# Analytic continuation of a real function

I know that for $U \subset _{open} \mathbb{C}$, if a function $f$ is analytic on $U$ and if $f$ can be extended to the whole complex plane, this extension is unique.

Now i am wondering if this is true for real functions. I mean, if $f: \mathbb{R} \to \mathbb{R}$, when is it true that there is an analytic $g$ whose restriction to $\mathbb{R}$ coincides with $f$ and also when is $g$ unique.

Surely $f$ needs to be differentiable but this might not be sufficient for existance of such $g$.

edit: I mean, is it easy to see that there is and extension of sine cosine and exponential real functions?

Thanks a lot.

• Wait--what? That's not true, is it? Are you talking about analytic functions $\mathbb C\rightarrow\mathbb C$? What about $z\mapsto 1/z$ on a small disc around $z=1$? Or $\log z$ in the same disc? – MPW Jun 20 '14 at 1:04
• This is not nearly a strong enough condition. You need that $f$ must be real analytic to even hope to analytically continue it. – Cameron Williams Jun 20 '14 at 1:05
• Your first sentence is terribly false. And, yes, you need $f$ to be real analytic to have a hope. But what do you do with $f(x)=\dfrac1{1+x^2}$? – Ted Shifrin Jun 20 '14 at 1:06
• @TedShifrin: Yes, agreed! That's what I meant in my comment. That can't possibly be correct. Analyticity is a rather rigid condition. – MPW Jun 20 '14 at 1:08
• Oooops sorry, I am editing it. What a terrible sentence – ThePortakal Jun 20 '14 at 1:08

Necessary and sufficient condition for existence of an entire function $g$ extending $f$ to the whole complex plane: $f$ is infinitely differentiable at $0$, and the power series for $f$ at the origin converges to $f$ on the real line.
A counterexample for something more is (as noted in a comment) $f(x) = 1/(1+x^2)$. It is real analytic on the real line, but cannot be extended analytically to any connected region containing both $0$ and either $i$ or $-i$. The power series at $0$ only has radius of convergence $1$.
• Is there an example of a function on $\mathbb{R}$ having a nonunique extension to $\mathbb{C}$? – John Fernley Apr 29 '16 at 14:29
• say if a holomorphic function is identically $0$ on $\mathbb{R}$ is it $0$ on the whole plane? The dirichlet problem implies this is true but I wasn't confident as $\mathbb{R}$ is closed – John Fernley Apr 29 '16 at 14:32