$H$ Hilbert space, $T$ symmetric bounded linear, when is $H=R(T) \oplus N(T)$?

Question

I just saw in an exercise that if I have a prehilbert space $H$ and $T$ a linear, bound and symmetric operator then $R(T)=N(T)^{\perp}$. Now I was asking myself whether $H=R(T) \oplus N(T)$. On wiki I saw that if $X$ is closed in $H$ then $H=X \oplus X^{\perp}$. So the question is, is $R(T)$ or $N(T)$ closed in general? Is $N(T)$ closed if it is separable?

Answer 1

The kernel of a bounded linear operator is always closed (it's the preimage of the closed set $\{0\}$ under a continuous map). The range may or may not be closed. In fact, the closedness of $R(T)$ is equivalent to $R(T)=N(T)^\perp$, which you wrongfully claim to hold in general.

In my experience, the most useful criterion for closedness of $R(T)$ comes in the form of an abstract Poincaré inequality: The range of $T$ is closed if and only if there exists $c>0$ such that $$ \|Tx\|\geq c\|x\| $$ for all $x\in N(T)^\perp$.