Analytic maps on Banach spaces: analyticity upgrade

Question

Consider the following problem.

Let $E,F,G$ be real or complex Banach spaces, such that $F\subset G$ with continuous embedding. Let $U\subset E$ an open set and $$ f:U\to G $$ an analytic map, such that $f(E)\subset F$. (To be clear, we require that $f$ can be written locally around each point as a power series that converges in $G$ with positive radius of convergence).

Question: Is $f$ analytic as a map $f:U\to F$?

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Background.

1. If the map $f$ is linear, the answer is affirmative, that is, $f$ is a bounded linear map from $E$ to $F$. This is a simple, but remarkable consequence of the closed graph theorem (see the post by Jan Bohr). It also extends to the case of Fréchet spaces.
2. It is not difficult to show that the statement also holds true if $f$ is a polynomial (a finite sum of monomials, which are restrictions to the diagonal of bounded, multilinear operators from $E$ to $G$). The idea is that if $f$ is a finite sum of monomials, one can first show that all monomials map $E$ to $F$ by plugging $\lambda x$ inside $f$ for various values of lambda, and then use the closed graph theorem as in the linear case to conclude that each monomial is bounded.
3. There is a positive answer to the above question in the complex case (that is, for general holomorphic maps) assuming in addition that $f$ is locally bounded from $U$ to $F$ (see my answer to the aforementioned post). This also proves that asking for the analyticity of $f$ is equivalent to asking for the continuity of $f$. The proof does not make use of the closed graph theorem at all, and in fact the kind of result is different from the previous points as it requires local boundedness.

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It is easy to suspect that the local boundedness might not be necessary, exactly like in the linear case, that is why I posted this question.

I thought about this problem because often in PDE applications one has data-to-solution maps which are analytic between two Banach spaces, and often one can prove some additional properties for the solutions, so it is natural to ask whether the map is still analytic when considered with the new, restricted range (to be fair, one can usually prove this kind of things by hand looking at the specific problem, without abstract tools, so the interest in the above question is mainly motivated by curiosity).

I would guess that more than one person knows the answer to the above question, but it also looks a little far from the well-known facts about analytic/holomorphic maps between Banach spaces. If anyone could point me to a solution, a counterexample, or useful references, it would be great.

Answer 1

I will summarize what has been said in the comments (thanks to Jochen Glueck for all his help).

The answer to the question is no, in general. What is going on is actually very simple.

Theorem. Let $E,F,G$ be complex Banach spaces, such that $F\subset G$ with continuous embedding. Let $U\subset E$ open (edit: it might require some assumptions on $U$, see my first comment). The following are equivalent:

1. All analytic maps $$ f:U\to G $$ such that $f(E)\subset F$ are also analytic as maps $f:U\to F$.
2. $F\subset G$ is closed.

In other words, if $E,F,G$ are complex Banach spaces, the analyticity upgrade works for every function $f$ as above (i.e., the answer to the question is "yes") if and only if $F\subset G$ is closed.

The case where $E=\mathbb C$ is proved in this article by Arendt and Nikolsky. More precisely, Theorem 1.6 proves the non-trivial implication "$1\implies 2$" by constructing a counterexample (see Theorem 1.5), while the easy one (i.e., the analyticity upgrade holds if $F\subset G$ is closed), follows from one of the previous theorems of the papaer, or can be proved easily by hand.

The counterexample when $E$ has higher dimension can be obtained by the previous case simply by extending the 1D counterexample trivially in a suitable codimension 1 subspace of $E$ (recall that finite-dimensional spaces are close and complemented in Banach spaces, so everything works nicely).

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The real case is not covered by the cited paper. I am confident that the same result holds for real Banach spaces, and a proof might follow from the complex case by complexification of the spaces, but I did not check carefully this part. Again, it is not hard to verify that analyticity upgrade holds for all functions $f$ whenever $F\subset G$ is closed, so the other implication is the one left to be verified (I don't have time now to verify; comments are welcome on this).