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Let $d\in\mathbb N$ and $\varphi_n:\mathbb R^d\to\mathbb C$ be a continuous function$^1$ for $n\in\mathbb N$.

How can we show that

  1. $(\varphi_n^n(x))_{n\in\mathbb N}$ is convergent for all $x\in\mathbb R^d$ and $$\varphi(x):=\lim_{n\to\infty}\varphi_n^n(x)\;\;\;\text{for }x\in\mathbb R^d$$ is continuous at $0$;
  2. $(n(\varphi_n(x)-1))_{n\in\mathbb N}$ is convergent for all $x\in\mathbb R^d$ and $$f(x):=\lim_{n\to\infty}n(\varphi_n(x)-1)\;\;\;\text{for }x\in\mathbb R^d$$ is continuous at $0$

are equivalent and if they hold, then $\varphi=e^f$?

Addendum: Assuming $\varphi_n$ is actually the characteristic function of a probability measure, how do we see that if 1. (hence 2.) holds, then $\varphi$ is the characteristic function of a probability measure?

We can show that $$\frac12|z-1|\le\left|\ln z\right|\le\frac32|z-1|\;\;\;\text{for all }z\in\mathbb C\text{ with }|z-1|<\frac12\tag1$$ from which we immediately deduce that if $(z_n)_{n\in\mathbb N}\subseteq\mathbb C$, then $$\limsup_{n\to\infty}n|z_n-1|<\infty\Leftrightarrow\limsup_{n\to\infty}n\left|\ln z_n\right|<\infty\tag2.$$ Moreover, $$\lim_{n\to\infty}n(z_n-1)=\lim_{n\to\infty}n\ln z_n\tag3,$$ if either of the limits exists.

However, I'm struggling to see how we can use this result in order to show the desired claim. For example, assuming that (2.) holds, we clearly want to choose $z_n=\varphi_n(x)$ for an arbitrary fixed $x\in\mathbb R^d$. By (2.) and the former result, $$\lim_{n\to\infty}n(z_n-1)=\lim_{n\to\infty}n\ln z_n\tag4$$ and hence, by continuity of $\exp:\mathbb C\to\mathbb C$, $$\lim_{n\to\infty}e^{n(z_n-1)}=e^{\lim_{n\to\infty}n(z_n-1)}=e^{\lim_{n\to\infty}n\ln z_n}=\lim_{n\to\infty}\left(e^{\ln z_n}\right)^n\tag5.$$ However, in order to simplify the right-hand side, it seems like we need to use that $$e^{\ln z}=z\;\;\;\text{for all }z\in\mathbb C\setminus\{0\}\tag6;$$ but in order to use this, we need to know assume $z_n\ne 0$ ... am I missing something?

Regarding the addendum: This must clearly be a consequence of Lévy's continuity theorem, but how do we need to apply it?

$^1$ In my application, $\varphi_n$ is actually the characteristic function of a probability measure, but I've omitted this assumption at this stage, since I don't know whether this is really relevant and I don't want to prevent anybody which is not familiar with probability theory to provide an answer.

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  • If $\lim_n n(z_n-1)=c\in\mathbb{C}$, then $\lim_n\big(1+\frac{n(z_n-1)}{n}\Big)^n=e^c$. This means that $\lim_nz^n_n=e^c$. This is based on a well known result in complex variables: Lemma: If $c_n\xrightarrow{n\rightarrow\infty}c$ in $\mathbb{C}$, then $$\lim_{n\rightarrow\infty}\Big(1+\frac{c_n}{n}\Big)^n=e^c$$ A proof can be found in this MSE posting). – Mittens Sep 09 '21 at 19:39
  • If $z_n = \varphi_n(x) = 0$, then $f(x) = \lim_{n\to \infty} n(0 - 1) = -\infty$, and therefore condition 2 does not hold, as the sequence fails to converge. – Paul Sinclair Sep 10 '21 at 04:02

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