Does $\left\|u_n-u\right\|_{L^2(\Omega)}\stackrel{n\to\infty}{\to}0$ imply $u\in L^2(\Omega)$?

Question

Let $\Omega\subseteq\mathbb{R}^n$ be bounded, $u\in C^0(\Omega)$ and $(u_n)_{n\in\mathbb{N}}\subseteq C_0^0(\Omega)$ with $$\left\|u_n-u\right\|_{L^2(\Omega)}\stackrel{n\to\infty}{\to}0\tag{1}$$ Can we conclude $u\in L^2(\Omega)$ from $(1)$?

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Note: $C_0(\Omega)$ is the set of all functions $\Omega\to\mathbb{R}$ that vanish at infinity.

Answer 1

Assuming that you mean $C_0^0 = C_c (\Omega)$ (i.e. compactly supported, continuous functions), this is true.

Namely, you have $u_n \in C_c (\Omega) \subset L^2 (\Omega)$ for all $n$. Furthermore, since $\Vert u_n - u\Vert_2 \to 0$, there is some $n$ for which $\Vert u_n - u \Vert_2 < \infty$. Hence,

$$ u = u_n - (u_n -u) \in L^2 (\Omega) $$ as a linear combination of functions in $L^2 (\Omega)$.

Note that we did not use completeness of $L^2$ for this.