When reading A course in functional analysis, Conway, I meet a problem about normal operator on Hilbert space. It says that If $N$ is a normal operator on Hilbert space $H$, then range of $N$ ($\operatorname{Ran}N$) is closed if and only if $0$ is not a limit point of $\sigma(N)$, which is the spectrum of operator $N$.
I try to decompose $H$ as follows: $$ H=\operatorname{Ker}N\bigoplus X. $$ where $X$ is a closed subspace which is perpendicular with $\operatorname{Ker}N$. Then $\operatorname{Ran}N$ is closed if and only if there is $\varepsilon_0>0$ s.t. $$ \|Nx\|\geqslant\varepsilon_0\|x\|,\quad \forall x\in X. $$ If $0\notin\sigma(N)'$, we can find $\{\lambda_n\}\subset\sigma(N)$ and $\lambda_n\to 0$.
For $\lambda_n\in\sigma_p(N)$, we can find $y_n\in H$, $y_n=x_n+z_n$, where $x_n\in X$ and $z_n\in\operatorname{Ker}N$, s.t. $$ Ny_n=\lambda_ny_n,\quad\Longrightarrow\quad Nx_n=\lambda_n(x_n+z_n). $$ Then $$ \|Nx_n\|^2=|\lambda_n|^2\|x_n+z_n\|^2=|\lambda_n|^2(\|x_n\|^2+\|z_n\|^2). $$ I stopped here since I have no idea about what to do next.
Since $N$ is normal, $X$ is an invariant subspace for $N$, so the spectrum of $N$ is $\{0\} \cup \sigma(N|_X)$. Hence, $0$ is a limit point of $\sigma(N)$ iff $0$ is a limit point of $\sigma(N|_X)$. By definition, $N|_X$ is injective, so by either the spectral theorem or functional calculus, we see that $0$ is either a limit point of $\sigma(N|_X)$, or it is not in $\sigma(N|_X)$ altogether. Hence, $0$ is a limit point of $\sigma(N)$ iff $0 \in \sigma(N|_X)$.
Now, if $0$ is not a limit point of $\sigma(N)$, then $0 \notin \sigma(N|_X)$, i.e., $N|_X$ is invertible, so its range (which is the same as the range of $N$) must be the entirety of $X$, which is closed. Conversely, suppose the range of $N$ is closed. We note that in that case it must be $X$, since its orthogonal complement is the kernel of $N^*$, which, again because $N$ is normal, is the same as $ker(N)$. Thus, $N|_X$ is bijective, so by the open mapping theorem, it is invertible, i.e., $0 \notin \sigma(N|_X)$, so $0$ is not a limit point of $\sigma(N)$.