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Let $Z$ be a standard normal. Now define \begin{align} g(t)= E[ f(Z-t)] \end{align} where $f(x)$ is a real-analytic function and $|f(x)| \le x^4$.

Question: Is it true that $g(t)$ is also a real analytic function?

If this is true can we show this without using complex analysis tools?

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  • $\begingroup$ @ChristianRemling But Morera is a tool from complex analysis, right? $\endgroup$
    – Boby
    Commented Aug 9, 2017 at 20:44
  • $\begingroup$ If $|f(x)| \leqslant |x|^4$ holds only on the real line, then I doubt the answer to the question is indeed "yes". At least, Morera's theorem cannot be applied directly: consider, for example, $f(x) = \exp(i \exp(x^2))$, which is bounded on the real line, but behaves badly in any strip $\{|\operatorname{Im} x| < \varepsilon\}$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 9, 2017 at 21:40
  • $\begingroup$ @MateuszKwaśnicki: Yes, this is a good point, we will need control on $f$ on a strip or at least an open set containing $\mathbb R$. $\endgroup$
    – Christian Remling
    Commented Aug 9, 2017 at 22:06

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$E(f(Z-t)) = \int f(x) e^{-\frac {x^2} 2} e^{tx - \frac {t^2} 2} dx =e^{ \frac {t^2} 2} \int f(x) e^{-\frac {x^2} 2} e^{tx} dx $. $$$$ $e^{ \frac {t^2} 2}$ is an analytic function of t, and the integral pretty clearly is too, even complex analytic even if f is only real analytic. This clearly has not much to do with f, only that it is not too large.

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  • $\begingroup$ You can not choose you $f$ to be like this. First, you have to specify $f(x)$ and then you plug it into the expectation by setting $x=Z-t$. You can not do this with the function you picked. $\endgroup$
    – Boby
    Commented Aug 10, 2017 at 14:32
  • $\begingroup$ This argument seems fine to me (modulo a missing factor of $\frac{1}{\sqrt{2\pi}}$ and a sign error). The expectation ${\bf E} f(Z-t) = \frac{1}{\sqrt{2\pi}} \int f(x-t) e^{-x^2/2}\ dx$ can be rewritten as $\frac{1}{\sqrt{2\pi}} \int f(x) e^{-(x+t)^2/2}\ dx$ after a change of variables $x \mapsto x+t$. One now has complex analyticity from Morera's theorem; one could also proceed in a real variable fashion by Taylor expansion and Fubini. $\endgroup$
    – Terry Tao
    Commented Oct 9, 2017 at 15:42

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