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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 …

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 …

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 …

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 …

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 …

G. H. Hardy, Divergent series (Oxford, Clarendon Press, 1956) discusses and proves this theorem, see p. 79 and 190-191.

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 …

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 …

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 …

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 …

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- …

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/ …

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 …