Show $N$ normal there is a sequence of invertible normal operators that converges to $N$.

Question

This is a question from Conway, a course in Functional Analisys. Background is the Spectral theorem, which states that for a normal operator $N$ there is a unique spectral measure $E$ on the Borel subsets of $\sigma(N)$ such that $N=\int zdE(z).$

I am lost on how to use this for the question: "Show for $N$ normal there is a sequence of invertible normal operators that converges to $N$."

Also this (The set of invertible normal operator is dense in the set of normal operator) relates to my problem, but I am not sure how to use it. Can I just say my sequence exists?

My own attempt: If $N$ itself is invertible, take any sequence $\{\alpha_i\}\subset\mathbb{C}$ that converges to 1. Then $N_i=\alpha_iN$ suffices. Is this true? What happens when $N$ is not invertible? Should I rather look at convergence of the spectral measures instead?

Any help or hint in the right direction will be greatly appreciated.

Answer 1

Let $\left\{f_n\right\}_{n\in \mathbb{N}}$ be a sequence of Borel functions which are bounded on the spectrum $\sigma(N)$ of $N$. Then the integral $$N_n:=\int f_n(z)\,dE(z) $$ defines a bounded linear operator. This operator is normal, since $$N_n^*=\int \overline{f_n(z)}\,dE(z),\qquad N_nN_n^*=\int f_n(z)\overline{f_n(z)}\,dE(z)=\int |f_n(z)|^2\,dE(z)=N_n^*N_n $$ Also, if $f_n$ is such that $f_n\neq 0$ in $\sigma(A)$, then $N_n$ is invertible, with $N_n^{-1}=\int{\frac{1}{f_n(z)}\,dE(z)}$. Indeed we have $$N_n N_n^{-1}=\int f(z)\frac{1}{f(z)}\,dE(z)=\int 1\,dE(z)=I=N_n^{-1}N_n $$

We now need to construct $\left\{f_n\right\}$ so that $N_n\to N$ strongly. By the properties of integrals w.r.t. projection-valued measures, this holds if

• $f_n(z)\to z$ $E$-a.e. in $\sigma(A)$;
• $\left\{f_n\right\}$ is uniformly bounded in the supremum norm.

For instance, a sequence which satisfies these properties is $$f_n(z)=\begin{cases}z & |z|\geq 1/n\\ 1/n & |z|<1/n \end{cases} $$ notice that $f_n(z)\neq 0$ for all $z$ and so each $N_n$ is invertible.

EDIT: As the other answer shows, we may also prove that $\|N_n-N\|\to 0$.

Answer 2

If $N$ is bounded and normal with spectral measure $E$, then $$ N_{\epsilon}=\int_{|\lambda|\ge\epsilon}zdE(z)+\int_{|\lambda| < \epsilon}\epsilon dE(z) $$ is invertible for $\epsilon > 0$ with inverse $$ N_{\epsilon}^{-1}=\int_{|\lambda|\ge \epsilon}\frac{1}{z}dE(z)+\frac{1}{\epsilon}\int_{|\lambda| < \epsilon} dE(\lambda). $$ And, $$ \|N-N_{\epsilon}\|=\|\int_{|\lambda|<\epsilon}(\epsilon-z)dE(z)\| \le 2\epsilon \|E\{ z: |z|<\epsilon\}\| \le 2\epsilon. $$