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Taking the complex argument of the complex number $2$ as $0$, I've computed for complex numbers $s=x+iy$ $$1-2^{1-s}=1-2^{1-x}(\cos(y\log 2)-i\sin(y\log 2)),$$ in the equation $$\frac{\eta(s)}{\zeta(s)}=1-2^{1-s}$$ that I believe is valid in the (critical) strip $0<\Re s<1$ for Dirichlet eta function $\eta(n)$, and Riemann zeta function $\zeta(s)$, see for example [1].

Then from $$\frac{\eta(s)}{\zeta(s)}=u(x,y)+iv(x,y)$$ where $u(x,y)=1-2^{1-x}\cos(y\log 2)$ and $v(x,y)=2^{1-x}\sin(y\log 2)$, I've computed the Cauchy-Riemann equations for such functions.

Question. Can I claim thus that $$\frac{\eta(s)}{\zeta(s)}$$ is a complex differentiable function, that is a holomorphic function, in all critical strip $0<\Re s<1$ except in the points such that $\zeta(s)=0$? Or are there more points in doubt? Thanks in advance.

After I've read the section Physical interpretation in [2], I ask to me if it is possible to prove that the gradient $$\nabla u=\frac{\partial u}{\partial x}\mathbf{i}+\frac{\partial u}{\partial y}\mathbf{j}$$ has bounded norm in previous cited strip.

Question. Has sense/It is possible find a bound for the norm of the gradient of the real part $u(x,y)$ in the strip $0<\Re s<1$? Or is undounded? Thanks in advance.

My goal is obtain from your answer easy facts and computations that I've can to learn.

References:

[1] Wikipedia, Dirichlet eta function https://en.wikipedia.org/wiki/Dirichlet_eta_function

[2] Wikipedia, Cauchy-Riemann equations https://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations

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  • $\begingroup$ I don't understad if for second question, this is find a possible bound on the cited unbounded open set, there is a problem in the points such that $\zeta(s)=0$. I want refresh mathematics with your answer. Thanks. $\endgroup$ – user243301 Dec 28 '15 at 10:47
  • $\begingroup$ There is a typo in the second question. The right question is Or is unbounded? $\endgroup$ – user243301 Jan 1 '16 at 10:35
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    $\begingroup$ Hi JuanLG, you should study complex analysis first : understand the Cauchy integral formula, power series representation of holomorphic functions and their congervence, analytic continuation, removable singularities, etc. Honestly, spend 1 week to really understand in depth the proof of Cauchy's integral formula and you'll nearly be a master of complex analysis (you'll be able to read any complex analysis course). After that, study for 1 year Laplace/Fourier transform and you'll be done for $\zeta(s)$ $\endgroup$ – reuns Jan 2 '16 at 23:03
  • $\begingroup$ Thanks @user1952009 I think that were my circumstances and my lack of courage to do the things in a right way. Today is this or nothing, I believe that I can improve, but it is very difficult to me, today scheduling alone a task and go on with it. Thanks. $\endgroup$ – user243301 Jan 3 '16 at 8:41
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I assume that you know that $\zeta$ is holomorphic on $\Bbb C \setminus \{1\}$, having a simple pole in $s = 1$; also, $1 - 2^{1-s}$ is holomorphic on the whole $\Bbb C$, and since $\eta = (1 - 2^{1-s}) \zeta$, then $\eta$ will be holomorphic on $\Bbb C \setminus \{1\}$. What happens in $s = 1$? In order to study this, we shall use Riemann's theorem on removable singularities:

$$\lim \limits _{z \to 1} (z-1) \eta (z) = \lim \limits _{z \to 1} (z-1) (1 - 2^{1-z}) \zeta (z) = 0$$

because $\lim \limits _{z \to 1} (z-1) \zeta (z) = 1$ (this is a classic result), whence it follows that we may extend $\eta$ by continuity to $\Bbb C$ and the extension will be holomorphic on the whole $\Bbb C$ (i.e. it will be entire). (This conclusion is also briefly stated in the introduction section of the Wikipedia page of the $\eta$ function.)

Since we have seen that $\eta$ has no poles, then the quotient $\frac \eta \zeta$ that you are investigating will be defined and holomorphic on $\Bbb C$ out of which you must extract the zeros and poles of $\zeta$ (where the quotient cannot be defined). Since the single pole of $\zeta$ is $s=1$ and this is not contained in the critical strip, your conclusion is correct: in the critical strip $\frac \eta \zeta$ is holomorphic on $\{z \in \Bbb C \mid 0 < \Re z < 1, \ \zeta (z) \ne 0 \}$.

Concerning your second question, you want to study whether $\| \nabla u \|$ is finite in the critical strip, where the norm is the usual Euclidean norm on $\Bbb R ^2$. To this end, note that

$$\frac {\partial u} {\partial x} = 2^{1-x} (\log 2) \cos(y \log 2), \quad \frac {\partial u} {\partial y} = 2^{1-x} \sin(y\log 2) (\log 2) ,$$

so, using the fact that $\sin^2 + \cos^2 = 1$,

$$\| \nabla u \| = \sqrt {\left( \frac {\partial u} {\partial x} \right)^2 + \left( \frac {\partial u} {\partial y} \right)^2} = \sqrt {2^{2-2x} (\log 2)^2 \cos ^2 (y \log 2) + 2^{2-2x} (\log 2)^2 \sin ^2 (y \log 2)} = \ \sqrt {2^{2-2x} \log^2 2} = 2^{1-x} \log 2 .$$

Since $0 < x < 1$, then $1 < 2^{1-x} <2$, so $\log 2 < \| \nabla u \| < 2 \log 2$, which shows that the gradient of $u$ is bounded in the critical strip.

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  • $\begingroup$ Very thanks much @AlexM $\endgroup$ – user243301 Dec 31 '15 at 17:26

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