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 $M$.

  • $\begingroup$ In general you can trade «continuity» for «ample boundedness», which here should translate as plain local boundedness. That being said, have you any reason to doubt Mujica's theorem? Does the proof seem gappy to you? $\endgroup$ Mar 24, 2016 at 8:03
  • $\begingroup$ @LoïcTeyssier Mujica's theorem is originally stated for general complex Banach spaces $E,F$ in place of $\mathbb C^N , \mathbb C^M$, and I do not doubt it. My question is if, in this very special case, the hypotesis can be weakened. I ask this because I am reading an old paper (1954) and the author says (more or less) that he doesn't know if what I am now asking is true, but it may be. $\endgroup$
    – Ervin
    Mar 24, 2016 at 8:44
  • $\begingroup$ Ok, sorry, I misunderstood your question (read it too fast actually). I don't know the answer. You might find it in the book "Barroso, Jorge Alberto, mrnumber = "779089", Introduction to holomorphy, North-Holland Mathematics Studies, vol 106, 1985". $\endgroup$ Mar 24, 2016 at 11:01
  • $\begingroup$ What is the argument with the assumption of continuity? $\endgroup$
    – Ben W
    Mar 25, 2016 at 18:54
  • $\begingroup$ @anonymous I added my attempt. $\endgroup$
    – Ervin
    Mar 25, 2016 at 19:03

2 Answers 2


No. There are non-holomorphic functions $f:\mathbb D\to \ell^2$ such that all components $f_n=\pi_n\circ f$ are holomorphic. This follows from a general result of Arendt and Nikolski [Vector-valued holomorphic functions revisited. Math. Z. 234 (2000), no. 4, 777–805]:

Theorem 1.5 Let $X$ be a Banach space and $W$ a subspace of $X'$ which does not determine boundedness. Then there exists a function $f : \mathbb D \to X$ which is not holomorphic such that $\varphi \circ f$ is holomorphic for all $\varphi\in W$.

Here, a subspace $W\subseteq X'$ is said to determine boundedness if, for all subsets $B\subseteq X$, the condition $\sup\lbrace |\varphi(x)|: x\in B\rbrace <\infty$ for all $\varphi\in W$ implies that $B$ is bounded. This theorem is applied to $X=\ell^2$ and the linear span $W$ of all projections $\pi_n$. It does not determine boundedness by considering $B=\lbrace ne_n:n\in\mathbb N\rbrace$ with the unit vectors $e_n\in\ell^2$.


I assume that $C^\infty=\ell^2(\mathbb C)$ means the space of square summable complex-valued sequences to you. Let me also stress that by definition validity of your conjecture in the case $N>1$ is equivalent to validity in the case $N=1$, so I will focus on this simpler case.

As a corollary of Vitali's theorem, the following criterion for holomorphy of vector-valued functions is known (Theorem A.7 in Arendt-Batty-Hieber-Neubrander, Vector-valued Laplace Transforms and Cauchy Problems, Birkhäuser 2011)

Let $\Omega\subset \mathbb C$ be open and connected, and let $f:\Omega \to X$ be a locally bounded function. Assume that $W \to X^*$ is a separating subspace such that $x^∗\circ f$ is holomorphic for all $x^∗ \in W$. Then f is holomorphic.

(If you are interested in the definition of a separating subspace you can check the book, but if $X$ is a separable Hilbert space as in your case, then $X\simeq X^*$ itself is certainly separating.)

So this proves your conjecture.

Let me also quote the following remark from the same book, which seems to suggest that your caution in not just carrying over Mujica's assertion to the infinite dimensional case was justified after all:

Usually, Vitali’s theorem is proved with the help of Montel’s theorem which is only valid in finite dimensions. A vector-valued version is proved in the book of Hille and Phillips by a quite complicated power-series argument going back to Liouville. The very simple proof given here is due to Arendt and Nikolski who also proved Theorem A.7.

  • $\begingroup$ Is the (not closed) subspace of $\mathbb C^\infty$ generated by the $\pi^j$ separating? In this case you have proved that it suffices to assume $f^j$ holomorphic and $f$ locally bounded. I think this locally boundedness and continuity are interchangeable in this case, so what if we do not suppose $f$ continuous? $\endgroup$
    – Ervin
    Mar 26, 2016 at 15:33
  • $\begingroup$ Can you give me an example of a discontinuous function each of whose coordinates are holomorphic? $\endgroup$ Mar 26, 2016 at 15:58
  • 1
    $\begingroup$ This is exactly what I am asking for (or a proof that it does not exist). $\endgroup$
    – Ervin
    Mar 26, 2016 at 16:02

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