# separate vs joint real analyticity

Let $$f(x,y) := xy\exp\left(-\frac{1}{x^2+y^2}\right),$$ if $(x,y)\neq (0,0)$ and $f(0,0):=0$.

I read the claim that $f$ is

(a) separately real analytic on $\mathbb{R}\times\mathbb{R}$ (i.e. for each fixed $y$ the map $x\mapsto f(x,y)$ is real analytic and for each fixed $x$ the map $y\mapsto f(x,y)$ is real analytic),

(b) $C^\infty$ on $\mathbb{R}\times\mathbb{R}$,

(c) not jointly real analytic near $(0,0)$.

I see how to prove (b), but am not sure how to go about (a) and (c). Any hints would be much appreciated.

• Due to the symmetry, it suffices to see that for every fixed $y$ the function $x \mapsto f(x,y)$ is real-analytic. If $y = 0$, then $x\mapsto 0$ is plainly real-analytic. If $y\neq 0$, look at the function. Regarding (c), you know that if it were jointly real-analytic, then it would be the restriction of a holomorphic function, don't you? Aug 7 '14 at 16:04
• Another way to solve (c): The function is $o((x^2+y^2)^N)$ for every $N$. Suppose it is real-analytic at zero, and look at the lowest degree of monomials in its Taylor series to get a contradiction.
– user147263
Aug 7 '14 at 16:12
• @DanielFischer Can you give a reference for that fact? Aug 7 '14 at 16:13
• @JulienClancy Like for single-variable functions, expand it into a power series, the domain of convergence of the power series in $\mathbb{C}^n$ is open and furnishes a local holomorphic extension. The identity theorem does the rest. Aug 7 '14 at 16:27
• You're right regarding (a). Regarding (c), if you let $x$ and $y$ be complex numbers, can you see a serious problem with the argument of $\exp$ in every neighbourhood of $(0,0)$? Aug 7 '14 at 23:36

I gave a brief explanation of (c) elsewhere, but here is a different argument. Suppose $f(x,y) =\sum_{m,n=0}^\infty a_{m,n}x^m y^n$ in a neighborhood of $(0,0)$. Then $$f(x,x) = \sum_{m,n=0}^\infty a_{m,n}x^{m+n} = \sum_{k=0}^\infty c_k x^k$$ where $c_k=\sum_{m+n=k}a_{m,n}$. (Simply put, the restriction of a real-analytic function to a line is real-analytic.) Take the smallest $k$ such that $c_k\ne 0$; it must exist since $f$ is not identically zero in any neighborhood of $0$. Then $$c_k = \lim_{x\to 0}x^{-k} f(x,x) =\lim_{x\to 0}x^{2-k}\exp(-1/x^2) = 0,$$ a contradiction. (The last limit can be evaluated, for example, by substituting $u=\exp(1/x^2)$ and using L'Hôpital's rule.)