Denote by $\mathbb C^\infty $ the Hilbert space $\ell^2 (\mathbb C)$. Fix $1\leq N,M \leq \infty$, and let $U$ be an open subset of $\mathbb C^N $. Following Mujica's book "complex analysis in Banach spaces", a function $f:U\to \mathbb C ^M $ is called $holomorphic$ if for every $p \in U$ there is a bounded linear map $A:\mathbb C^N \to \mathbb C^M$ such that

$\displaystyle \lim_{h\to 0} \frac{||f(p+h)-f(p)-Ah||}{||h||}=0$.

In Mujica's book, theorem 8.12 says that a function $f:U\to \mathbb C^M $ is holomorphic if and only if, for every bounded linear map $\psi : \mathbb C^M \to \mathbb C$, the function $\psi \circ f$ is holomorphic.

Denote $\pi_j:\mathbb C^M \to \mathbb C$ the map that to every $z\in \mathbb C^M$ associates its $j$-th component. Then $\pi_j $ is a bounded linear map, so $f^j := \pi_j \circ f$ is holomorphic for every $j$.

My question is: is the converse true? I.e. is it true that if every $f^j$ is holomorphic then $f$ is holomorphic?

The answer is obviously yes if $M$ is finite. I think that, if $f$ is continuous, then the answer is yes also in the infinite-dimensional case. Can we drop the continuity hypotesis?

Thank you.

EDIT2: I add my proof attempt of the fact that it suffices to assume $f$ continuous. So, let $f:U\to \mathbb C^\infty$ (with $U\subseteq \mathbb C^N$ and $N\leq \infty$) continuous such that, for every $j$, $f^j$ is holomorphic. Let $\psi:\mathbb C^\infty \to \mathbb C$ be a bounded linear map: then $\psi$ acts like $z\mapsto <z,u>$ for some $u\in \mathbb C^\infty$, so $f$ is holomorphic if and only if the map $z\mapsto <f(z),u> $ from $U$ to $\mathbb C$ is holomorphic for every $u\in \mathbb C^\infty$. Now, select $u\in \mathbb c^\infty$ and denote $g_n (z) = \sum _{j=1}^{n} f^j (z) \overline u^j $ and $g=\lim_n g_n $: since every $g_n $ is holomorphic because every $f^j$ is, it suffices to prove that $g_n \to g$ uniformly on compact subsets. But, if $K$ is a compact of $U$, then $\|g_n (z) -g(z)\|\leq \|<f(z),u^{>n}>\|\leq \|f(z)\|\|u^{>n}\|\leq R_K \|u^{>n}\|$ for $z\in K$, where $u^{>n}$ is the vector of $\mathbb C^\infty$ with $j$-th component equal to $0$ if $j\leq n$, and equal to $u^j$ if $j>n$, and $R_K$ is a positive constant that bounds $\|f\|$ on $K$. Since $\|u^{>n}\|\to 0$ for $n\to \infty$, by generality of $u$, we have the thesis.

EDIT3: following this question, I found a counterexample in the case $N=\infty$ and $M=\infty$. Define $f:\mathbb C^\infty \to \mathbb C^\infty$ by setting

$f^1 (z) = z^1$;

$f^2 (z) = f^3 (z) = \frac{1}{\sqrt 2} (z^2 + z^3)$;

$f^4 (z) = f^5 (z) = f^6 (z) = \frac{1}{\sqrt 3} (z^4 + z^5 + z^6)$;

and so on. Then the Jacobian matrix of $f$ is the one given in this example, and does not represent a bounded linear operator.

However, the question still remains unanswered for finite $N$ and infinite $M$.

2more comments