Prove small-oh with big-oh - algorithm

Question

If I have already known f(n) is O(g(n)). From the definition of little-oh, how to prove that f(n) is o(n * g(n))?

Answer 1

Given: f(n) is in O(g(n)).
Using the definition of big-O notation, we can write this as:
f(n) is in O(g(n))
=> |f(n)| ≤ k*|g(n)|, for some constant k>0 (+)
for n sufficiently large (say, n>N)
For the definition of big-O used as above, see e.g.
https://www.khanacademy.org/computing/computer-science/algorithms/asymptotic-notation/a/big-o-notation
Prove: Given (+), then f(n) is in o(n*g(n)).
Lets first state what little-o notation means:
Formally, f(n) = o(g(n)) (or f(n) ∈ o(g(n))) as n → ∞ means that
for every positive constant ε there exists a constant N such that
|f(n)| ≤ ε*|g(n)|, for all n > N (++)
From https://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation.
Now, using (+), we can write
|f(n)| ≤ k*|g(n)|, som k>0, n sufficiently large
<=> { n > 0 } <=> n*|f(n)| ≤ k*n*|g(n)|
<=> n*|f(n)| ≤ k*|n*g(n)|
<=> |f(n)| ≤ (k/n)*|n*g(n)| (+++)
Return to the definition of little-o, specifically (++), and let, without loss of generality, k be fixed. Now, every positive constant ε can be described as
ε = k/C, for some constant C>0 (with k fixed, k>0) (*)
Now, assume, without loss of generality, that n is larger than this C, i.e., n>C. Then, (*) and (+++) yields
|f(n)| ≤ (k/n)*|n*g(n)| < (k/C)*|n*g(n)| = ε*|n*g(n)| (**)
^ ^
| |
since `n>C` (*)
Since we're studying asymptotic behaviour, we can choose to to assign a lower bound to n to any value larger than C (in fact, that's in the definition of both big-O and little-o, "n sufficiently large"), and hence---by the definition of little-oh above---, we have:
- As shown above, (+) implies (**)
- By the definition of little-o, (**) shows that f(n) is in o(n*g(n))
- Subsequently, we've shown that, given (+), then: f(n) is in o(n*g(n))
Result: If f(n) is in O(g(n)), then f(n) is in o(n*g(n)), where these two relations refer big-O and litte-O asymptotic bounds, respectively.
Comment: The result is, in fact, quite trivial. The big-O and little-o notation differ only in one of the two constants used in proving the upper bounds, i.e., we can write the definitions of big-O and little-O as:
f(n) is said to be in O(g(n)) if we can find a set of positive constants (k, N), such that f(n) < k*g(n) holds for all n>N.
f(n) is said to be in o(g(n)) if we can find a positive constant N, such that f(n) < ε*g(n) holds for all n>N, and for every positive constant ε.
The latter is obvious a stricter constraint, but if we can make use of one extra power of n in the left-hand-side of f(n) < ε*g(n) (i.e., f(n) < ε*n*g(n)), then even for infinitesimal values of ε, we can always choose the other constant N freely to be sufficiently large for ε*n to provide us any constant k that can be used to show that f(n) is in O(g(n)) (as, recall, n>N).