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Based on your comment below the original post, you are mainly interested in the case of $f(s)=\zeta(s)$. Let me restrict to that case. Then, for a given $t$, your inequality is implied by Conjecture …

I suspect the answer is no in general, but if you assume local boundedness then (1) is true and, for (2), convergence is locally uniform. This is a special case of the Vitali-Porter theorem, see here. …

The phenomenon you observe is a special case of a theorem of Frobenius (1880): If a series is Cesaro summable then it is also Abel summable, and the Cesaro limit is the same as the Abel limit. …

$\newcommand\eps\varepsilon$ We want to show that, under $R\to\infty$ and $\eps\to 0+$, we have $$\int_{(-R,-\eps)\cup(\eps,R)} \frac{1-e^{itu}}{e^{itu}-1-it}\,\frac{dt}t=\pi i\,\frac u{1-u}+o(1).$$ E …

Actually this problem is from Chapter 13 (at least in my edition, see p.266). Here is a sketch. By the conditions $f$ and $g$ do not have any zeros in $K-\Omega$. In particular, zeros do not accumulat …

The answer is no. Take, for example, $a_n:=n^{-it}$ with any fixed $t\neq 0$. Then $f(s)$ converges absolutely to $\zeta(s+it)$ for any $s>1$, and $\lim_{s\downarrow 1}f(s)=\zeta(1+it)$ exists, but $f …

As Dan Petersen anticipated, any function satisfying the requirements is constant. Indeed, the function defined as $g(z):=f(z)$ for $|z|\leq 1$ and $g(z):=\overline{f(1/\overline{z})}$ for $|z|\geq 1$ …

An affirmative answer follows from (9) in this paper by Ritt.

The following argument is based on Christian Remling's proof (given in a comment), but is more elementary. Let us examine the behavior of $f(1/z)$ as $z\to 0$. The function $f(1/z)$ is algebraic over …

The inequality in the post can be rewritten as $$\left|F(z)e^{\frac{b+c}{2}z^2}\right|\leq ae^{\frac{c-b}{2}|z|^2},\qquad z\in\mathbb{C}.$$ For $c<b$ it follows that $F=0$, and for $c=b$ it follows th …

Let $\rho=1/2+iy$ be the first zero of $\zeta(s)$. Then $\zeta'(\rho)$ has nonzero real and imaginary parts, and $$\zeta(\rho+h)\sim\zeta'(\rho)h\qquad\text{as}\qquad h\to 0.$$ That is, if $x>0$ is su …

Your inequality does not make sense, since the RHS has $n$ in it, while the LHS does not. What Titchmarsh claims is the bound $$\int_{x}^{x+i\infty} (\cot(\pi z)+i)z^{-s} \mathrm{d}z \ll \frac{x^{-\si …

Well, one definitely needs to work a bit to justify the convergence of the integral at the endpoints. Apart from that, the comment by mathworker21 is on point. Namely, let us consider the holomorphic …

Theorem 1. Let $f\in\mathbb{C}[[t]]$. Then $f(zw)\in \mathbb{C}[[z]]\otimes_{\mathbb{C}}\mathbb{C}[[w]]$ if and only if $f\in\mathbb{C}[t].$ Proof. The "if" part is obvious. For the "only if" part as …

Something similar is true. Let $$ A(x):=\sum_{n\leq x}f(n),\qquad x>0,$$ then the first condition is equivalent to $$ \forall\sigma>\sigma_c:\ A(x)\ll_\sigma x^\sigma. \tag{1}$$ This clearly implies $ …