Is $\frac{\sigma(n)}{{2n}}$ submultiplicative or supermultiplicative?

Question

This question is an offshoot of this earlier post.

It is known that the abundancy index $$I(n) = \frac{\sigma(n)}{n}$$ is a multiplicative function. (Note that the divisor sum $\sigma$ is also multiplicative.)

How about $$J(n) = \frac{\sigma(n)}{2n}?$$

MY ATTEMPT

Suppose that $\gcd(x,y)=1$. Then $$J(xy) = \frac{\sigma(xy)}{2xy} = \frac{\sigma(x)\sigma(y)}{2xy} \neq \frac{\sigma(x)}{2x}\cdot\frac{\sigma(y)}{2y}.$$

In general, we have $$2J(x)J(y) = \frac{\sigma(x)\sigma(y)}{2xy} = J(xy).$$

This implies that $$J(xy) = 2J(x)J(y) > J(x)J(y),$$ for all $x,y$ such that $\gcd(x,y)=1$. This means that $$J(n) = \frac{\sigma(n)}{2n}$$ is supermultiplicative.

QUESTION

Is my understanding of the notion of a supermultiplicative function correct?

Answer 1

The argument is correct and can be generalized. Suppose $\mu(x)$ is any non-negative multiplicative function and $0\leq a\leq1$. Then we have $a\geq a^2$, and therefore $$a\mu(xy)\geq a^2\mu(xy)=a^2\mu(x)\mu(y)=a\mu(x)\,a\mu(y).$$ In other words, $J(x)=a\mu(x)$ is supermultiplicative. By analogous argument, it is submultiplicative when $a\geq1$.