If $\Omega$ is a space of finite measure, then it is well known that $L^{N}(\Omega) \subseteq L^{2}(\Omega)$ for $N \in [2, \infty]$. I want to know if the image of this embedding is closed in $\left(L^{2}(\Omega),\| \cdot \|_{2}\right)$.
My attempt is: if $u_{n} \rightarrow u$ in $\left(L^{N}(\Omega), \| \cdot \|_{2}\right)$ then $u_{n}(x) \rightarrow u(x)$ a.e. in $\Omega$. Therefore, by Fatou's lemma we have $$\int_{\Omega}|u(x)|^{N} = \displaystyle\int_{\Omega} \lim |u_{n}(x)|^{N}\leq \liminf \displaystyle\int_{\Omega}|u_{n}(x)|^{N} = \liminf \|u_{n}\|_{N}^{N} \ .$$
I'm stuck here, because what guarantees me that the sequence $(\|u_{n}\|_{N})$ is bounded? So, if someome can prove my statement or give a counterexample I'll be grateful.
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 $L^N(\Omega)$.