Holomorphic and analytic functions.

I read this today (source: Wikipedia):

The fact that all holomorphic functions are complex analytic functions, and vice versa, is a major theorem in complex analysis.

Is there a similar result in real analysis?

• It is pretty freaky if you think about it. The fact that the first (complex) derivative exist implies 1) That all derivatives of any order exists and that 2) The taylor series converges to the value of the function at every point. It's insane! :D Even more so if you compare it to real analysis, where there are all kinds of issues with existance of derivatives, being analytical, etc.
– Ant
Jul 10 '17 at 14:21
• There are actually a huge number of these "freakishly strong" theorems in complex analysis (e.g. the Picard theorems, the Residue theorem, etc.). The field is arguably much easier than real analysis, in the sense that it's "easier" to prove very strong and interesting results. Another way of looking at it is that complex differentiability is a much stronger statement than real differentiability (because the former is two-dimensional and the latter is one-dimensional), so complex analysis is focusing on a much smaller and more "well behaved" set of functions than real analysis. Jul 10 '17 at 16:42

No. "Holomorphic" means "complex differentiable in a neighborhood of a point", "analytic" means "Taylor series at a point converges to the function in a neighborhood of that point".

For real analysis the analogue for holomorphic would be "real differentiable in a neighborhood of a point", but this condition is far too weak to imply the latter. E.g. there are functions which are only once differentiable but not twice, and even infinitely differentiable functions differentiable everywhere yet analytic nowhere.

It's worth adding to the above answer that there's a theorem which says that "holomorphic" can be ambiguous between "differentiable once" and "differentiable infinitely often" (i.e. in complex analysis the above identity of holomorphic and analytic also means that complex-differentiable-in-a-neighborhood implies complex-smooth-on-that-neighborhood).

However there also exist functions which are $C^\infty(\mathbb R^n, \mathbb R)$ and are not analytic. The most common example is $$f(x) = \exp(-1/x)\text{ if } x > 0\text{ else } 0.$$By induction the derivatives of $\exp(-1/x)$ have the form $e^{-1/x} P_n(1/x)$ for polynomial $P_n$, since we start out with that form $P_0 = 1$ and the derivative gives a simple recursive formula to calculate the next polynomial $P_{n+1}(u) = u^2~P_n(u) + P_n'(u)$ for a new polynomial. But $\lim_{u\to\infty} e^{-u}P_n(u) = 0$ and hence $\lim_{x\to0^+}f^{(n)}(x) = 0$ for all $n$, so all derivatives are continuous at the origin and this is $C^\infty(\mathbb R, \mathbb R)$ at $x=0$. However it is not analytic at $x=0$ precisely because the Taylor series at $x=0$ is the zero function, and $f$ is not the zero function.

The whole point of having a theorem like this is that there isn't a (true) real analogue.

The supposed thing you're probably looking for is something like

smooth if and only if real-analytic,

which of course is false (because of the 'only if' direction).

Also, look at the term 'real-analytic': why don't we have 'complex-analytic'?