# Topological characterization of the range of a bounded normal operator

Let $T$ be a bounded normal operator on a Hilbert space $H$. I want to prove the following statement: $\text{ran}(T)$ is closed if and only if 0 is not a limit point of $\sigma(T)$. I tried to use the spectral theorem for bounded normal operators, but I think I did not find the right function to plug in.

Suppose $\lambda$ is an isolated point of the spectrum of $T$. Let $F$ be a continuous function that is identically $1$ near $\lambda$, and is $0$ on the remaining part of the spectrum $\sigma(T)\setminus\{\lambda\}$. Then $F$ is either $1$ or $0$ on the spectrum, which makes $P=F(T)$ an orthogonal projection. Clearly $(T-\lambda I)P=0$ because $H(z)=(z-\lambda)F(z)$ vanishes on $\sigma(T)$.
Let $Q=I-P$. Then $Q(T-\lambda I)=(T-\lambda I)$, which means that the range of $T-\lambda I$ is contained in the range of $Q$. Define $$G(z) = \left\{ \begin{array}{ll} 0, & z = \lambda \\ \frac{1-F(z)}{z-\lambda}, & \mbox{otherwise} \end{array}\right.$$ Then $(z-\lambda)G(z) = 1-F(z)$, which gives $$(T-\lambda I)G(T) = Q.$$ So the range of $T-\lambda I$ equals the range of $Q$, which is a closed subspace.
Conversely, suppose $T$ has a closed range. Then $$X = \mathcal{R}(T)\oplus\mathcal{N}(T)$$ because $\mathcal{N}(T)=\mathcal{N}(T^{\star})$ for a normal operator $T$. Both subspaces are invariant under $T$. The restriction $T : \mathcal{R}(T)\rightarrow\mathcal{R}(T)$ is then a bijection, which means that $T$ is continuously invertible on this closed subsapce $\mathcal{R}(T)$. That means $(T-\lambda I)$ is also invertible on $\mathcal{R}(T)$ for all $\lambda$ near $0$, with inverse $Q_{\lambda}$. If $P$ is the orthogonal projection onto $\mathcal{R}(T)$, then it is easy to verify that $T-\lambda I$ is invertible for $\lambda$ is a punctured disk centered at $0$, with inverse. $$Q_{\lambda}P -\frac{1}{\lambda}(I-P)$$ Therefore $0$ is an isolated point of the spectrum for $T$ if $T$ has closed range.
• @user159517 : The inverse is $(T-\lambda I)^{-1}$ on $PX$ and $T=0$ on $(I-P)X$, which gives $(T-\lambda I)^{-1}=-\frac{1}{\lambda}I$ on $(I-P)X$. Nov 8 '15 at 23:29