Let $X$ be a complex Banach space, and let $f : \Omega \subset \mathbb{C} \to X$ be a holomorphic function on an open set $\Omega$. I'll denote by $X^\ast$ the (topological) dual of $X$.
When $X = \mathbb{C}$ and we go back to the standard theory of complex-valued holomorphic functions, then it is a consequence of the open mapping theorem (OMT) that if $f$ is injective, then $f'$ has no zeros on $\Omega$, and even that $f$ has a holomorphic inverse $f^{-1} : f(\mathbb{C}) \to \Omega$.
For a reference, see Rudin's Real and Complex Analysis (RCA), Chapter $10$, last Corollary before the end of the part titled "The Open Mapping Theorem". The result is meant for connected domains but applying it on all discs contained in $\Omega$ and glueing the inverses using the overall injectivity to make one inverse works.
I don't know yet about anything in the theory of holomorphic functions with a Banach space as a domain rather than its codomain, even in the case of $\mathbb{C}^n$, so I'm not interested yet about the latter part about the inverse being holomorphic, but the goal of this question would be to (dis)prove the following claim:
If $f$ is injective, then $f'$ has no zeros on $\Omega$.
EDIT: this is the only "big" change I'll do to this question: if a counterexample is found but it's of constant norm, I would then like to examine the very closely related claim:
If $f$ is injective and of non-constant norm, then $f'$ has no zeros in $\Omega$.
However I will not force nor expect anyone to do this additional check. A counterexample is still a counterexample, and it would still be quite enlightening!
In regards to my research:
While the maximum modulus principle can be adapted to Banach-valued functions, by replacing the "is constant" conclusion in "$|f|$ has a local maximum implies $f$ is constant" by "is of constant norm" (and can't be improved since there are spaces with nonconstant constant-norm holomorphic functions, e.g. $(\mathbb{C}^2, \| \cdot \|_\infty)$), $f : z \mapsto (1, z)$), I haven't seen any proof of anything resembling the minimum modulus principle, which the OMT's standard proof relies on. To prove the minimum modulus principle, you would use the maximum modulus principle on $\frac{1}{f}$ for $f$ a holomorphic function that has no zeros, but there is no notion of "$\frac{1}{f}$" in a generic Banach space as far as I know. I only briefly looked in Chapter $18$ of RCA about Banach algebras by curiosity, but I saw there that a complex Banach algebra with unit where all nonzero elements have a multiplicative inverse is isometrically isomorphic to $\mathbb{C}$, so if my basic understanding is correct that would get rid of the notion of a general $\frac{1}{x}$ in all Banach spaces with dimension over $2$.
In turn, this means that the OMT and thus the aforementioned corollary is probably not viable either (big question mark?), since the proof for complex-valued functions uses the minimum modulus principle in a seemingly inescapable way, at least in RCA. I can't seem to be able to find online nor prove myself any meaningful variant of the OMT in this context, though I'm sure it's probably just me being bad at googling such things for now. The best thing I have is that if $\Omega$ is connected and $f(U)$ for any $U$ open connected subset of $\Omega$ is compact then $f$ is constant on $\Omega$ (by contradiction assume it's nonconstant. Then there's a $\Lambda \in X^\ast$ for which $\Lambda(f(U))$ is not reduced to a singleton thus open in $\mathbb{C}$ by applying the standard OMT on $\Lambda \circ f$. Yet it's compact since continuous images of compact sets are compact and $f(U)$ is assumed to be compact, absurd since $\mathbb{C}$ is connected), but talk about a most likely useless replacement of the OMT...
As for my own efforts on the question itself:
I had the idea of attempting to look at the open discs $D$ such that $\bar{D} \subset \Omega$, and trying to make a finite-intersection-property type proof on each $\bar{D}$ compact to show that if the $\Lambda \circ f$, $\Lambda \in X^\ast$ are all non-injective on $\bar{D}$, then $f$ is non-injective on $D$. The reason is that it would mean that if $f$ is injective, then at least one $\Lambda \circ f$ would be injective, which would imply that $(\Lambda \circ f)' = \Lambda \circ (f')$ has no zeros on $D$, which would prevent $f'$ from having any zeros either. Finally, if $f'$ has no zeros on any such disc, then it has no zeros on $\Omega$, and we would be done. However as you'll see I still have some issues to work out...
First, define the following function: $$g : (z,w) \in \Omega^2 \mapsto \begin{cases} \displaystyle\frac{f(z) - f(w)}{z - w} \,&\text{if } z \neq w \\ f'(z) \,&\text{if } z = w\end{cases} \in X$$
$g$ can be shown to be continuous just like in the complex-valued case.
Now let $\mathcal{P}_{fin}(X^\ast)$ be the set of all finite subsets of $X^\ast$, and pick $L \in \mathcal{P}_{fin}(X^\ast)$. Define the following set:
$$E_L := \left\{(z,w) \in \bar{D}^2 \mid \forall \Lambda \in L,\, \Lambda(g(z,w)) = 0\right\}$$
Each $E_L$ is closed by continuity of the $\Lambda$s and of $g$, and contained in $\bar{D}$ compact. Moreover, we have, for $(L_1, \dots, L_d) \in \mathcal{P}_{fin}(X^\ast)^d$:
$$\begin{cases}\bigcap_{k = 1}^d E_{L_k} = E_{\bigcup_{k = 1}^d L_k} \\ \bigcup_{k = 1}^d L_k \in \mathcal{P}_{fin}(X^\ast)\end{cases}$$
Hence, it would only remain to show that all $E_L$s are non-empty to have that $(E_L)_{L \in \mathcal{P}_{fin}(X^\ast)}$ has the finite intersection property in $\bar{D}$ compact, and thus the intersection of all the $E_L$s would be non-empty. But that intersection is exactly the set of the zeroes of $g$ on $\bar{D}^2$, therefore $g$ would have a zero on $\bar{D}^2$.
The two main issues with this approach are:
- I'm stumped on how to show that the $E_L$s are non-empty, if it's even possible;
- More importantly, this would only prove that $f$ is non-injective OR that $f'$ has a zero, hence the current contrapositive would require $f'$ to not have any zeros which is counter-productive...
I couldn't overcome the second issue as changing any aspect would result in either defining the $E_L$s in the set $\bar{D}^2 \setminus \left\{(z,z) \mid z \in \bar{D}\right\}$ which would make them potentially not closed, or changing from "$\Lambda(g(z,w)) = 0$" to "$\Lambda(f(z) - f(w)) = 0$" which would trivially give a non-empty intersection of the $E_L$s since any $(z,z)$, $z \in \bar{D}$ would be in it...
Any help is appreciated, and feel free to edit or re-tag if/when necessary.
Consider $f(t)=(t^2, t^3)$, $t\in {\mathbb C}$.