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Motivation of my question: Let $A$ be a bounded selfadjoint operator with spectral measure $E$ and $x$ a vector. Then it is easily seen that the closed linear span of all $A^nx$ ($n\in\mathbb N$) coincides with that of all $E(\Delta)x$, where $\Delta$ runs through all Borel sets. I would like to know if the same holds for normal operators. The inclusion "$\subset$" holds trivially. Just approximate $z^n$ uniformly by simple functions on $\sigma(A)$ (the spectrum of $A$) and use the spectral theorem. The question for the converse inclusion is whether to a disc $D$ one can find a sequence of polyniomials $(p_n)$ which converges at least pointwise to the characteristic function of $D\cap\sigma(A)$ on $\sigma(A)$. If so, $p_n(A)x$ converges to $E(D)x$ and the original claim follows from simple measure theoretic arguments.

Does anyone know an answer to my question?

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  • $\begingroup$ The natural generalization to normal operators would be polynomials $p(z,\overline{z})$ in $A, A^*$, and Stone-Weierstrass takes care of that. There is a huge literature on related issues, for example on when $1/z$ is in the span of $1,z,z^2, \ldots$ in $L^2(S,\mu)$. Try googling something like Szego's Theorem. $\endgroup$ – Christian Remling Dec 16 '15 at 0:39
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Yes, using Runge's theorem. I'll assume wlog your $D$ is the unit disk $\{z: |z| \le 1\}$. Given positive integer $n$, take $$ K_n = \{z: (|z| \le 1 \ \text{or}\ 1 + 1/n \le |z| \le n)\ \text{and} (\text{Im}(z) \le 0 \ \text{or}\ \text{Im}(z) \ge 1/n)\}$$ Note that $\mathbb C \backslash K_n$ is connected, and there is a function $g_n$ analytic in a neighbourhood of $K_n$ that is $1$ on $K_n \cap D$ and $0$ on $K_n \backslash D$. By Runge's theorem, there is a polynomial $p_n$ such that $|p_n - g_n| < 1/n$ on $K_n$. Since every point of $\mathbb C$ is eventually in $K_n$, $p_n$ converges pointwise to the characteristic function of $D$.

CAUTION: the convergence is highly non-uniform, and $p_n$ might have very large values on the complement of $K_n$. This makes measure theoretic arguments tricky.

In fact the statement you're trying to prove is false. Let $A$ be multiplication by $z$ on $L^2$ of the unit circle $\mathbb T$, which is a unitary operator whose spectrum is $\mathbb T$. If $f$ is in the closed linear span of $A^n 1 = z^n$, $n \in \mathbb N$, then $\oint_{\mathbb T} f(z)\; dz = 0$. But if $D \cap \mathbb T$ is, say, the right half-circle, then $\oint_{\mathbb T} E(D) 1 \; dz = 2 i \ne 0$.

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  • $\begingroup$ Thanks so much, Robert. You are right. That means that the $p_n$'s must blow up in the small gaps of your construction. I guess that means (as also Christian indicates above) that I have to add the polynomials of $A^*$ to get equality. $\endgroup$ – Friedrich Philipp Dec 16 '15 at 1:48

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