I want to show that:
$$\psi(x;q,a) \asymp x$$
i.e. I want to show that $\psi(x; q, a) \ll x$ and $\psi(x; q, a) \gg x$
$\psi(x; q, a)$ is the Chebyschev's Second function in residue class a mod q.
In other words, $$\psi(x, q, a) = \sum_{n \leq x}\Lambda(n).\mathbf{1}_{q,a}(n)$$
where, $$\mathbf{1}_{q,a}(n) = \begin{cases} 1 &\quad\ n = a \text{ mod }q \\ 0 &\quad\ n \neq a \text{ mod }q \\ \end{cases}$$
and
$\Lambda$ is the von mangoldt function.
On pages 83-84 of the notes: Notes
$\psi(x) \asymp x$ has been proven. Therefore we know $\psi(x) \ll x$ and $\psi(x) \gg x$
Now,
it is simple to see that:
$$\psi(x) - \psi(x; q, a) \geq 0 \implies \psi(x; q, a) \leq \psi(x) \ll x \implies \psi(x; q, a) \ll x$$
Now, I do not know how to show $\psi(x; q, a) \gg x$
I have tried hard to solve this but I am unable to.
Any pointer towards a possible direction would be very helpful.
Thank you.
I believe that this lower bound is quite a bit more difficult than the corresponding lower bound for $\psi(x)$ itself; by way of analogy, both lower bounds imply that there are infinitely many primes being counted, but it is far harder to prove that there are infinitely many primes congruent to $a$ (mod $q$) than just to prove that there are infinitely many primes.
Unless I'm missing something, I don't think there's a substantially easier way to prove $\psi(x;q,a) \gg x$ then to prove Dirichlet's theorem (on the Dirichlet density of such primes)—for example, equation (6.8) on page 181 of the (excellent) notes you linked—and then use partial summation starting from $\psi(x;q,a) = \int_1^x t \,dS_{a,q}(t)$.