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I am self studying number theory from Tom M Apostol Dirichlet Series and Modular functions in number theory and I have a doubt in theorem 8.20 of the book. I am attaching images of relevant results.

The Theorem-- enter image description here

enter image description here

My doubt is in last paragraph of proof how by integral analog of Landau theorem function on left is analytic in half plane $\sigma$ > 1/2 .

Landau theorem - enter image description here

What I could deduce using Landau theorem--> The function on right is analytic. So, the function on left converges for $ \sigma$ >1 , which is not similar to result deduced by Apostol.

Can somebody please help to make deduction given in the book .

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  • $\begingroup$ Do you know Landau's Theorem? $\endgroup$
    – Conrad
    Commented Dec 16, 2019 at 21:45
  • $\begingroup$ @Conrad it's statement is given in question . Is Landau theorem something else? $\endgroup$
    – user775699
    Commented Dec 17, 2019 at 4:58
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    $\begingroup$ Landau's theorem is the statement that a Dirichlet series/integral with coefficients (integrating function) of constant sign must have a singularity at the real value on the abscissa of convergence; it is usually used in the counterpositive (no singularity at a real value and constant sign coefficients implies abscissa is strictly to the left) or to show omega results (we know the abscisssa and no singularity, hence coefficients/integrating function must change sign indefinitely, hence something cannot be always greater than something else, but they must switch order indefinitely) $\endgroup$
    – Conrad
    Commented Dec 17, 2019 at 12:50

2 Answers 2

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  • The first part is to assume that for some $\Re(s) > 1$, $L_n(s)=\sum_{n\le x}\lambda(n)n^{-s}= 0$, then take a sequence $t_k$ such that for $p\le x$, $p^{-s-it_k}\to -p^{-s}$ which implies that $n^{-s-it_k}\to\lambda(n)n^{-s}$ and hence $\zeta_n(z+s+it_k)\to L_n(z+s)$ uniformly around $z=0$ which means that for $k$ large enough $\zeta_n(s+it_k+z)$ has a zero near $z=0$.

  • The last part is that $\forall x > X,\sum_{n\le x} \lambda(n)n^{-1}\ge 0$ implies $F(s)=\int_X^\infty (\sum_{n\le x}\lambda(n)n^{-1})x^{-s}dx$ has a singularity at its abscissa of convergence $\sigma$.

    Proof : if $F$ is analytic then $F(\sigma-\epsilon) = \sum_{k\ge 0}\frac{\epsilon^k}{k!} \int_X^\infty (\sum_{n\le x}\lambda(n)n^{-1})x^{-\sigma} (\log x)^k dx$ if it is finite then we can invert $\sum,\int$ obtaining that $\int_X^\infty (\sum_{n\le x}\lambda(n)n^{-1})x^{-\sigma+\epsilon}dx$ is finite.

    The analyticity of the LHS shows $\sigma=1/2$.

I don't know about the converse, assuming the RH what can we say about the zeros of $\zeta_n$ and the sign of $\sum_{n\le x}\lambda(n)n^{-1}$ ?

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  • $\begingroup$ my doubt was how to deduce that function on left is analytic everywhere in half plane $\sigma$ >1/2 using integral analog of Landau 's theorem . But you are perhaps explaining something else. $\endgroup$
    – user775699
    Commented Dec 17, 2019 at 5:15
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    $\begingroup$ The $\ge 0$ condition implies the RHS has a singularity on the real axis at its abscissa of convergence $\sigma$. The LHS is analytic on $(1/2,\infty)$ thus it must be $\sigma=1/2$. Once the RHS converges for $\Re(s) > 1/2$ then the LHS is analytic for $\Re(s) > 1/2$ (ie. the RH is true). $\endgroup$
    – reuns
    Commented Dec 17, 2019 at 5:17
  • $\begingroup$ I understand why LHS is analytic on (1/2, $\infty $) . But why condition that C(x) is nonnegative implies that RHS has a singularity on real axis at its abcisaa of convergence? $\endgroup$
    – user775699
    Commented Dec 17, 2019 at 5:27
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    $\begingroup$ Expand $F(\sigma-\epsilon)$ in a Taylor series in $\epsilon$, every term is non-negative and their formal sum is $\int_{n_0}^\infty C(x)x^{-\sigma+\epsilon}dx$ which has to be finite $\endgroup$
    – reuns
    Commented Dec 17, 2019 at 5:30
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    $\begingroup$ It is obvious that the RHS is complex differentiable, holomorphic, analytic on its half-plane of convergence. The equality between LHS and RHS is true everywhere both sides are analytic (this is called analytic continuation) $\endgroup$
    – reuns
    Commented Dec 17, 2019 at 5:32
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This is a famous result that the partial sums $\zeta_{n}(s)$ has zeros even for $\sigma > 1$, and it must be true since many have already read the argument and bought it.

I guess the author of this question is concerned about what an analog of Landau's Theorem is.

One person has already given details using power series, and in fact the nonnegativity of the sum is used (namely, one can switch sum and integral in power series) so I may just add to it.

By singularity, there are many meanings in it.

The Dirichlet series \[ \sum \frac{1}{(\log n)^2 n^{s}} \] converges for $s = 1$, and yet viewed as a function on a disk containing a point s = 1, it has a singularity of some sort.

That is what made me puzzled when I read the same book.

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