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Let $(X_n)_{n\ge 0}$ be a series of real random variables and $\phi_n(\lambda)=\mathbb E(e^{\lambda X_n})$, where $\lambda$ is a complex number. Consider the following two types of convergence:

(1) There exists an open neighborhood $U$ of $0$ on the real line such that $\phi_n(\lambda)$ converges to $\phi_0(\lambda)$ uniformly for $\lambda \in U$;

(2) There exists an open neighborhood $U$ of $0$ on the complex plane such that $\phi_n(\lambda)$ converges to $\phi_0(\lambda)$ uniformly for $\lambda \in U$;

(3) There exists an open strip-shaped neighborhood $U=I\times \mathbb R$ of $0$ on the complex plane such that $\phi_n(\lambda)$ converges to $\phi_0(\lambda)$ uniformly for $\lambda \in U$;

Clearly $(3)\Rightarrow (2) \Rightarrow (1)$. Are they actually equivalent (the first implies the third)?

No One
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  • I'm not sure you get uniformly in (3) coming back from (1), but you get local uniform convergence in a strip (along the same line as #990086) which is probably good enough? – user10354138 Oct 16 '18 at 21:04

1 Answers1

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I believe this one might be solved using Cauchy-Schwartz. Indeed, you have, for $\lambda = a + ib$, $a, b \in \mathbb{R}$,

$$ \mathbb{E}[e^{(a + ib) X_n}] \leq \mathbb{E}[e^{2a X_n}]^{1/2} $$ where I used that $|e^{i2bX_n}| = 1$ for the second factor of Cauchy-Schwartz. Hence if $a$ is small enough, you can use (1) to conclude that (3) holds.

Gâteau-Gallois
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  • I think in this case we need to look at $\mathbb{E}[e^{(a + ib) X_n}-e^{(a+ib)X_0}]$. How do you apply the inequality to remove $ib$ in this case? – No One Oct 15 '18 at 21:22