What is the exact statement involving the Liouville Lambda function, which is equivalent to Riemann Hypothesis, and true iff RH is true? Can anyone cite the sources for it and/or outline its proof in brief?
Just curious, because hardly any literature available regarding it, unlike other equivalent problems of RH. Thanks in advance!
For ${\rm Re}(s) > 1$ we have $\sum_{n \geq 1} \lambda(n)/n^s = \zeta(2s)/\zeta(s)$, which is analogous to $\sum_{n \geq 1} \mu(n)/n^s = 1/\zeta(s)$ for the Mobius function. Set $L(x) = \sum_{n \leq x} \lambda(n)$, which is analogous to the function $M(x) = \sum_{n \leq x} \mu(n)$.
There is a link between the functions $\lambda$ and $\mu$: $\lambda(n) = \sum_{d^2 \mid n} \mu(n/d^2)$ for all $n \geq 1$, where the sum runs over squares dividing $n$. Show that implies $L(x) = \sum_{d \leq \sqrt{x}} M(x/d^2)$ for all $x \geq 1$, which is a link between $L$ and $M$ that lets us convert some properties of $M$ into similar properties of $L$.
Let's show RH is equivalent to the bound $L(x) = O(x^{1/2 + \varepsilon})$ for all $\varepsilon > 0$, where the $O$-constant is allowed to depend on $\varepsilon$. This is the $L$-analogue of RH being equivalent to the bound $M(x) = O(x^{1/2 + \varepsilon})$ for all $\varepsilon > 0$, which is more widely accessible in the literature.
To show RH implies $L(x) = O(x^{1/2 + \varepsilon})$ for all $\varepsilon > 0$, use the equation $L(x) = \sum_{d \leq \sqrt{x}} M(x/d^2)$ to show that if $M(x) = O(x^{1/2 + \varepsilon})$ for an $\varepsilon > 0$, then $L(x) = O(x^{1/2 + \varepsilon})$ for the same $\varepsilon$. It should be easier for you to find a proof in the literature that RH implies the estimate $M(x) = O(x^{1/2 + \varepsilon})$ for all $\varepsilon > 0$, so RH implies $L(x) = O(x^{1/2 + \varepsilon})$ for all $\varepsilon > 0$ too.
To show, conversely, that the bound $L(x) = O(x^{1/2 + \varepsilon})$ for all $\varepsilon > 0$ implies RH, start with the equation $$ \frac{\zeta(2s)}{\zeta(s)} = \sum_{n \geq 1} \frac{\lambda(n)}{n^s} = s\int_1^\infty \frac{L(x)}{x^{s+1}}\,dx $$ when ${\rm Re}(s) > 1$. We already know $\zeta(s)$ is meromorphic on $\mathbf C$, so $\zeta(2s)/\zeta(s)$ is meromorphic on $\mathbf C$. When $L(x) = O(x^{1/2 + \varepsilon})$ for an $\varepsilon > 0$, show the integral on the right side is absolutely convergent when ${\rm Re}(s) > 1/2 + \varepsilon$, which implies the right side is analytic for such $s$. Thus $\zeta(2s)/\zeta(s)$ is analytic when ${\rm Re}(s) > 1/2 + \varepsilon$. If there is an $s_0$ where $\zeta(s_0) = 0$ and ${\rm Re}(s_0) > 1/2 + \varepsilon$ then $\zeta(2s)/\zeta(s)$ has a pole at $s_0$ because $\zeta(2s)$ is analytic and nonvanishing when ${\rm Re}(s) > 1/2$. We have contradicted analyticity of $\zeta(2s)/\zeta(s)$ when ${\rm Re}(s) > 1/2 + \varepsilon$, so the bound $L(x) = O(x^{1/2 + \varepsilon})$ implies the zeta-function has no zeros with real part greater than $1/2 + \varepsilon$. So if $L(x) = O(x^{1/2 + \varepsilon})$ for all $\varepsilon > 0$, then $\zeta(s)$ has no zeros with ${\rm Re}(s) > 1/2$, which is RH.
If you have ever seen a proof that the bound $M(x) = O(x^{1/2 + \varepsilon})$ for all $\varepsilon > 0$ implies RH, you should recognize that the previous paragraph is essentially the same argument, with the role of the displayed integral identity above for ${\rm Re}(s) > 1$ replacing the identity $$ \frac{1}{\zeta(s)} = \sum_{n \geq 1} \frac{\mu(n)}{n^s} = s\int_1^\infty \frac{M(x)}{x^{s+1}}\,dx $$ for ${\rm Re}(s) > 1$.
The reason you probably don't see much written directly about relationships between RH and the Liouville function is that formulating and proving such results at a basic level is always just an exercise in adapting what is more widely written about $\mu(n)$ or $M(x)$, as I illustrated above.
Here are two things to try yourself.
Locate a proof that the prime number theorem is equivalent to $M(x) = o(x)$ and adapt that method to prove the prime number theorem is equivalent to $L(x) = o(x)$.
Locate a proof that RH is equivalent to convergence of $\sum_{n \geq 1} \mu(n)/n^s$ for ${\rm Re}(s) > 1/2$ and adapt that method to prove RH is equivalent to convergence of $\sum_{n \geq 1} \lambda(n)/n^s$ for ${\rm Re}(s) > 1/2$. This is a second equivalence between RH and the Liouville function. (The series $\sum_{n \geq 1} \lambda(n)/n^s$ and $\sum_{n \geq 1} \mu(n)/n^s$ turn out not to converge anywhere on ${\rm Re}(s) = 1/2$, which is analogous to $\sum_{n \geq 1} 1/n^s$ not converging anywhere on ${\rm Re}(s) > 1$ and is not as technical to prove as the RH equivalences.)