Turán proof that constant sign of Liouville function implies RH In Mat.-Fys. Medd. XXIV (1948) Paul Turán gives what he says is a proof of the statement that if the summatory $L(x) = \sum_{n\leq x} \lambda(n)$ of the Liouville function $\lambda(n) = (-1)^{\Omega(n)}$ eventually has constant sign for $x \geq x_0,$ then the Riemann Hypothesis follows. Where $\Omega(n)$ is the number of not necessarily distinct prime factors of $n.$
It appears to me that Turán first gives the Dirichlet series of $\lambda$ as $D\lambda(s) = 
\zeta(2s)/\zeta(s)$ (though in terms of argument $s+1$ in place of $s$), whenever Re$(s) > 1,$ which is correct.
Then, to paraphrase, he applies the theorem by Landau (Math. Ann. 61(1) (1905)) which in this case implies that if the summatory $L$ has constant sign for $x \geq x_0,$ then $D\lambda(s)$ is defined (and regular) for all $s$ with Re$(s) > \sigma_0,$ if only $D\lambda(\sigma)$ is defined for all $\sigma > \sigma_0.$
He seems to continue by arguing that since $\zeta(2s)/\zeta(s)$ is defined for all real values $s$ between $1/2$ and $1,$ then $D\lambda$ is also defined in the same real interval. And then he concludes $\sigma_0 = 1/2.$ So that the $\zeta$ function cannot have any zeros $s$ for which Re$(s) > 1/2.$
I have seen the argument reproduced in different forms in several articles since then. But I cannot figure out how Landau's theorem can be used in this way, since the identity
$D\lambda(s) = \zeta(2s)/\zeta(s)$ is only valid for Re$(s) > 1,$ and cannot be applied in the case Re$(s) < 1,$ even if restricted to real-valued arguments.
Can anybody confirm whether there is a problem here or not?
 A: A comment that got too long: to make @reuns comments clearer note that Landau theorem says that if an analytic function $f$ is given on some half plane, by a Dirichlet series or integral with (eventually) constant sign coefficients or integrand, then if $\alpha \in \mathbb R$ is the abscissa of convergence, then $f$ has a singularity as an analytic function at $\alpha$.
Now $\zeta(2s)/\zeta(s)$ is analytic outside the zeroes of $\zeta$ which are non-real when $\Re s >0$ and of $1/2$, the pole of the numerator and it is given by a Dirichlet integral with abscissa $0 \le \alpha \le 1$ since we know it converges absolutely precisely for $\Re s >1$ and we know hence that the abscissa of convergence which is at most and within $1$ of the abscissa of absolute convergence is between $0$ and $1$; if its integrand was eventually of constant sign, the first singularity (going right to left) on the real axis would be $\alpha$ and since said singularity is known apriori to be $1/2$ that would give $\alpha=1/2$ hence $\zeta(2s)/\zeta(s)$ analytic for $\Re s >1/2$ by general Dirichlet series theory, hence RH true.
Edit later - since there seems to be some confusion still, let's note that if $f(s)=\sum a_n n^{-s}$ with abscissa $\alpha_0 \ge 1$ of absolute convergence and if $A(x)=\sum_{n \le x}A(n)$ then integrating by parts shows that $f(s)=s\int_{1^-}^{\infty}\frac{A(x)dx}{x^{s+1}}$ and the representation is valid wherever the integral on RHS converges; but now that integral has an abscissa of convergence $\alpha$ and if $A(x) \ge 0, x \ge x_0$ and $\alpha > 0$ Landau's theorem implies that $f$ (which is then defined and analytic on $\Re s >\alpha$ by general theory) has a singularity at $\alpha$ (note that if $\alpha=0$ we only know that $f(s)/s$ has a singularity at $0$)
So if we apriori know $f$ analytic on a larger domain than $\Re s > \alpha_0$ and if we know that $A(x) \ge 0, x \ge x_0$ we can find the abscissa of convergence of the Dirichlet integral by looking at the singularities of $f$ on the real axis, so, in particular, $f$ must be analytic and representable by the Dirichlet integral up to the first such singularity (or to $0$ if there is no singularity on the real axis as in the case of $\eta$ which has nonnegative summary function too but no singularities, however $\eta(s)/s$ does have such at $0$ so no contradiction with Landau theorem)
There is also a subtle point, namely that even if the Dirichlet integral above $f(s)=s\int_{1^-}^{\infty}\frac{A(x)dx}{x^{s+1}}$ converges on $\Re s >\alpha$, it doesn't necessarily follows that the Dirichlet series $\sum a_n n^{-s}$ converges on $\Re s >\alpha$ since that requires also $A(x)/x^{\alpha+\epsilon} \to 0$ by summation by parts.
Edit 2 per comments: $f(s)/s=\zeta(2s)/(s\zeta(s))$ is analytic on some domain $U$ and is given by a Dirichlet integral on a maximal open half-plane $V=\Re s >\alpha, V \subset U$.
Apriori we know that $U$ is the plane minus the zeroes of $\zeta$ and $0,1/2$ and $V$ contains the half plane $\Re s >1$ and is contained in $\Re s >1/2$ (since $f$ singular at $1/2$), and we also know that the first real point from the right that is not in $U$ is $1/2$.
We know that the Dirichlet integral converges to $f(s)/s$ on $\Re s > \alpha$ where $\alpha$ is the abscissa of the convergence of the integral and we know that $f(s)/s$ has a singularity at $\alpha$ by Landau.
If $\alpha >1/2$ then it would follow that $f(s)/s$ is defined by the Dirichlet integral on $\Re s >\alpha$ AND has a singularity at $\alpha$, SO $\alpha \notin U$ which we know it's not true.
This means that $\alpha =1/2$ so $V={\Re s >1/2}$ and $U$ contains $V$
