Proof verification of non vanishing of $ ~L(1, \chi) \neq 0~$ for real valued character

Question

I am self studying analytic number theory from Tom M Apostol introduction to analytic number theory and I am asking for solution verification for a part of Theorem 6.20 of Apostol.

I am adding it's image -

I have only doubt in proving that $L(1, \chi) \neq 0$ .

My attempt - Assuming Part (b) , if I put $x\to\infty$ in part (b) , then if $ L(1, \chi ) =0 $ for some $n$, then $B(x) \to \infty = O(1)$ , which is a contradiction.

Is my proof right? Can someone please verify.

Edit 1 -> This is the proof given in Apostol Introduction to analytic number theory. Please note that he doesn't proves how L(1 , $\chi ) \neq $ 0 . That's why I am asking for verification of my argument.

Answer 1

Yes, assuming (b), if L(1,X)=0 then B(x) = O(1), and this is in contradiction with (a). So given these results (a) and (b) it follows that L(1,X) is not zero.

[Edit: Originally I mistyped this as "B(x) = O(x)"]

Answer 2

Why is $B(x)=O(1)$ a contradiction. What can you say about $A(n)$ (hint : look at $A(p^k)$ or at the Euler product of $\sum_n A(n)n^{-s}$, you'll find that $A(n)\ge 0$ and for $p> q$, $A(p^2)\ge 1$, since $\sum_p \frac1p \approx \lim_{s\to 1} \log \zeta(s)=\infty$ we get $\sum_n \frac{A(n)}{n^{1/2}}\ge \sum_{p> q} \frac{A(p^2)}{p}=\infty$ )

Also it is worth noticing that for complex Dirichlet characters $\bmod q$, if $L(1,\chi)=0$ then so does $L(1,\overline{\chi})=0$ thus $\prod_{\chi\bmod q} L(s,\chi)$ has at least a simple zero at $s=1$ which is a contradiction because its $\log $ is $\sum_{p^k} \frac{p^{-sk}}{k} \sum_{\chi\bmod q} \chi(p^k)=\varphi(q)\sum_{p^k\equiv 1\bmod q} \frac{p^{-sk}}{k} $ which is non-negative at $s=1+\epsilon$.