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$s= \sigma + it$ is any complex number with real part $> 0$.

This came up because $L(s,\chi) = \zeta(s)\prod_{p | q} (1-p^{-s})$ and I have a bound for zeta I want to change to a bound for $L$ where $\chi$ is the principal character mod $q$.

Let $q$ be squarefree then $$\prod_{p | q} (1-p^{-s}) < d(q)$$ for all $s > 0$.

$d$ is the number of divisors of $q$.

I tried doing 1/ to both sides and then change the product to sum $\sum_{n\text{ is a product of primes in }q} n^{-s} > 1/d(q)$ but this makes the inequality look trivial because the LHS is $> 1$. I must be missing or misunderstanding something. Thank you for any suggestion.


im trying to get this theorem: $$\left| L(s,\chi_0) - \frac{s}{s-1}\prod_{p|q}(1-p^{-s})\right| \le d(q)\frac{|s|}{\Re(s)}$$ from the bound on the zeta function $$\left| \zeta(s) - \frac{s}{s-1} \right| \le \frac{|s|}{\Re(s)}$$ if that helps make this clearer.


As I said I already showed $L(s,\chi) = \zeta(s)\prod_{p | q} (1-p^{-s})$, show multiply both sides the zeta inequality by $\prod_{p | q} (1-p^{-s})$ and we get $$\left| L(s,\chi_0) - \frac{s}{s-1}\prod_{p|q}(1-p^{-s})\right| \le \left[ \prod_{p | q} (1-p^{-s}) \right]\frac{|s|}{\Re(s)}$$ therefore if I can just show $$\prod_{p | q} (1-p^{-s}) \le d(q)$$ I get the theorem.

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@anon, I added a third edit showing how my original question is connected to that. –  user58512 Feb 3 '13 at 16:19
    
@anon, yeah! that's what I'm confused about! I wonder why my notes have d(q) in it, and I couldn't find this theorem in my book. –  user58512 Feb 3 '13 at 16:23
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Actually, the $s$ is no longer real, so $|1-p^{-s}|<1$ may not hold in general even assuming $\mathrm{Re}(s)>0$. When you withheld the background information originally, your question made it seem like you were only interested in real $s$, which was misleading. –  anon Feb 3 '13 at 16:25
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up vote 2 down vote accepted

If $\Re(s)>0$ then $\lvert p^{-s}\rvert<1$, so $\lvert 1-p^{-s}\rvert < 2$.

Note that $d(n) \ge 2^{\omega(n)}$ where $\omega(n)$ is the number of unique prime factors of $n$. They are actually equal when $n$ is squarefree, but this is not a necessary assumption for the inequality you seek.

By the way, you misstated the inequality. What you want is $$\left| \prod_{p | q} (1-p^{-s}) \right| \le d(q),$$

since the quantity inside the absolute value is not necessarily real (neither is $s$ so "$s>0$" makes little sense).

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