}^{N_i}$ for $i=1,2$ and noticing that $B \subset B^{N_1} \times B^{N_2}$, one has $$ \int_B \frac{1}{f_\alpha(x,y)} d(x,y) \leq \int_{B^{N_1} \times B^{N_2}} \frac{1}{f(x,y)} d(x,y) = \int_{B^{N_1}}\int_{B^{N_2}} |x|^{-\alpha} dy dx = C_1 \int_{B^{N_1}} |x|^{-\alpha} dx = C_2 \int_0^1 r^{-\alpha + N_1 - 1} dr = C_2 \frac{1}{(-\alpha + N_1)}. $$ I know how to prove that $\frac{u}{f_\alpha} \in L^1_{loc}(B)$ for some restriction on $\alpha$. This is not the problem. My questions are:
- Given $u \in L^2(\Omega)$, then $\frac{u}{f_\alpha} \not\in L^1_{loc}(B)$ for some $\alpha > 0$ ?
- Under which condition on $u \in L^2(B)$ can we have $\frac{u}{f_\alpha} \in L^1_{loc}(B)$ for all $\alpha > 0$ ?