# Analytic continuation on the disk

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?
Help

• I think you perhaps mean $$\limsup_{n \rightarrow \infty} \root{n}\of{|\frac{f^{(n)}(1)}{n!}|} \le \frac{1}{4}$$
– Hmm.
Commented Jun 11, 2016 at 12:04
• Define, $$g(z) = \cases{f(z) \qquad ,Imz \geq0 \\ \overline f( \overline{z}) \qquad, Imz \leq 0}$$
– Hmm.
Commented Jun 11, 2016 at 12:06
– user99914
Commented Jun 12, 2016 at 10:09
• @JohnMa , just did. Sorry Commented Jun 12, 2016 at 10:41

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 $W$.

• 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 $V$ Commented Jun 11, 2016 at 13:22
• and you need to use the continuity of $g$ at $Im(z) = 0$ when letting $\epsilon \to 0$ on that side of the contour Commented Jun 11, 2016 at 13:29
• Oh... that's Morera's Theorem right? Beautiful! Commented Jun 11, 2016 at 13:44
• @UriaMor not exactly, but now that you mention it, Morera's theorem seems particularly suited to prove $g$'s analycity. Commented Jun 11, 2016 at 15:41
• 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! Commented 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.