Convergence to a function in $W^{1,N}_{0}$ after multiplying with other function and taking integral

Question

Let $\Omega \subset \mathbb{R}^{N}$ be a bounded domain, $ N \geq 2.$

Define $R:= \ ^{\text{sup}}_{x \in \Omega} |x|$. Let $u \in W^{1,N}_{0}(\Omega), u_{n} \in C^{\infty}_{0}$ such that $u_{n} \rightarrow u \ \text{in} \ W^{1,N}_{0}(\Omega)$.

I would like to show that the following convergence holds: $$ \int_{\Omega} \frac{|u_{n}|^N}{|x|^N (\text{log R/|x|})^N} \text{d}x \rightarrow \int_{\Omega} \frac{|u|^N}{|x|^N (\text{log R/|x|})^N} \text{d}x $$ In other words, we still have convergence after inserting $u_{n}$ into the above integral.

For the purpose of this proof, we can assume $u$ is such that $\frac{|u|^N}{|x|^N (\text{log R/|x|})^N}$ is actually integrable.

Please let me know how I can start to tackle this problem!

Answer 1

I'm afraid the answer lay in information I neglected to post in this question. We have the following inequality for all $n \in \mathbb{N}$: $\int_{\Omega} \frac{|u_{n}|^{N}}{|x|^{N} (log R/|x|)^{N}} \leq \int_{\Omega} |\frac{x}{|x|} \cdot \nabla u_{n}|$. I was able to show that the right hand side does converge easily to $u \in W_{0}^{1,N}$ and thus use Fatou's Lemma to conclude cnovergence on the left hand side as well.

Answer 2

Is the convergence of the sequence of integrals you desire pointwise or in the $W_0^{1,N}$ norm?

If its pointwise and assuming this definition for$ W_0^{1,N}$, then here is one strategy that could work.

Suppose we knew that the operator \begin{align} I(v) := \int_{\Omega} \frac{|v|^N}{|x|^N (\text{log R/|x|})^N} \text{d}x \end{align}

was bounded in the operator norm on $W_0^{1,N}$, i.e, $|| I(u) ||_{W_0^{1,N}} \leq k || u||_{W_0^{1,N}}$ for an absolute constant $k$.

Could you then conclude that if $u_n \rightarrow u$ in $W_0^{1,N}$ that $I(u_n) \rightarrow I(u)$? Assume further that $u_n \rightarrow 0$ (since you can always just correct to $u'_n := u_n- u \rightarrow 0$ anyway).

$I(u)$ is not linear, because of the power of $N$, but you might be able to show that the triangle inequality is enough, i.e. that with $| I(u) - I(v) | \leq I(u) + I(v)$ and the boundedness of $I$ that $I(u_n) \rightarrow I(0)$.