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Suppose $\mathcal A$ is a unital C*-algebra and $\omega$ is a positive linear functional on $\mathcal A$. Consider a $1$-parameter group $(\tau_t)_{t\in\mathbb R}$ of strongly continuous *-automorphisms of $\mathcal A$.

For every $A,B \in \mathcal A$, if we know that there is always an analytic function $z\mapsto F(A,B;z)$ on the strip $ S_\beta := \{ z\in \mathbb C : 0 \lt \mathcal{Im}(z)\lt \beta \}$ for $\beta \in \mathbb R$, and continuous on $\overline{S}_\beta$ such that on $\mathbb R,$ $F(A,B;t)= \omega(A\tau_t(B))$, can we conclude that the function $z \mapsto \omega(A\tau_z(B))$ is also analytic on $ S_\beta$ and continuous on $\overline{S}_\beta$? In particular, does the function $F$ induce an analytic continuation of $\mathbb R \to \omega \circ \tau_t(\mathcal A)$ to the strip $ S_\beta$?

Once that is settled, can you please tell me if the following proposition is true?

If the set of zeros of a complex analytic function $f$ has an accumulation point inside its domain, then $f$ is zero everywhere on the connected component containing the accumulation point.

This would imply that $z \mapsto \omega(A\tau_z(B)) - F(A,B;z)$, which is zero on $\mathbb R$, is zero on the whole of $S_\beta \cup S_{-\beta}$ (extend to the lower half-plane by Schwarz reflection).

To summarize, I would like to know why $z \mapsto \omega(A\tau_z(B))$ is analytic on $ S_\beta$ and continuous on $\overline{S}_\beta.$ Furthermore, I would like to understand the above proposition (perhaps by means of a proof).

Reference: KMS States: Proof of Theorem $24$

Nanashi No Gombe
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1 Answers1

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Principle of Permanence or Unique Continuation Theorem

Given any analytic function $f(z)$ defined on an open (and connected) subset $U$ of the complex numbers $\mathbb C$, one can expand $f$ in exactly one way in a power series (in the largest open ball $B \subseteq U$) around $z_0 \in B$.

$$ f(z) = \sum_{k=0}^\infty a_k (z-z_0)^k \quad \text{for } z \in B \tag1$$

$$ \quad \Big(\Rightarrow f^{(n)}(z_0)=0 \quad \text{if and only if} \quad a_n=0\Big) \,.$$

Define the order of the zero as the largest $n \in \mathbb N\cup\{\infty\}$ such that

$$ f(z_0) = f'(z_0)=\cdots=f^{(n-1)}(z_0)=0 \tag 2$$

$$ \Rightarrow n = \sup \ \{m \in \mathbb N\cup \{\infty\} : a_0=a_1=\cdots=a_{m-1}=0\} \,.$$

Now it is easy to see, that if $f$ has a zero of order $n \lt \infty$, then it can be rewritten as

$$ f(z) = (z-z_0)^ng(z) \,,$$

where $g(z)$ is an analytic function on $U$ with $g(z_0) \ne 0$. Therefore, if and only if $f$ has a zero of infinite order, it vanishes identically near the zero (i.e. these zeroes are accumulation points).

By connectedness of $U$ and continuity of $f$ and all of its derivatives, we conclude that it is zero everywhere in the domain $U$. Note that continuity of each of the derivatives implies that the set of all zeroes of infinite order is closed, but this set is open as well because the accumulation points have open neighborhoods filled with zeroes of infinite order, and hence, connectedness implies that this clopen set has to be the whole of $U$ (or the empty set).

In summary, the theorem reads:

Given any analytic function $f(z)$ defined on an open (and connected) subset $U \in \mathbb C$, the zeroes of $f$ have no limit points in $U$, unless $f$ is identically zero.

Why $z \mapsto \omega(A\tau_z(B))$ is analytic?

It is not. Note that the linked chapter assumes that $B \in \mathcal A_\tau$, the set of all analytic elements in $\mathcal A$ for the dynamics $\tau$, in which case, of course, it is.

Foshiba
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