Is $L^{N}(\Omega) $ closed in $L^{2}(\Omega)$?

Question

If $\Omega \subset \mathbb{R}^{N}$ is a bounded set,then is well know that $L^{N}(\Omega) \subset L^{2}(\Omega)$. I want to know if the linear space $\left(L^{N}(\Omega) \cap L^{2}(\Omega),||.||_{2}\right)$ is closed in $\left(L^{2}(\Omega),||.||_{2}\right)$. My attemp is:

If $u_{n} \rightarrow u$ in $\left(L^{N}(\Omega) \cap L^{2}(\Omega),||.||_{2}\right)$ then $u_{n}(x) \rightarrow u(x)$ a.e in $\Omega$. Therefore, by the Fatou's Lemma we have $$\displaystyle\int_{\Omega}|u(x)|^{N} = \displaystyle\int_{\Omega} \lim |u_{n}(x)|^{N}\leq \liminf \displaystyle\int_{\Omega}|u_{n}(x)|^{N} = \liminf ||u_{n}||_{N}^{N}$$.

I'm stuck here, because what guarantess me that the sequence $(||u_{n}||_{N})$ is bounded?

So, if someome can proof my statement or give a counterexample I'll be gratefull.

Answer 1

Simple functions are in the intersection and they are dense in the full space. So if it were closed and dense it would be the full space, which it is not unless $N=2$.

To see this concretely, we can consider $\Omega = [0,1]$ and let $N-2>0$. We know that $x^{-p}\in L^1(\Omega)$ iff $p<1$, so the function $x^{-1/2+\epsilon}\in L^2(\Omega)$ for all $\epsilon>0$. Because $N>2$ (otherwise we don't have the inclusion) we have $1/N<1/2$ so we can choose $\epsilon$ so that $1/N + \epsilon < 1/2\;$ so that $-1/2+\epsilon<1/N$ and so for this $\epsilon$ we have that $x^{-1/2+\epsilon}\in L^2(\Omega)$ but is not in $L^N(\Omega)$.