Let $\{c_k\} \subset \mathbb C$ sch that $|c_k|\leq e^{-|k|}$ $(k\in \mathbb Z).$ We also assume that $f(\mathbb R) \subset \mathbb R.$

Put $f(x):= \sum_{k\in \mathbb Z} c_k e^{ikx}$ $x\in \mathbb R.$

(We note that $f$ is periodic function on $\mathbb R,$ and its Fourier series is absolutely convergent. Thus, $f$ is a continuous periodic on $\mathbb R.$

Question: Is $f$ is real analytic at some point on $\mathbb R$? Is $g(x)= f(\arcsin (x/r))$ ($|x/r|<1$) is real analytic in the neighborhood of origin?

My thought: (1) We note that series $\sum_{k\in \mathbb Z} c_{k}e^{i k(s+it)}$... (A) converges absolutely for $|t|<1, s \in \mathbb R$? Can we say the sum of series (A) is analytic extension of $f$? (2) We also observe that Fourier coefficient of $f:$ $|\hat{f}(k)|= |c_k| \leq e^{-|k|}$, have very nice decay.

  • $\begingroup$ According to the definition both in your link and the Wikipedia's article, it seems that real-valuedness is required for the notion of real-analyticity. But if we choose $c_k = \operatorname{sign}(k) e^{-|k|}$ so that $$ f(x) = i \frac{2e\sin x}{1 - 2e\sin x + e^2}, $$ then $f$ cannot be real-valued on any non-empty open subset of $\Bbb{R}$. $\endgroup$ Mar 23, 2017 at 13:51
  • $\begingroup$ See here: math.stackexchange.com/questions/801114/… , the first answer. $\endgroup$ Mar 23, 2017 at 13:56
  • $\begingroup$ @SangchulLee: Thanks. I agree. I have edited my question. $\endgroup$
    – abcd
    Mar 23, 2017 at 14:28
  • $\begingroup$ @uniquesolution: Thanks. From that answer, I think the answer is positive. But I do not know how to prove that exercise by the book of Katznelson. Any idea? thanks.. $\endgroup$
    – abcd
    Mar 23, 2017 at 14:44
  • $\begingroup$ @SangchulLee on wikipedia, that the coefficients $a_n$ are real is a mistake. They can of course be complex $\endgroup$
    – reuns
    Mar 23, 2017 at 15:26

1 Answer 1


If $|c_k| < e^{-|k|}$ then $\sum_{k=0}^\infty c_k z^k$ and $\sum_{k=0}^\infty c_{-k} z^k$ are complex analytic on $\{z \in \mathbb{C}, |z| < e\}$ so that $\displaystyle F(z) =\sum_{k=-\infty}^\infty c_k z^k$ is complex analytic on $\{z \in \mathbb{C}, 1/e <|z| < e\}$.

Thus $f(x) = \sum_{k=-\infty}^\infty c_k e^{i kx} = F(e^{ix})$ is complex analytic on $\{ x \in \mathbb{C}, Im(x) \in (-1,1)\}$

(complex-analytic is much stronger than real-analytic)

If you start from a continuous $2\pi$ periodic function $g : \mathbb{R} \to \mathbb{C}$ such that $\hat{g}(k) = c_k$, it means that $\forall x \in \mathbb{R}, \ g(x) = f(x)$ and $g$ is real-analytic and extends to a complex analytic function $\{x \in \mathbb{C}, Im(x) \in (-1,1)\} \to \mathbb{C}$

I'm not sure why you want to look at $f(\arcsin (x/r))$, you should look at the composition of analytic functions.

  • $\begingroup$ Thanks. But $f:\mathbb R \to \mathbb R$, how can we talk of complex analytic function? $\endgroup$
    – abcd
    Mar 23, 2017 at 15:43
  • $\begingroup$ To conclude real analyticity of $f,$ do I need to go to some function $\tilde{f}$ defined on $\mathbb C$? $\endgroup$
    – abcd
    Mar 23, 2017 at 15:44
  • $\begingroup$ What is $m(x)$? $\endgroup$
    – abcd
    Mar 23, 2017 at 15:56
  • $\begingroup$ @abcd $Im(x)$ : imaginary part of complex number $x$. And see en.wikipedia.org/wiki/Radius_of_convergence explaining how a real analytic function on $(a,b)$ extends to a complex analytic function $\endgroup$
    – reuns
    Mar 23, 2017 at 16:03
  • $\begingroup$ @user192009: Thanks. Here $x\in \mathbb R, $ so $Im (x)=0$. Then why do we need $Im(x)$? $\endgroup$
    – abcd
    Mar 23, 2017 at 16:18

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