Let $V = \big\{z: |z|<5,\text{Im}(z)>0 \big\}$. Let $f$ analytic in $V$, continuous in $\overline{V}$ and suppose $$\forall x \in \left[ -5,5\right]:\ f\left( x\right) \in \mathbb{R}$$ Show that $$\limsup_{n \rightarrow \infty} \root{n}\of{\frac{f^{(n)}(1)}{n!}} \le \frac{1}{4}$$

I tried expanding to power series near $z=1$ but it's impossible since $1\in \partial V$, then I tried device an analytic expansion of $f$ to the disc $|z| \leq 5$ by defining $$g(z) = \cases{f(z), \qquad &\text{Im}(z) \geq0 \\ -f( \overline{z}), \qquad &\text{Im}(z) < 0}$$ But I cant show that $g$ is analytic in the whole disc. How can I show that $f$ is holomorphicaly extendible to the whole disk of radius 5?

  • $\begingroup$ I think you perhaps mean $$\limsup_{n \rightarrow \infty} \root{n}\of{|\frac{f^{(n)}(1)}{n!}|} \le \frac{1}{4}$$ $\endgroup$
    – Hmm.
    Jun 11, 2016 at 12:04
  • $\begingroup$ Define, $$g(z) = \cases{f(z) \qquad ,Imz \geq0 \\ \overline f( \overline{z}) \qquad, Imz \leq 0}$$ $\endgroup$
    – Hmm.
    Jun 11, 2016 at 12:06
  • $\begingroup$ You might consider answering your question instead of posting the answer in the question. $\endgroup$
    – user99914
    Jun 12, 2016 at 10:09
  • 1
    $\begingroup$ @JohnMa , just did. Sorry $\endgroup$
    – Uria Mor
    Jun 12, 2016 at 10:41

2 Answers 2


Let $V'$ be the complex conjugate of $V$, $\gamma$ the path following the border of $V$ in a direct orientation, $\gamma'$ the path following the border of the $V'$ also in a direct orientation, and $W$ the open ball of radius $5$ centered at $0$ (so that $\overline W = \overline V \cup \overline {V'}$)

Then we can define an continuous extension $g$ of $f$ to $\overline W$ by $g(z) = \overline {f(\overline z)}$ for $z \in \overline {V'}$.

Since $g$ is continuous on $\overline W$ and analytic on $V$ and $V'$, for any $z \in V$ you have by the residue theorem the equalities $2i\pi g(z) = \int_\gamma g(w)dw/(w-z)$ and $0 = \int_{\gamma'} g(w)dw/(w-z) $.

Adding the two integrals you get that $2i\pi g(z) = \int_C g(w)dw/(w-z)$ where $C$ is the circle of radius $5$.

The same result is true if $z \in V'$, and because both sides extend continuously to $W$, the result is also true for $z \in W$, and this shows that $g$ is analytic on $W$.

  • $\begingroup$ Maybe you could write that $\int_{\gamma_\epsilon} +\int_{\overline{\gamma_\epsilon}} = \int_{|z| = 5-\epsilon}$ and say that $\gamma_\epsilon$ is $\epsilon$ smaller than the border of $V$ $\endgroup$
    – reuns
    Jun 11, 2016 at 13:22
  • $\begingroup$ and you need to use the continuity of $g$ at $Im(z) = 0$ when letting $\epsilon \to 0$ on that side of the contour $\endgroup$
    – reuns
    Jun 11, 2016 at 13:29
  • $\begingroup$ Oh... that's Morera's Theorem right? Beautiful! $\endgroup$
    – Uria Mor
    Jun 11, 2016 at 13:44
  • 1
    $\begingroup$ @UriaMor not exactly, but now that you mention it, Morera's theorem seems particularly suited to prove $g$'s analycity. $\endgroup$
    – mercio
    Jun 11, 2016 at 15:41
  • $\begingroup$ Well, our dear prof. Gluskin used some pretty similar notions to prove Morera in class, hence my confusion. Nevertheless, an accurate and elegant solution! Thank you! $\endgroup$
    – Uria Mor
    Jun 11, 2016 at 16:03

This is a slightly different approach to solve it using Morera's Theorem.

Of course the point in this exercise was to find an analytic continuation of $f$ to the disk, and once that is done, one can simply expand $f$ around $z=1$, to a power series $\sum \frac{f^{(n)}(1)}{n!}(z-1)^n$ and since $f\in Hol(D(0,5))$ we get $f\in Hol(D(1,4))$ which implies that $\frac{f^{(n)}(1)}{n!}(z-1)^n$ converges uniformly in a disc of radius $4$ around $1$, and by Cauchy-Hadamard rule: $$\limsup_{n \rightarrow \infty} \root{n}\of{\frac{f^{(n)}(1)}{n!}} \le \frac{1}{4}$$
So, as discussed with mercio, the "obvious" way to exted $f$ is by defining: $$g(z) = \cases{f(z), \qquad &\text{Im}(z) \geq0 \\ \overline{f( \overline{z})}, \qquad &\text{Im}(z) < 0}$$

and the "classic" way to show $g\in Hol(D(0,5))$ is using Morera's Theorem:

Let $f(z)$ be continuous function on domain D. If $\int_{\partial R}f(z)dz = 0 \ $ for every closed rectangle $R$ contained in $D$ with sides parallel to the coordinate axes, then $f$ is analytic on $D$
Complex Analysis, T.W Gamelin, chapter IV.6

Now let $R \subset D(0,5)$ closed rectangle with sides parallel to the coordinate axes.
case 1: If $R$ contained in one of the upper/lower half of the plane, than by Cauchy's theorem for $f(z)$ and $\overline{f(\overline{z})}$ on the matching domains we get what we need.
case 2: If $R$ contains both positive and negative values of the imaginary axis, we write $\partial R$ = $\partial R_1 + \partial R_2$ where $\partial R_1$ is the counter clockwise directed curve on the boundary of $R \cap \{ z | Imz \geq 0 \}$, and $\partial R_2$ is the counter clockwise directed curve on the boundary of $R \cap \{ z | Imz \leq 0 \}$. And by case 1 we get: $$\int_{\partial R}g(z)dz = \int_{\partial R_1}g(z)dz +\int_{\partial R_2}g(z)dz =0$$
Hence by Morera - $g\in Hol(D(0,5))$ and on the real axis $g \equiv f$. Getting the bound for the limit is trivial.


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