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Here is the definition of $\xi(s,\chi)$:

$\xi(s,\chi)= \left(\frac{s(s-1)}{2} \right)^{1_{\chi=1}} (q/\pi)^{\frac{s+a}{2}} \Gamma \left( \frac{s+a}{2} \right) L(s,\chi)$

Here is the definition of the Hadamard product applied to $\xi(s,\chi)$:

Since it is an entire function of order one, there exists constants $A_\chi, B_\chi \in \mathbb{C}$ such that $\xi(s,\chi)=e^{A_\chi+B_\chi s} \prod_{p} \left(1-\frac{s}{p} \right) e^{s/\rho} $

with the product running over every non-trivial zero of $L(s,\chi)$.

Why does this imply there are infinitely many non-trivial zeros for $L(s,\chi)?$

FYI: I know there non trivial zeros are in one to one correspondence with the zeros of $\xi(s,\chi)$

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The existence of the Hadamard product by itself doesn't show the function has any zeroes (it might just be an exponential!) -- you need a little more.

The idea is to study the logarithmic derivative, which by your two expressions above satisfies ($\delta_\chi = 1$ for the principal character):

$$ -\frac{\xi'}{\xi}(s;\chi) = \delta_\chi\left(\frac{1}{s}-\frac{1}{1-s}\right)-\frac{1}{2}\log\frac{q}{\pi} -\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{s+a}{2}\right) -\frac{L'}{L}(s;\chi) $$ $$ = -B_\chi - \sum_\rho \left(\frac{1}{s-\rho}+\frac{1}{\rho}\right)\,.$$

Now plug in $s=2+iT$ and examine the terms as $T\to\infty$. In the first row, all terms are bounded except for the logarithmic derivative of the Gamma function, which grows like $\log T$. On the second row, individual terms are bounded in $T$ so there must be infinitely many.

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