# Is this true :$|\zeta(z)| \leq |z|$ for $z$ a complex number with $Re(z)$ and $Im(z)$ >0?

I would like to know more about the behavior of Riemann zeta function and it's

lower bound , after some calculations which i performed in wolfram alpha I got this result :

Result: $|\zeta(z)| \leq |z|$ for $z =\alpha+i\beta$ where $(\alpha , \beta) >$0.

My question here : Is the above result true and if it is true how do i can show it ?

Thank you for any help !!!!

• At most you may show that $$\left|(z-1)\,\zeta(z)\right|\leq \frac{1}{2}+|z|$$ holds in a neighbourhood of the positive real line, but your inequality cannot hold for trivial reasons: the harmonic series is divergent, hence $\zeta(s)$ is unbounded in a neighbourhood of $s=1$. – Jack D'Aurizio Oct 26 '16 at 0:01

Over the region $\text{Re}(s)>0$ we may define the zeta function through

$$\begin{eqnarray*}\zeta(s) = \left(1-\frac{2}{2^s}\right)^{-1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} &=&\color{green}{\left(1-\frac{2}{2^s}\right)^{-1}\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}\,dx}{e^x+1}}\\&=&\color{blue}{\frac{4^s}{2(2^s-2)\,\Gamma(s+1)}\int_{0}^{+\infty}\frac{x^s\,dx}{\cosh^2 x}}\tag{1}\end{eqnarray*}$$ where $\frac{1}{\Gamma(s)}$ and $\int_{0}^{+\infty}\frac{x^{s-1}\,dx}{e^x+1}$ are both bounded in a neighbourhood of $s=1$, but $\left(1-\frac{2}{2^s}\right)^{-1}$ is clearly not, so your inequality cannot hold. Additionally, by $(1)$ we have $\lim_{s\to 0^+}\zeta(s)=-\frac{1}{2}$.

• note that with the Abel summation formula : $\zeta(s) = \frac{s}{s-1}- \frac{1}{2}-s\int_1^\infty (x-\lfloor x \rfloor-\frac{1}{2}) x^{-s-1}dx$ where $(x-\lfloor x \rfloor-\frac{1}{2}) x^{-s-1}dx$ converges for $Re(s) > -1$ – reuns Oct 25 '16 at 23:40
• @user1952009: of course, but the OP is only interested in the behaviour over the region $\text{Re}(s)>0$. Over such a region we have a pretty concise integral representation, hence we do not really need Abel's formula to prove or disprove the claim. – Jack D'Aurizio Oct 25 '16 at 23:42

It is not true. Of course there is the pole at $s=1$ where $\zeta(s)\to \infty$, and as $s \to 0 : \frac{\zeta(s)}{s}$ is unbounded too.

Now with the Abel summation formula you get for $Re(s) > 1$ : $$\zeta(s) = \sum_{n=1}^\infty n^{-s} = s \int_1^\infty \lfloor x \rfloor x^{-s-1}dx = \frac{s}{s-1}- s\int_1^\infty (x-\lfloor x \rfloor) x^{-s-1}dx$$

Note that the RHS is analytic for $Re(s)> 0$, therefore $\zeta(s) = \frac{s}{s-1}- s\int_1^\infty (x-\lfloor x \rfloor) x^{-s-1}dx$ stays true for $Re(s) > 0$ and

$$\left|\frac{\zeta(s)}{s}\right| = \left|\frac{1}{s-1}-\int_1^\infty (x-\lfloor x \rfloor) x^{-s-1}dx\right|$$ $$\le \frac{1}{|s-1|} +\int_1^\infty x^{-Re(s)-1}dx$$ $$= \frac{1}{|s-1|} + \frac{1}{Re(s)}$$

• I think for $Re(s) >1$ :$\frac{1}{|s-1|} + \frac{1}{Re(s)} \leq 1$ and that don't disprove what i claimed above!!!!! – zeraoulia rafik Oct 25 '16 at 23:52
• @user51189: eh? Both proofs clearly show that $\zeta(s)$ is unbounded in a neighbourhood of $s=1$, hence $\left|\zeta(x)\right|\leq K|x|$ cannot hold in a neighbourhood of $x=1$, no matter what $K$ is. – Jack D'Aurizio Oct 25 '16 at 23:57
• And we are actually just stating that the harmonic series is divergent, hence your claim cannot be true. – Jack D'Aurizio Oct 25 '16 at 23:58
• @user51189 We will be glad to help you if you explain what you don't understand. Here you have 3 representations of $\zeta(s)$ that let you understand its behavior on $Re(s) > 0$. Also, take a look at en.wikipedia.org/wiki/Riemann_zeta_function – reuns Oct 26 '16 at 0:05