Notation: all little-o notations are as $x\to\infty$. All "$p$" denote primes. All "$\log x$" mean natural log of $x$. Given an arithmetic function $f: \mathbb N \to \mathbb C$, the (formal) Dirichlet series is $\mathcal D f(s) := \sum_{n=1}^\infty \frac{f(n)}{n^s}$.
I'll keep adding to this later, but for now, here's what I have:
The "OG" PNT
(PNT): $\sum_{p\leq x} 1 =:\pi(x) = (1+o(1)) \frac{x}{\log x}$
(PNT-dyadic): $\sum_{x<p\leq 2x} 1 =: \pi(2x)-\pi(x) = (1+o(1))\frac{x}{\log x}$
Clearly (PNT) $\implies$ (PNT-dyadic). The converse direction can be done according to Dyadic sum of $\frac{x}{\ln x}$ (i.e. dyadic asymptotic for prime number theorem).
Bringing in von-Mangoldt
- (PNT-dyadic-logweighted): $\sum_{x<p\leq 2x} \log p = (1+o(1))x$.
This is clearly equivalent to (PNT-dyadic) by the sandwich inequalities $\log x \sum_{x<p\leq 2x} 1 \leq \sum_{x<p\leq 2x} \log p \leq \log(2x) \sum_{x<p\leq 2x} 1$.
Recall that the von-Mangoldt function $\Lambda(n)$ equals $\log p$ for prime $n=p$. One can manage the prime power contributions, as e.g. I discuss in these slides (see Heuristic 1) to show that (PNT-dyadic-logweighted) is equivalent to
- (vM-PNT-dyadic): $\sum_{x<n\leq 2x} \Lambda(n) = (1+o(1))x$.
And by summing (geometric series) this is equivalent to
- (vM-PNT): $\sum_{n\leq x} \Lambda(n) = (1+o(1))x$.
Bringing in Mobius
In these blog notes, Terry Tao proves that (vM-PNT) is equivalent to
($\mu$-meanzero): $M(x)=\sum_{n\leq x} \mu(n) = o(x)$.
($\mathcal D \mu (1)$-zero): $\sum_{n\leq x} \frac{\mu(n)}n = o(1)$.
The/A key formula (for the equivalence ($\mu$-meanzero) $\iff$ ($\mathcal D \mu (1)$-zero) ) is $1=\sum_{d \leq x} \mu(d)\left\lfloor \frac xd \right\rfloor = \sum_{d\leq x} \mu(d) \frac xd + \sum_{d\leq x} \mu(d) \{\frac xd\}$ (which is very fundamental and I discuss it in these slides under the name "Mobius cancellation"; also proved Prove $\sum_{d \leq x} \mu(d)\left\lfloor \frac xd \right\rfloor = 1 $).
The key identities to go between these statements and (vM-PNT) are
- $(\frac 1{\zeta})' = \frac 1{\zeta} \cdot (-\frac{\zeta'}{\zeta}) \iff [\mu \cdot \log ] = \mu \star \Lambda$ [Terry's equation (11)]
- $-\zeta' \cdot (\frac 1{\zeta}) = -\frac{\zeta'}{\zeta} \iff \log \star \mu = \Lambda$ [Terry's equation (12)]
- $\zeta = \frac{1}{\zeta} \cdot \zeta^2 \iff 1 = \mu \star \tau$ [Terry's equation (13)].
It is fairly easy to see (Liouville function and PNT but for $\mu$ instead of $\lambda$ + partial summation nonsense, for which I've recently found a really nice geometric explanation for on MSE! Intuitive explanation of proof of Abel's limit theorem) to see that if $\sum_n \frac{\mu(n)}n$ converges at all, then it must converge to $0$. I.e. ($\mathcal D \lambda (1)$-zero) is equivalent to
- ($\mathcal D\mu(1)$-converges): $\sum_{n\leq x} \frac{\mu(n)}n$ converges (conditionally) as $x\to\infty$.
Also, as Terry says, it may be "more natural" to prove everything is equivalent to ($\zeta(1+it)$-nonvanishing) below. But looking at the proof of Terry's Notes 2 Prop. 31 I cite below, only one direction seems easy to transfer, i.e. the meanzero statements $\implies$ (vM-PNT). The other direction, at least in the proof of Prop. 31, require formulas rather tailored to $\Lambda$.
Bringing up Liouville
($\lambda$-meanzero): $L(x):=\sum_{n\leq x} \lambda(n) = o(x)$.
($\mathcal D \lambda (1)$-zero): $\sum_{n\leq x} \frac{\lambda (n)}n = o(1)$.
($\mathcal D\lambda(1)$-converges): $\sum_{n\leq x} \frac{\lambda(n)}n$ converges (conditionally) as $x\to\infty$.
Instead of using the above exact Mobius cancellation ($1=\sum_{d \leq x} \mu(d)\left\lfloor \frac xd \right\rfloor$) formula, we have a similar Liouville cancellation formula $\sum_{d\leq x} \lambda(d)\lfloor \frac xd \rfloor= \lfloor \sqrt{x} \rfloor$, proven in much the same way $\sum_{n=1}^N\lambda(n)[N/n]=[\sqrt{N}]$ Identity involving Liouville Lambda function. I think that running Terry's above proof of ($\mu$-meanzero) $\iff$ ($\mathcal D \mu (1)$-zero), basically word for word, one gets ($\lambda$-meanzero) $\iff$ ($\mathcal D \lambda (1)$-zero).
And for the equivalences between these and (vM-PNT), use similar Dirichlet convolution/zeta function identities as bulleted above, but replace $\frac 1{\zeta(s)}$ with $\frac{\zeta(2s)}{\zeta(s)} \cdot \frac{1}{\zeta(2s)}$ (and note that $\zeta(2s)$ corresponds with the arithmetic function $1_\square$, i.e. the indicator function of squares; and $\frac{1}{\zeta(2s)} \leftrightarrow [\mu(\sqrt n) \cdot 1_{\square}(n)]$, and $\frac{\zeta(2s)}{\zeta(s)} \leftrightarrow \lambda$, and $\frac{d}{ds} \zeta(2s) \leftrightarrow -\log \cdot 1_\square$, and $(\frac{\zeta(2s)}{\zeta(s)})' = \frac{\zeta(s) [\frac{d}{ds} \zeta(2s)] - \zeta(2s) \zeta'(s)}{\zeta(s)^2} \leftrightarrow \mu \star [-\log \cdot 1_\square] + [\lambda \star \Lambda]$). Actually not sure this works.... :(
Ok: using the formula $L(x)=\sum_{d\leq x} M(\frac x{d^2}) = \sum_{d\leq \sqrt x} M(\frac x{d^2})$ from Liouville function and PNT (coming from $\frac{\zeta(2s)}{\zeta(s)} = \frac 1{\zeta(s)} \cdot \zeta(2s) \leftrightarrow \lambda = \mu \star 1_\square$), and assuming ($\mu$-meanzero) i.e. $\forall \epsilon>0$ exists $X_\epsilon$ s.t. $x\geq X_\epsilon \implies |M(x)| \leq \epsilon x$ (and $x<X_\epsilon$ we have trivially $|M(x)|\leq x$), get (for $x$ sufficiently large so that $x^{1/4}>X_\epsilon$),
$$\sum_{d < x^{1/4}} |M(\frac x{d^2})| \leq \sum_{d < x^{1/4}} \epsilon \cdot \frac{x}{d^2} \leq \frac{\pi^2}6 \epsilon x.$$
And
$$\sum_{x^{1/4} < d\leq x^{1/2}} |M(\frac x{d^2})| \leq \sum_{x^{1/4} < d\leq x^{1/2}}\frac{x}{d^2} \leq x \cdot \left( \sum_{x^{1/4}<d\leq x^{1/2}} \frac 1{d^2} \right).$$
For $x$ sufficiently large, because $\sum \frac 1{n^2}<\infty$, we get the coefficient of $x$ on the RHS is $< \epsilon$. So in total we get: $\forall \epsilon>0$, there exists a threshold $X_\epsilon'$ past which
$$|L(x)| \leq \sum_{d\leq x} |M(\frac{x}{d^2})| \leq 10\epsilon x,$$
and we are done (proving that ($\mu$-meanzero) $\implies$ ($\lambda$-meanzero) ).
Furthermore, the identity $\frac{\zeta(2s)}{\zeta(s)} \cdot \frac{1}{\zeta(2s)} = \frac{1}{\zeta(s)} \leftrightarrow \mu = \lambda \star [\mu(\sqrt \bullet) \cdot 1_\square]$ (recalling what I wrote above $\frac{1}{\zeta(2s)} \leftrightarrow [\mu(\sqrt n) \cdot 1_{\square}(n)]$) means that
$$\mu(n) = \sum_{d\mid n} 1_\square (d) \mu(\sqrt d) \lambda(\frac nd) = \sum_{d^2 \mid n} \mu(d) \lambda(\frac n{d^2})$$
so summing $\sum_{n\leq x}$ one gets
$$$$
$$\begin{aligned}
M(x) &= \sum_{n\leq x} \mu(n) = \sum_{n\leq x} \sum_{d^2 \mid n} \mu(d) \lambda(\frac n{d^2}) \\
&= \sum_{d\leq \sqrt x} \sum_{n\leq x: d^2 \mid n} \mu(d) \lambda(\frac n{d^2}) = \sum_{d\leq \sqrt x} \mu(d) \sum_{k \leq x/d^2} \lambda(k) \\
&= \sum_{d\leq \sqrt x} \mu(d) L(\frac x{d^2}),
\end{aligned}$$
and applying the triangle inequality, we get
$$|M(x)| \leq \sum_{d\leq \sqrt x} |L(\frac x{d^2})|,$$
which is exactly what we used above (with $M,L$ switched). That is, using this bound instead with the above proof, we see that ($\lambda$-meanzero) $\implies$ ($\mu$-meanzero). And so indeed everything above is equivalent.
Entering the Complex Domain
Prop. 31 (equivalences to PNT) in these notes of Terry Tao is the claim that (vM-PNT) is equivalent to
- ($\zeta(1+it)$-nonvanishing): all the zeroes of $\zeta$ have real part strictly $<1$.
Measure Theoretic Formulations
As I explain/motivate in these slides, (vM-PNT-dyadic) is equivalent to the statement
- (vague-conv-to-Leb): the translates $\tau_h \mu$ converge in the vague topology to Lebesgue measure $m$ as $h \rightarrow \infty$ (notated $\tau_h \mu \rightharpoonup m$ ), WHERE $\mu, \tau_h$ defined as follows:
Definition 2.1: Mertens1 measure
We denote
$$
\mu:=\sum_{n=1}^{\infty} \frac{\Lambda(n)}{n} \delta_{\log n}
$$
Picture this as a bunch of sticks supported on essentially $\{\log p: p \text{ prime}\} \subseteq(0, \infty)$, with the sticks getting shorter and shorter as you go right, but also denser and denser. Measuring $[u, u+\varepsilon]$ against this for $u \rightarrow \infty$ can be thought of as having a fixed interval of width $\varepsilon$ and shifting $\mu$ (i.e. the sticks) leftward.
Definition 2.2: Left shift
For any $h \in \mathbb{R}$, let $\tau_h \mu$ denote the left translate of $\mu$ by $h$ units. That is,
$$
\int_{\mathbb{R}} G(t) d \tau_h \mu(t)=\int_{\mathbb{R}} G(t-h) d \mu(t) \quad \forall G \in C_{\mathrm{c}}(\mathbb{R}) .
$$
It is this formulation that admits a very nice proof using the theory of Banach algebras. Also I think this is essentially your (B3).
