Suppose we have a sequence of functions in $L^2(\Bbb R^n)$, $u_n$, that are (if necessary, compactly supported) and such that $u_n \xrightarrow{L^2} u$, where $u$ is comapctly supported in $K \subseteq \Bbb R^n$.
My question is:
can we make a replacement $u'_n$ such that each $u'_n$ are supported in $K$ and $u'_n \rightarrow u$ in $L^2$ norm.
My thoughts are simply let $u'_n:= u_n1_K$. But I could not actually spell the argument out.
Yes, this is correct. In general, if you know that $u_n\to u$ in $L^2(\mathbb{R}^n)$, then it is true that $u_n|_K\to u|_K$ in $L^2(K)$ for any bounded, measurable subset $K$, just because $$ \|u_n|_K-u|_K\|_{L^2(K)}\le \|u_n-u\|_{L^2(\mathbb{R}^n)}\to 0. $$ and, of course, $\|u_n\mathbb{1}_K-u\mathbb{1}_K\|_{L^2(\mathbb{R}^n)}=\|u_n|_K-u|_K\|_{L^2(K)}$