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Dobner in his paper defines (https://arxiv.org/abs/2005.05142) some complicated function $\Phi_F$ (Eq. 8) and then a page after defines $H_t(z) = \int_{-\infty}^{\infty} e^{tu^2} \Phi_F(u) e^{izu} du$. Then he defines the set $Z$ to be the set of those real $t$ such that all the roots of $H_t$ are real. Then claims the following:

Note that Z is closed because the roots of H_t vary continuously in t by Hurwitz's theorem.

I searched the web including many posts in MSE related to Hurwitz's theorem and also there is this explanation: https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(complex_analysis) . I doubt that this is the theorem the author is citing or otherwise I can't figure out how I can prove the mentioned claim but this theorem in the Wiki page.

My question is how Z is closed?

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    $\begingroup$ if $t_n \to t$ it follows that $H_{t_n} \to H_t$ locally uniformly, so Hurwitz theorem tells us that since $H_t$ non zero, every zero of it is a limit point of zeroes of $H_{t_n}$ hence is real if those are real. This shows the required closure property $\endgroup$
    – Conrad
    Commented Mar 18 at 21:41
  • $\begingroup$ @Conrad, thanks a lot. Sorry one more question about the paper: at page 12 it is written "so by Rouche's theorem $h$ has a zero inside the shifted circle $C + i \tau_m$." I can't see any connection between Rouche's theorem of Wiki (en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem) and the claim in the paper. Can you please guide me with that? (I know this is a separate question but the reason I am asking this in here is because this is a kind of question that from my previous experience I am sure I will not receive any hint/answer for this if I post it). Thank you in advance. $\endgroup$
    – Ali
    Commented Mar 19 at 20:12
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    $\begingroup$ just read the line before which gives you the inequality $|h(s+i\tau_m)-F_t(s)| < |F_t(s)|$ on the circle $C$ which by Rouche means that $h(s+i\tau_m)$ has as many zeroes inside $C$ as $F_t$ so precisely one zero by the choice of $C$ some lines before, hence $h$ has precisely one zero in the shifted circle $C+i\tau_m$ $\endgroup$
    – Conrad
    Commented Mar 19 at 20:18
  • $\begingroup$ @Conrad, Thank you very much :) $\endgroup$
    – Ali
    Commented Mar 19 at 20:39

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