If $T: X \to Y$ be a bounded linear operator between Hilbert spaces. Let $P$ be a dense subset of $X$. Then how to prove that $\overline{R(T)} = \overline{R(T')}$ where $T'$ is restriction of $T$ over $P$. $\overline{A}$ denotes closure of the set $A$.
Now one thing is obvious $\overline{R(T')} \subseteq \overline{R(T)}$. How to prove the other inclusion?
Let $T(x) \in \overline{R(T)}$. We want to show that $T(x) \in \overline{R(T')}$.
By definition, there exists $x_n \to x$, where $x_n \in P$. Note that by boundedness/continuity of $T$, $T(x_n) \to T(x)$, but then $T(x_n) \in R(T')$,which implies that $T(x) \in \overline{R(T')}$, since $T(x)$ is a limit point of $R(T')$.
The other way is obvious, hence we are done.