Can the Lagarias inequality be written as a "kernel inequality"?

Question

The Lagarias inequality, which is equivalent to the Riemann hypothesis, is:

$$\sigma(n) \le H_n + \exp(H_n) \log(H_n) =:L(n)$$

for all natural numbers $n$, where $\sigma=$ sum of divisors, $H_n=n$-th harmonic number. Lagarias inequality is equivalent to:

$$\sigma\left(\frac{ab}{\gcd(a,b)^2}\right) \le L\left(\frac{ab}{\gcd(a,b)^2}\right)$$

for all natural numbers $a,b$. (We see this by plugging in $n=\frac{ab}{\gcd(a,b)^2}$ in one direction, and by setting $a=n,b=1$ in the other direction.)

We write:

$$k(a,b) = \frac{1}{L\left(\dfrac{ab}{\gcd(a,b)^2}\right)}$$

and

$$K(a,b) = \frac{1}{\sigma\left(\dfrac{ab}{\gcd(a,b)^2}\right)}$$

Then the Lagarias inequality is equivalent to: For all natural numbers $a,b$ we have:

$$k(a,b) \le K(a,b)$$

My question is, if $k(a,b),K(a,b)$ are positive definite (obviously symmetric) functions over the natural numbers? (Positive definite kernel: https://en.wikipedia.org/wiki/Positive-definite_kernel)

This would allow to interpret the Lagarias inequality as "kernel inequality" in the same way the abc-conjecture is conjectured to be a kernel inequality (see: The abc-conjecture as an inequality for inner-products?)

Sagemath computations to back up the conjecture above

Thanks for your help!

Edit: It seems that it is better to define: $$k(a,b) = \frac{1}{L\left(\dfrac{ab}{\gcd(a,b)}\right)}$$

and

$$K(a,b) = \frac{1}{\sigma\left(\dfrac{ab}{\gcd(a,b)}\right)}$$

Then if we let:

$$\phi(n) = \frac{1}{\sigma(n)} \sum_{d|n} \sqrt{d} e_d$$ where $e_d$ is the $d$-th standard basis vector, we get:

$$<\phi(a),\phi(b)> = \frac{\sigma(\gcd(a,b))}{\sigma(a) \sigma(b)}$$

which should be equal to:

$$\frac{1}{\sigma\left(\dfrac{ab}{\gcd(a,b)}\right)}$$

but I have no proof for this and have not seen this equality in any textbook. I am sure the proof is not difficult, but I can not come up with a proof for this at the moment, so any help is appreciated.

"$e_d$ is the $d$-th standard basis vector" in which vector space? — Gerry Myerson, May 02 '21 at 00:15
@GerryMyerson: In the $\mathbb{R}$ Hilbert space of sequences. — mathoverflowUser, May 02 '21 at 07:14
If you are thinking that ${(1,0,0,\dots),(0,1,0,0,\dots),\dots}$ is a basis for the Hilbert space of square-summable real sequences, it isn't. Elements of that space can't generally be expressed as finite linear combinations of those vectors. — Gerry Myerson, May 02 '21 at 07:18
I meant Schauder basis: https://en.wikipedia.org/wiki/Schauder_basis. — mathoverflowUser, May 02 '21 at 07:22
@GerryMyerson: How would you call the $e_d$ in this context? — mathoverflowUser, May 02 '21 at 07:26

score 2 · Accepted Answer · answered May 02 '21 at 00:14

There are several questions being asked here. I'll just deal with the easiest one, showing that $${\sigma(a)\sigma(b)\over\sigma(\gcd(a,b))}=\sigma\left({ab\over\gcd(a,b)}\right)$$

Let $p$ be a prime, let $p^c\|a$ (meaning $a$ is a multiple of $p^c$ but not of $p^{c+1}$), let $p^d\|b$, let $r$ be the larger, and $s$ the smaller, of $c$ and $d$. Note that $r+s=c+d$, $p^{r+s}\|ab$, $p^s\|\gcd(a,b)$, $p^r\|{ab\over\gcd(a,b)}$. Thus, the contribution of the prime $p$ to $\sigma\left({ab\over\gcd(a,b)}\right)$ is ${p^r-1\over p-1}$.

The contribution of $p$ to ${\sigma(a)\sigma(b)\over\sigma(\gcd(a,b))}$ is $${{p^c-1\over p-1}\cdot{p^d-1\over p-1}\over(p^s-1)/(p-1)}={p^r-1\over p-1}$$ and we're done.

(+1) thank you for your detailed answer. – mathoverflowUser May 02 '21 at 07:11 — mathoverflowUser, May 02 '21 at 07:11

Can the Lagarias inequality be written as a "kernel inequality"?

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