Nevanlinna/Herglotz functions are analytic functions defined on the upper half complex plane and have non-negative imaginary part there.
A remarkable theorem asserts that every such function has a unique integral representation of the form $$f(z) = a + bz + \int_{\mathbb{R}} \left( \frac{1}{\lambda-z} -\frac{\lambda}{\lambda^2+1} \right) d\mu(\lambda),$$
where $\mu$ is a positive Borel measure satisfying $\int (\lambda^2+1)^{-1} d\mu(\lambda) < \infty$, $a \in \mathbb{R}$ and $b \geq 0$.
The measure $\mu$ is recovered from $f$ by the Stieltjes inversion formula $$\mu(( \lambda_1,\lambda_2]) = \lim \limits_{\delta \downarrow 0} \lim \limits_{\epsilon \downarrow 0} \frac{1}{\pi} \int_{\lambda_1+\delta} ^{\lambda_2+\delta} \text{Im} \ (f(\lambda+i\epsilon)) d\lambda.$$
I am interested in understanding the relationship between the support of the measure $\mu$ and the restriction of $f$ to the real line. To this end, I define the Domain of $f(x)$ to be $$\text{Dom}(f) := \{ x \in \mathbb{R} : f(x) \ \text{is defined there and} \ f(x) \ \text{is a finite real number} \}.$$
My question to you :
Is it true in general that for all Nevanlinna functions, $\text{Dom} (f) \cup \text{supp} \ \mu = \mathbb{R}$ and $\text{Dom} (f) \cap \text{supp} \ \mu = \emptyset$ ?
Intuitively, from the Stieltjes formula $ \text{Dom} (f) \cap \text{supp} \ \mu = \emptyset$ makes sense...but I'd like to hear from the experts. Thanks !
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Some examples of Nevanlinna functions :
1) $f(z) = z$. Then $\text{Dom}(f) = \mathbb{R}$, $\mu = 0$ and $\text{supp} \ \mu = \emptyset$.
2) $f(z) = -z^{-1}$. Then $\text{Dom}(f) = \mathbb{R}\setminus \{0\}$, $\mu = 1_{\{0\}}$ and $\text{supp} \ \mu = \{0\}$.
3) $f(z) = \log(z)$. Then $\text{Dom}(f) = (0,+\infty)$, assuming the standard branch cut, $\mu = 1_{(-\infty,0]}$ and $\text{supp} \ \mu = (-\infty,0]$.
4) $f(z) = \tan(z)$. Then, denoting $\Sigma := \{ (n+1/2)\pi : n \in \mathbb{Z}\}$, we have $\text{Dom}(f) = \mathbb{R} \setminus \Sigma $, $\mu = 1_{\Sigma}$ and $\text{supp} \ \mu = \Sigma$.