Existence of a distinguished continuous version of the logarithm of a continuous function

Question

Let $E$ be a $\mathbb R$-Banach space and $\varphi\in C^0(E,\mathbb C\setminus\{0\})$ with $\varphi(0)=1$.

I want to show that there is an unique $\psi\in C^0(E,\mathbb C)$ with $\psi(0)=0$ and $$\varphi=e^\psi\tag1.$$

If $E=\mathbb R^d$ for some $d\in\mathbb N$, a proof of this claim can be found in Lévy Processes and Infinitely Divisible Distributions by Sato (Lemma 7.6). As far as I can see, the proof given there should generalize to the setting of this question.

However, I've read that we can also show the claim by using the results provided in Foundations of Modern Analysis by Dieudonneé (Appendix IX.2), which I will briefly reproduce:

Let $f\in C^0(E,U)$, where $U:=\{z\in\mathbb C:|z|=1\}$. Say $f$ is inessential if $f=e^{{\rm i}g}$ for some $g\in C^0(E,\mathbb R)$; otherwise it is called essential.. It's easy to see that

1. if $f$ is inessential, then $1/f=\overline f$ is inessential;
2. if $f_i$ are inessential, then $f_1f_2$ is inessential;
3. if $f_1$ is essential and $f_2$ is inessential, then $f_1f_2$ and $f_1/f_2$ are essential.

Moreover, any $f\in C^0(E,U)$ with $f(E)\ne U$ is inessential. And if $f_i\in C^0(E,U)$ satisfy $f_1(x)\ne-f_2(x)$ for all $x\in E$ and $f_1$ is inessential (resp. essential), then $f_2$ is inessential (resp. essential).

And last but not least, if $E$ is compact and $f\in C^0(E\times[0,1],S^1)$ is such that $E\ni x\mapsto f(x,0)$ is inessential (resp. essential), then $E\ni x\mapsto f(x,1)$ is inessential (resp. essential).

He particularly concludes that any continuous function from a closed ball in $\mathbb R^d$ into $U$ is inessential.

How are we able to show the desired claim using these results? I can't see how we need to apply them.

Just in case it is important: I've read that we can use the results of Dieudonneé instead of the result of Sato in a footnote referenced in the proof of Corollary 2.5 of these lecture notes. The author doesn't mention it, but maybe it's important that the setting considered there is $E=\mathbb R^d$ and $$\varphi(t)=\int\mu({\rm d}x)e^{{\rm i}\langle t,\:x\rangle}\;\;\;\text{for all }t\in\mathbb R^d$$ for some probability measure $\mu$ on $\mathbb R^d$ (this can be generalized to $E$ being the dual of a $\mathbb R$-Banach space $\tilde E$ and $\mu$ being a probability measure on $\tilde E$).

This particular choice of $\varphi$ satisfies $$\inf_{t\in\overline B_r(0)}\varphi(t)>0\;\;\;\text{for all }r>0\tag2.$$ In light of the result about about closed balls in $\mathbb R^d$ mentioned above, this might be relevant. But again, I don't see how it could be used to show the desired claim. But if it's really the crucial ingredient, I'd be interested in generalizing it to more general spaces then $\mathbb R^d$. The only crucial property which of the closed balls in $\mathbb R^d$ which is being used should be the compactness.

Answer 1

The result (1) follows from the lifting property of covering spaces: Consider $f=\varphi/|\varphi|: E\to S^1$ and the standard covering map $p:\mathbb R\to S^1$, $t\mapsto e^{it}$. Since Banach space are path connected, locally path connected, and simply connected (the fundamental group is trivial) it follows from Proposition 1.33 in Allan Hatcher's Algebraic Topology (which is freely available on his homepage) that $f$ has a unique lift $g:E\to \mathbb R$ with $g(0)=1$, which means $f=p\circ g$. Then $\psi=\log(|\varphi|)+ig$ satisfies $e^\psi=\varphi$.

The compactness argument you mention will not easily carry over to infinite dimensional Banach spaces which are never locally compact.