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. – Kevin Jul 10 '17 at 16:42

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.