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This is going to be completely obvious, but I can't seem to get a satisfying answer on my own. Any help would be much appreciated.

I thought I understood the definitions of complex vs real analytic functions, but an example I found has proved me wrong. What does it mean for a function to be a real analytic non-holomorphic function from $\mathbb{C} \rightarrow \mathbb{C}$. It seems to imply having a series expansion but I don't understand how that doesn't make it holomorphic.

baltazar
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    Real-analyticity ignores complex structure. $z\mapsto \bar z$ is real analytic -- partial derivatives of all orders exist -- but not holomorphic. – John Brevik Mar 06 '18 at 15:52
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    @JohnBrevik: No, what you defined is $C^\infty$, not real-analytic. – Moishe Kohan Mar 06 '18 at 15:54
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    Recall that you can identify $\mathbb{C}$ with $\mathbb{R}^2$. Choose a function from $\mathbb{R}^2$ to itself that can be written as a power series in $(x,y) \in \mathbb{R}^2$ and that does not satisfy the Cauchy-Riemann equations. Extra credit if you can pick a really simple function. – Hans Engler Mar 06 '18 at 15:54
  • @HansEngler So then a function that can be written a series isn't necessarily holomorphic. It was my understanding that complex analytic refers to function that can be written as a power series and holomorphic refers to those that are complex differentiable , and that complex analytic can be shown to be equivalent to holomorphic. But if there is a function that has a power series expansion and is not holomorphic than the equivalence above doesn't hold and we only have one implication holomorphic $\implies$ analytic. Is this accurate? – baltazar Mar 06 '18 at 16:45
  • What is the counter-example you found? A real analytic function, for example $f(x)=\sum a_nx_n$ defined on $(x_0-\delta,x_0 +\delta)$, can always be extended to a holomorphic function on the disk of radius $\delta$ centered at $x_0$: just use the same power series, the radius of convergence is the same whether you consider the argument to be real or complex. This is different from asking whether a power series of two real variables is always given by a power series of one complex variable; but this is not what people usually mean when asking this question. – juan arroyo Mar 06 '18 at 17:15
  • @baltazar - a function that can be written as a power series in the complex variable $z$ is indeed holomorphic (assuming the series converges etc.). But that's not the same as "real analytic" and in fact it is much more restrictive. – Hans Engler Mar 06 '18 at 21:06

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A function $f:\mathbb{R}^n \to \mathbb{R}^m$ is real-analytic if each component $f_i$ ($i = 1,\ldots,m$) can be written as a power series in the variables $x_j$ ($j = 1,\ldots,n$) with non-zero radius of convergence. The function $f:\mathbb{R}^2 \to \mathbb{R}^2$ given by $$ f(x,y) = (x,-y)$$ is hence real-analytic on account that \begin{align} f_1(x,y) = x = &\sum_{r,s = 0}^\infty a_{rs} x^r y^s \quad \text{with} \quad a_{10} = 1, \,a_{rs} = 0 \quad\text{for}\quad rs \neq 10\\ f_2(x,y) = -y = &\sum_{r,s = 0}^\infty b_{rs} x^r y^s \quad \text{with} \quad b_{01} = -1, \,b_{rs} = 0 \quad\text{for}\quad rs \neq 01 \end{align} are convergent sums for all $x,y$.

A function $g : \mathbb{C} \to \mathbb{C}$ is complex-analytic if it can be written as a power series in $z$ with non-zero radius of convergence. The function $g:\mathbb{C}\to \mathbb{C}$ given by $$g(z) = \bar{z}$$ is not complex-analytic – for instance, it doesn't satisfy the Cauchy-Riemann equations.

Finally, note that under the identification $\mathbb{R}^2 \simeq \mathbb{C}$, we have $f \simeq g$.

gj255
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