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.

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  • 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 01:04
  • 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 01:05
  • 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 01:06
  • @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 01:08
  • Oooops sorry, I am editing it. What a terrible sentence – ThePortakal Jun 20 '14 at 01:08
  • Isn't this just the identity theorem for real functions? I.e http://math.stackexchange.com/questions/739476/the-identity-theorem-for-real-analytic-functions – YoTengoUnLCD Apr 29 '16 at 00:27
  • @YoTengoUnLCD ... Uniqueness of the extension (if it exists) is the identity theorem. But he is asking about existence of the extension. – GEdgar Apr 29 '16 at 01:00
  • @GEdgar Oh, I see, my bad. Disregard my comment then :-P. – YoTengoUnLCD Apr 29 '16 at 01:20

2 Answers2


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 $1$.

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  • 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
  • @JohnFernley ... That is the "identity theorem" mentioned before. If two analytic funtions defined on a connected domain agree on a set with a limit point (for example, the real line is such a set), then they agree on the whole domain. – GEdgar Apr 29 '16 at 14:32
  • that's quite straightforward, thanks. I had forgotten that condition – John Fernley Apr 29 '16 at 14:33
  • Is the class of real-valued functions extensible to entire functions a good class to "approximate" the class of analytic functions? And for infinite differentiable functions? – Canjioh Dec 09 '20 at 21:23

I think that most mathematicians would say that the functions you mention are restrictions to the real line of functions more naturally defined on the complex plane in the first place. So--yes.

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