Divisor sums of multiplicative functions Suppose that $f$ is a multiplicative function, i.e., $f(ab)=f(a)f(b)$ whenever $\gcd(a,b)=1$. Are there any techniques for estimating 
$$\sum\limits_{d \mid n}f(d)$$
in terms of $f(n)$?
I am interested in general lower and upper bounds, aside from the obvious ones
$$\tau(n) \leq \sum\limits_{d \mid n}f(d) \leq \tau(n)f(n).$$
The ideal would be to find a function $g$ such that
$$\lim\limits_{n \to \infty}\frac{\sum_{d \mid n}f(d)}{g(f(n))}=1.$$
A particular case of interest to me is when $f$ is the divisor function $\tau$, i.e., the number of distinct divisors of an integer.
Many thanks.
 A: Let $F(n) := \sum\limits_{d \mid n} \tau(d)$ and note that $F$ is itself multiplicative. Then 
$$F(p^a) = \sum_{i=0}^{a}\tau(p^i) = \sum_{i=0}^{a}(i+1) = \frac{a^2+3a+1}{2} \geq \frac{(a+1)^2}{2}.$$ 
Thus if $n=p_1^{a_1} \dots p_k^{a_k}$, then 
$$F(n) \geq \prod\limits_{i=1}^{k}\frac{(a_i+1)^2}{2} = \left(\frac{1}{2}\right)^{\omega(n)}\tau^2(n).$$
Similarly one can take $c$ to be the smallest positive real that makes the inequality $$\frac{a^2+3a+1}{2}\leq c (a+1)^2$$ valid for all $a \geq 1$ and 
deduce the upper bound 
$$F(n)\leq c^{\omega(n)}\tau^2(n).$$
I can't tell whether these bounds are in some sense best possible though. 
A: This does not answer your question but it did not fit in a comment. There is a large literature on maximal orders of multiplicative functions, for example, Wigert proved that there exists $c>0$ such that $d(n) \leq n^{c/\log \log n}$ for all large enough $n$, and he also found the best possible value for $c$. The literature gives bounds using calculus functions, not multiplicative functions. The function $F$ becomes very large especially when $n$ has many small divisors so any inequality that you want to prove should be first checked in the case that $n$ is the product of the first primes. To be precise let $n_x:=\prod_{p\leq x }p$ so that $F(n_x)=3^{\pi(x)}$. Then by the prime number theorem $$\log F(n_x) \sim (\log 3) \frac{x }{\log x },$$ as $x\to \infty $.
Note also that by the prime number theorem again we have $\log n_x=\theta(x) \sim x$, hence,  $$\log F(n_x) \sim (\log 3) \frac{\log n_x }{\log \log n_x }.$$ This means that for any $b<3$ there are infinitely many $n$ for which $$ F(n) >  n^{ \frac{\log b  }{\log \log n  }}.$$ This is in a way best possible. By the inequality $F(n)\leq \tau^2(n)$ and Wigert's result we also get that there exists $c>0$ such that for all $n $ the bound $$ F(n) \leq   n^{ \frac{c }{\log \log n  }}$$ holds.It would be nice to determine the best value for $c$. It would have to be at least $\log 3 $ from the previous argument and I would be surprised if it was strictly larger than $\log 3 $.
EDIT: this paper of Shiu shows that $c$ is indeed $\log 3$. Namely, it shows that if we have any $c>\log 3 $ then $ F(n) \leq   n^{ \frac{c }{\log \log n  }}$ holds for all large enough $n$. This can be seen by taking $\theta=1/2$ in his Theorem 1 and noting that his $M$ is achieved for $a=1$, thus, $\log M=\log 3 $.
