# A possible criterion for a function on a complex analytic space to be holomorphic

Let $$(X, \mathcal O)$$ be a reduced complex analytic space, and $$f:X \rightarrow \mathbb C$$ be a continuous function. Assume that for all holomorphic mappings $$F$$ from unit disc $$D \subset \mathbb C$$ to $$X$$ the function $$f \circ F$$ is holomorphic on $$D$$. Is it true that $$f$$ is holomorphic on $$X$$?

The inspiration for this question comes from a form of Ruckert's Nullstellensatz, that states that a for a reduced (i.e. without nilpotent elements) analytic algebra $$A$$ and a non-zero element $$s \in A$$ there exists a homomorphism $$\phi: A \rightarrow \mathcal O_1$$ such that $$\phi(s) \ne 0$$ (here $$\mathcal O_1$$ denotes the algebra of convergent power series of one variable). That is, if $$(X,\mathcal O)$$ is a complex space, which is reduced at a point $$x \in X$$, then a germ of analytic function at a point $$x$$ is completely determined by its compositions with holomorphic maps from a neighborhood of $$0$$ in $$\mathbb C$$ to $$X$$ that map $$0$$ to $$x$$. I already understood that this concept can lead to an easy proof of the maximum modulus principle: if absolute value of a holomorphic function has local maximum at $$x$$, then $$f$$ is constant in some neighborhood of $$x$$. Indeed, then compositions of $$f$$ with maps from $$D$$ to $$X$$ that map $$0$$ to $$x$$ are all contants (equal to $$f(x)$$) due to maximum modulus principle on $$D$$. Thus, the germ of $$f$$ at $$x$$ coincides with the germ of the constant function $$f(x)$$.

It is easy to see that this question (if the statement is indeed true) can lead to an easy proof of the fact that a limit of sequence of holomorphic functions that converges uniformly on compact sets is again holomorphic. The proof of this fact in the book of Gunning and Rossi is far more complicated.

Finally, it is clear that the statement obviously holds for a complex manifold due to the fact that a function on an open set in $$\mathbb C^n$$ that is holomorphic in each variable separately is holomorphic. However, I can't come up neither with a proof nor with a counterexample for this question.

• What is $f\circ F$? Did you mean to write $F\circ f$. But then how do you compose $f:X\to \mathbb{C}$ with a function defined only on $D$? In any case, my guess is that you want to use that $\mathcal{O}_X$ is a sheaf for the analytic topology. You are presumably looking at the collection of continuous maps $F|_{f^{-1}D} : f^{-1}D\to D\to \mathbb{C}$ for every open disc $D\subset \mathbb{C}$ and asking for these to be holomorphic? Is that right? (Another issue: what if your continuous map $f:X\to \mathbb{C}$ lands in the complement of the unit disc?) Commented Oct 10, 2023 at 21:19
• @AriyanJavanpeykar $F$ maps $D$ holomorphically to $X$ so it is possible to apply $F:D \rightarrow X$ and then $f:X \rightarrow \mathbb C$. I added missing information to the question (I forgot to write that $F$ maps $D$ into $X$). Commented Oct 11, 2023 at 8:00
• I suppose that my question can be formulated also in the form: if a continuous function on $X$ is holomorphic on all $1$-dimensional analytic subsets of open subsets of $X$, is it holomorphic on $X$? Commented Oct 11, 2023 at 8:02
• I see, that clarifies the question. Thank you. Question: I don't follow the last paragraph "Finally, it is clear that the statement obviously holds for a complex manifold due to the fact that a function on an open set in $\mathbb{C}^n$ that is holomorphic in each variable separately is holomorphic." Isn't every complex analytic space locally a closed subset of $\mathbb{C}^n$? Wouldn't then the same argument just apply to all complex analytic spaces? Commented Oct 11, 2023 at 8:20
• It is clear that a continuous function on an open $U \subset \mathbb C^n$ is holomorphic if for all $1$-dimensional affine subspaces $L \subset \mathbb C^n$ $f$ is holomorphic on $L \cap U$ as a $1$-variable function (more precisely, it is enough to consider only such $L$ that are parallel to one of the coordinate axis). If $X$ is a closed analytic subset of $U$, then $X$ has much less transparent $1$-dimensional subspaces than $U$ (in particular, affine $1$-dimensional subspaces $L \subset U$ usually will intersect $X$ by a $0$-dimensinal set). Commented Oct 11, 2023 at 8:33

I think this is false. Consider the analytic space $$X = \{ xy (x-y) = 0\} \subset\mathbb C^2,$$ which is the union of the three lines $$L_1 = \{x=0\}, L_2 = \{y=0\}$$ and $$L_3 = \{x=y\}$$. Let $$f:X \to \mathbb C$$ be defined by $$f|_{L_1} = 0, \quad f|_{L_2} = 0 \quad \text{and} \quad f|_{L_3} = x.$$ This $$f$$ is clearly continuous. Most curiously I claim:
Claim. The function $$f$$ is not holomorphic on $$X$$.
Proof. Suppose you could extend $$f$$ to a holomorphic map $$f: D^2 \to \mathbb C$$, where $$D^2 \subset \mathbb C^2$$ is a small open disc around $$0$$. As $$f$$ vanishes on $$L_1$$ and $$L_2$$, we may write $$f = xy g$$ for some holomorphic $$g$$. But then $$f|_{L_3} = x^2 g|_{L_3}$$ has zero of order $$\geq 2$$ in $$0$$.
Claim. For each holomorphic map $$F:D \to X$$, the composition $$f \circ F$$ is holomorphic.
Proof. Since $$D$$ is irreducible, $$F$$ factors as $$D \to L_i$$ for some $$i$$. As $$f$$ is holomorphic on each $$L_i$$, $$f \circ F$$ is also holomorphic.