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I have heard that a function $f: \mathbb R^2 \to \mathbb R$ can be analytic in each variable (i.e. $f(x,y_0) = \sum_{n=0}^{\infty} a_n x^n, \forall x \in \mathbb R$, and the same for $y$) without being jointly analytic (i.e. $f(x,y) = \sum_{i,j=0}^{\infty} a_i b_j x^i y^j, \forall x,y \in \mathbb R$). Is there some standard example of such a function?

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    $\begingroup$ Just for completeness: if we replace $\mathbb R$ by $\mathbb C$, the result is true! (Hartogs' theorem) $\endgroup$ Commented Jul 19, 2017 at 23:51

2 Answers 2

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$$ f(x,y) = \begin{cases} \dfrac{xy}{x^2+y^2} & \text{if $(x,y)\ne(0,0)$} \\ 0 & \text{if $(x,y) = (0,0)$} \\ \end{cases} $$ is analytic in each variable, but it is not continuous at $(0,0)$.

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  • $\begingroup$ Thanks! Indeed it was easy... but what if we add the requirement $f \in C^\infty(\mathbb R^2)$? $\endgroup$
    – a4d
    Commented May 23, 2016 at 20:12
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For completeness, I add a jointly $C^\infty$ example, taken from separate vs joint real analyticity: $$f(x,y) = xy\exp\left(-\frac{1}{x^2+y^2}\right),\qquad f(0,0)=0 \tag1$$ The restriction to each coordinate axis is identically zero. The restriction to other lines parallel to coordinate axes is real analytic since it's a composition of real analytic functions.

But $f$ is not jointly analytic, since it decays faster than any polynomial at $(0,0)$: $$\lim_{(x,y)\to (0,0)} (x^2+y^2)^{-N} f(x,y)=0 \qquad \forall N$$ (A function that is representable by a power series cannot do this, since it's asymptotic to the sum of the lowest-degree nonzero terms of the series.)


A relevant result was proved by J. Siciak in A characterization of analytic functions of n real variables, Studia Mathematica 35:3 (1970), 293-297: if a jointly $C^\infty$ function is real-analytic on every line segment, then it is jointly real-analytic. The function (1) fails the condition of Siciak's theorem, since it is not real-analytic on the line $y=x$.

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