Suppose $u$ is in $H^1(\Omega)$, where $\Omega$ is a unit square centered at the origin, such that it has singularity about the origin. As $C^1(\overline{\Omega}) $ is dense in $H^1(\Omega)$, we have a sequence $u_m$ such that $u_m \to u$ in the $H^1$ norm.
Now, from Rudins' Real and Complex Analysis, theorem $3.12$, we have a subsequence $u_{m_i}$ of $u_m$ such that $u_{m_i} \to u$ pointwise almost everywhere.
Q1. Show that $u_m$ is unbounded.
Attempt: $u$ is unbounded in the neighborhood of zero say around an open ball $B(0;\epsilon)$, where $\epsilon \to 0$. Now, let $u_{m_i} \to u$ on $K$. We then have $\mu(\Omega \setminus K)=0.$ Now as $\mu(B(0;\epsilon)) \neq 0$, choose $(x,y)\in K \cap B(0;\epsilon).$ Thus, we have by pointwise convergence there exists a $N \in \mathbb{N}$ such that $$ |u_{m_i}(x,y)-u(x,y)|< \epsilon \quad \forall i\ge N. $$ Thus we have, $u_{m_i}$ is an unbounded sequence.
Am I correct upto this?
Thanks!