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6
votes
Accepted
Analytic continuation to the Mittag-Leffler star using Mittag-Leffler summation
G. H. Hardy, Divergent series (Oxford, Clarendon Press, 1956)
discusses and proves this theorem, see p. 79 and 190-191.
6
votes
Functional equation of bounded analytic functions
Every bounded analytic function $h$ in the disk has the representation
$$h(z)=B(z)\exp(-P(z)),$$
where $B$ is a Blaschke product and $P$ has positive imaginary part. Applying this to $h=f^3=g^2$, we c …
10
votes
Accepted
What is the relationship between $\sum_{n=0}^\infty f(n) x^n$ and $-\sum_{n=1}^\infty f(-n) ...
First of all, to make sense of $f(-n)$ we need some assumptions about $f$. For example,
let
$$
\sum_{n=0}^\infty f(n)z^n \label{1}\tag{1}
$$ be a series with positive radius of convergence. Then there …
3
votes
Accepted
Quantitative analytic continuation estimate for functions small except on a small set
This conjecture is correct. Take $K=e$, and let $\gamma\leq 1/4$; we will fix $\gamma$ later.
First we give a crude estimate of $c_0$.
Let $g(z)=\sum_{1}^\infty c_nz^n.$ Since $|c_n|\leq e^n$,
we obta …
7
votes
Quantitative analytic continuation estimate for a function small on a set of positive measure
The answer depends on $C$. For example, for $C=1$ it is positive. Your estimate $|f^{(m)}(0)|\leq m!$ implies that $|f_n(z)|\leq 1/(1-|z|).$ Take $|z|=1/2$,
you conclude that $|f_n(z)|\leq 2,\; |z|<1/ …
2
votes
The monodromy in the proof of Little Picard via Klein's $J$
The usual proof of Picard's theorem along these lines used another
modular function which is called $\lambda$ and which is related to $J$
by
$$J=\frac{4}{27}\frac{(1-\lambda+\lambda^2)^3}{\lambda^2(1- …
8
votes
Accepted
On a variant of Carlson’s theorem
Condition
$$\lim_{n\to\infty}\frac{\log|c_n|}{n}=-\infty$$
is sufficient for $f=0$.
Since $f(z)=e^{-cz}$ and $c_n=e^{-cn}$ satisfy all
conditions, we see that this is best possible in certain sense.
T …
10
votes
Accepted
An extension of the Carlson's theorem in complex analysis
If $f$ is bounded on the imaginary line, (and has exponential type) then
$f$ has completely regular growth in the sense of Levin-Pfluger, with indicator
$c|\cos\theta|$. This implies that density of z …
4
votes
Does this method analytically continue gap series series?
For $k=1$ your formula indeed gives an analytic continuation, but for $n\geq 3$, it is known that your function $f$ has no analytic continuation (the unit circle is the natural boundary of your functi …
29
votes
An "analytic continuation" of power series coefficients
It is easy to see that for every convergent power series
$$f(z)=\sum_{n=0}^\infty a_nz^n$$
there exists an entire function of exponential type $F$ which interpolates
the coefficients: $F(n)=a_n$. This …
4
votes
Analytic continuation of holomorphic functions
To answer your specific questions, a) a maximal (by inclusion) domain does not have to exist. Consider $\sqrt{1-z}$ in the unit disk. That is actually the reason why Riemann introduced Riemann surface …
10
votes
Accepted
Good book on analytic continuation?
English books are Hardy, Divergent series,
and P. Dienes, Taylor series:
an introduction to the theory of functions of a complex variable. Dover Publications, Inc., New York, 1957. (The title is somew …
3
votes
Non-analyticity of convolution
Your function $g$ in $C_0^\infty$, if not identicaly equal to zero, certainly fails to be
real analytic at some point. Because a real analytic function cannot be in $C_0^\infty$.
Your convolution is i …