I just learned that real analytic functions (by real analytic, I mean functions $f: \mathbb{R} \to \mathbb{R}$ which admit a local Taylor series expansion around any point $p \in \mathbb{R}$) cannot be extended to complex entire function always. I believe functions with this extension property are called real entire functions in some books, and a function which is real analytic, but not real entire is $f(x) = \frac{1}{1 + x^2}$. Clearly, with this example, the problem with any extension happens around $\pm i$. See also this MSE post.

My question is, do real analytic functions admit extensions to a complex analytic function even locally? That is, given a real analytic function $f : \mathbb{R} \to \mathbb{R}$, can we find a complex analytic function $g : \Omega \to C$, such that $g|_\mathbb{R} = f$, and $\Omega$ contains a strip $\mathbb{R} \times (-\varepsilon, \varepsilon)$ around the real axis?

  • $\begingroup$ to understand all this you'll need the concept of analytic continuation : $\sum_{k=0}^\infty z^k$ is analytic on $|z| < 1$, but you can re-compute its Taylor series at $z = i/2$, which is analytic on $|z-i| < |i-1|$, this way from a disk to another, you'll get that the analytic continuation of $\sum_{k=0}^\infty z^k$ is $\frac{1}{1-z}$. and a real analytic function is complex analytic (on the real line) if and only if its Taylor series at some point is the analytic continuation of its Taylor series at another point. got it ? $\endgroup$
    – reuns
    Commented Apr 27, 2016 at 14:43

3 Answers 3


Yes, a real analytic function on $\Bbb R$ extends locally to a complex analytic function, except that (in my opinion) "locally" doesn't/shouldn't mean what you say it does.

If $f$ is real analytic on $\Bbb R$ then there exists an open set $\Omega\subset\Bbb C$ with $\Bbb R\subset\Omega$, such that $f$ extends to a function complex-analytic in $\Omega$. This is easy to show - details on request. But $f$ need not extend to a set $\Omega$ that contains some strip $\Bbb R\times(-\epsilon,\epsilon)$.

For example consider $$f(t)=\sum_{n=1}^\infty a_n\frac{1}{n^2(n-t)^2+1},$$where $a_n>0$ tends to $0$ fast enough. The extension will have poles at $n+i/n$, so $\Omega$ cannot contain that horizontal strip.

  • 1
    $\begingroup$ Ah, you beat me to it. We each provided the details to one of the claims we made (I explained how to construct the function, you explained the neighborhood is not necessarily a rectangle) so hopefully both answers will prove useful. $\endgroup$
    – Jason
    Commented Apr 27, 2016 at 14:29
  • $\begingroup$ @David: Is it obvious that $f$ is analytic? $\endgroup$ Commented Apr 28, 2016 at 4:26
  • $\begingroup$ @MartinArgerami Certainly. It's clear that "if $a_n\to0$ fast enough" then the series converges uniformly on compact subsets of $\Bbb C\setminus\{n\pm i/n\}$. Further, if $n$ is fixed the sum of all the terms other than the $n$-th term converges uniformly near $n+i/n$, showing that the sum indeed has a pole at $n+i/n$. $\endgroup$ Commented Apr 28, 2016 at 12:59
  • $\begingroup$ @David: are you claiming that every uniformly convergent series of functions is analytic? I don't think that's the case. $\endgroup$ Commented Apr 28, 2016 at 14:02
  • $\begingroup$ @MartinArgerami What? Of course the limit of a sequence of functions that converges uniformly on compact subsets of an open set in the plane is analytic! (No, I'm not claiming that a uniform limit of real-analytic functions is analytic - if you think I'm claiming that you should read my previous comment more carefully.) $\endgroup$ Commented Apr 28, 2016 at 14:10

We indeed have a local extension, given by the Taylor expansion. That is, given $x_0\in\mathbb R$ there exists $\epsilon>0$ such that $$\sum_{n=0}^\infty\frac{f^{(n)}(x_0)}{n!}(z-x_0)^n$$ converges for $|z-x_0|<\epsilon$, and we call the limit $g(z)$. This defines an analytic function $g$ on a neighborhood of $\mathbb R$ which clearly extends $f$. (We do need to verify that if $z$ is in the disk of convergence of two points $x_0$ and $\tilde x_0$ then the series above is equal to the same series with $x_0$ replaced by $\tilde x_0$, but this follows from the identity theorem.)

It is worth pointing out that the $\epsilon>0$ described above depends on $x_0$, and there does not necessarily exist such $\epsilon$ independent of $x_0$. So in general we cannot expect the neighborhood to be of the form $\mathbb R\times(-\epsilon,\epsilon)$.


As David C. Ullrich and Jason pointed out, the answer is not always affirmative. Let me quote a Theorem that gives a partial affirmative answer to your question, with somewhat different assumptions.

Theorem: Assume that $f$ satisfies the Fourier inversion formula, i.e. $f(x) = \int_{\mathbb{R}}\hat{f}(\xi)e^{2\pi i x\xi}\,d\xi$ and that $$|\hat{f}(\xi)| \le Ae^{-2\pi a|\xi|}$$ for some constants $a,A > 0$. Then $f$ is the restriction to $\mathbb{R}$ of a function $f(z)$ holomorphic in the strip $S_b = \{z \in \mathbb{C}: |\text{Im}(z)| < b\}$, for any $0 < b < a$.

I think this is an interesting result since it also tells how much you can push the $\epsilon$ is you question. You can find a proof of this theorem and further discussions on the possibility of extending functions on the real line to complex-analytic functions in Complex Analysis by E.M. Stein and R. Shakarchi.


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