Let $E$ be a complex Hilbert space with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$.
If $T\in\mathcal{L}(E)^+$ (i.e. $T^*=T$ and $\langle Tx\;, \;x\rangle\geq0$ for all $x\in E$). Consider $T_0=T|_{\overline{\mbox{Im}(T)}}$. Is $\mbox{Im}(T_0)$ a closed subspace of $\overline{\mbox{Im}(T)}$ with respect to $\|\cdot\|$?
No. If $T$ is compact and $\overline{im(T)}=E$ then $T=T_0$ and $im(T_0)$ is not closed.