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Heath-Brown (2016) proved that, for any $\varepsilon>0$, $$\zeta(\sigma+it)\ll_\varepsilon t^{\frac{1}{2}(1-\sigma)^{3/2}+\varepsilon},\qquad 0\leq\sigma\leq 1,\qquad t\geq 1.$$ The exponent $1/2$ can …

By definition, $\tilde{f}(0)$ exists if and only if $\int_0^\infty f(x)/x\,dx$ exists. As $f(x)$ is smooth and supported on $[0,2]$, we have that that $$|f(x)|=\left|\int_0^x f'(t)\,dt\right|\leq x\su …

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 …

Per the OP's request, I prove that if the Dirichlet series of $\zeta(s)F(s)$ converges absolutely in the half-plane $\Re z > c>1/2$, and the Riemann hypothesis is true, then the Dirichlet series of $F …

The question is somewhat broad, which makes it hard to answer it. At any rate, the following consequence of the residue theorem is standard and relevant. See also the "Added" section. Theorem. Let $f$ …

Since $$c_1=\zeta'(s_0)\qquad\text{and}\qquad c_1'=\zeta'(1-\overline{s_0})=\overline{\zeta'(1-s_0)},$$ the question concerns $|\zeta'(s_0)/\zeta'(1-s_0)|$. The functional equation for $\zeta(s)$ can …

If the Riemann Hypothesis fails, then there is no such lower bound. Indeed, if $\zeta(s_0)=0$ and $\Re(s_0)>1/2$, then $$\lim_{\delta\to 0+}\Re\frac{\zeta'(s_0-\delta)}{\zeta(s_0-\delta)}=-\infty.$$ N …

If a complex function $f(s)$ has a zero at $s_0$ of order $n\geq 1$, then for suitable $r>0$, $$\frac{f'(s)}{f(s)}=\frac{n}{s-s_0}+O(1),\qquad 0<|s-s_0|<r.$$ Writing $s=s_0+\rho e^{it}$ with $\rho\in( …

The result you mention is not due to Matiyasevich-Saidak-Zvengrowsk. Instead, it appeared in Sondow-Dumitrescu: A monotonicity property of Riemann's xi function and a reformulation of the Riemann hypo …

As Peter Humphries said in a comment, the best known bound for $\sigma=1/2$ (applying to all $\chi$) is due to Petrow and Young: $$L(1/2 + it,\chi) \ll_{\varepsilon} (q(|t| + 1))^{1/6 + \varepsilon}.$ …

There is a unique Hilbert function space with that property, and you can read about it here.

It is known that $L(V,s)$ is analytic and non-vanishing in $\Re(s)\geq 1$. Equivalently, $\log L(V,s)$ is analytic in $\Re(s)\geq 1$. Note that $\log L(V,s)$ is given by an absolutely convergent Diric …

Your series diverges for any $a<1$ and $b\in\mathbb{R}$. This is clear for $b=0$, so we can assume $b>0$ (in case of $b<0$, we replace $b$ by $-b$). We shall show that the series does not satisfy Cauc …

The functions $\xi(s)$ and $$f(s):=\xi\left(\frac{s-i}{2}\right) + \xi\left(\frac{i+s}{2}\right)$$ grow faster than exponentially on the positive axis, hence they do not satisfy the first bound. This …

The Lindelöf hypothesis yields the same bound for each derivative of $\zeta(s)$ via Cauchy's formula. Indeed, let $n\in\mathbb{N}$, $\sigma\in\mathbb{R}$, $T\in(1,\infty)$, $\varepsilon\in(0,1/2)$. Le …