If $f(x) = \Omega(n)$ and $g(x)= O(n)$, what would be the order of growth of $f(x) \cdot g(x)$ ?

First I figured it should $\Theta(n)$ , as two extremes would cancel each other and the order of growth will be same as $n$

But, where I came across this question, the answer given was $\Omega(n)$, and no proof was mentioned. Well, I didn't understand why, but intuitively I convinced myself as "you can't know for sure the upper limit of growth for $f(x) \cdot g(x)$ so you can't say it's $O(n)$, but you can be sure that it won't be lower than $\Omega(n)$"

Can someone help me in understanding this, in a more believable way?

  • 1
    $\begingroup$ The stated answer of $\Omega(n)$ is wrong (assuming $o(1)$ functions are allowed for $g$); $g(n) = 0$ would put $g$ safely in $O(n)$. $\endgroup$
    – G. Bach
    Feb 12, 2014 at 19:45

1 Answer 1


Basically, without further information you know nothing about the asymptotic growth of $f \cdot g$.

Let's unfold the definitions:

$\qquad f \in \Omega(n) \implies f(n) \leq cn$ and

$\qquad g \in O(n) \implies g(n) \geq dn$

for some $c,d \in \mathbb{N}$ and $n$ greater than some constant.

Now it's clear that you get no bound on $g \cdot f$; you have no upper bound for one factor and no lower for the other. In fact, you can get arbitrarily slowly and fast growing results:

  • For $f : n \mapsto n$ and $g_k : n \mapsto n^{-k}$, we get $f \cdot g_k : n \mapsto n^{-k+1}$. Note that the induced sequence of functions is properly decreasing in terms of asymptotic growth.

  • For $f_k : n \mapsto n^k$ and $g : n \mapsto n$, we get $f_k \cdot g : n \mapsto n^{k+1}$. Note that the induced sequence of functions is properly increasing in terms of asymptotic growth.

If you restrict yourself to (eventually) non-decreasing functions -- that is $f,g \in \Omega(1)$ -- you get one of the missing bounds and thus $f \cdot g \in \Omega(n)$.


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