Function $f(x,y)$ analytic with respect to both $x$ and $y$, is $\frac{\partial}{\partial y} f(x,y)$ analytic with respect to $x$?

Question

Assume $U$ is some open set in $\mathbb{R}$. Let $f: U \times U \rightarrow \mathbb{R}$ be a function such that for a fixed $y$ $$x \mapsto f(x,y)$$ is analytic, and for a fixed $x$ $$y \mapsto f(x,y)$$ is analytic. Is it true that for a fixed $y$ $$x \mapsto \frac{\partial}{\partial y} f(x,y)$$ is also analytic?

Answer 1

See edit before reading.

The answer is yes but requires some work. If $x \mapsto f(x,y)$ is analytic for any fixed $y \in U$ and $y \mapsto f(x,y)$ is analytic for any fixed $x\in U$ then by Hartogs' Theorem:

https://en.wikipedia.org/wiki/Hartogs%27s_theorem

$f(x,y)$ is analytic in the 2-variable sense, i.e. it admits a power series expansion in 2 variables for $(x,y) \in U \times U$. In particular, you can differentiate this power series with respect to $y$ and obtain a power series expansion for $x \mapsto \frac{\partial}{\partial y} f(x,y)$ for $x \in U$. Hence your desired map is analytic.

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Edit: I misread the question thinking that $U$ was a subset of $\mathbb{C}$. The answer to the question is in fact no and a counter example can be found in the link to the Wikipedia article above.