I was studying the paper "On Trudinger–Moser type inequalities with logarithmic weights" (by Martha Calanchi and Bernhard Ruf) when on page 1987 there was a sequence $(u_n)$ converging weakly to some $u$ in $H^1_{0,rad}(B(0,1),|\log(1/|x|)|)$ and they claimed that (up to subsequence) $u_n\to u$ in $L^1(B(0,1))$. I do not think it is necessary, but (in my case) the dimension is 2. How do I prove this?
My try:
In the article there is another weight: $|\log(e/|x|)|$. For this weight I note that $|\log(e/|x|)|\geq 1$ in $B(0,1)$. Then $H^1_{0,rad}(B(0,1),|\log(e/|x|)|)\hookrightarrow H^1_0(B(0,1))$ continuous. Using classic compact embedding, we have the result.
But for $|\log(1/|x|)|$ we have $|\log(1/|x|)|\to 0$ when $|x|\to 1$. So I can not proceed as above. Another attempt, since $u(x)=v(|x|)$ for some $v$, we have $v\in H^1((0,1),r|\log(1/r)|)$ with $v(1)=0$ (trace operator). I do not know how to progress here.