f ? O(g) says, essentially
For at least one choice of a constant k > 0, you can find a constant a such that the inequality 0 <= f(x) <= k g(x) holds for all x > a.
Note that O(g) is the set of all functions for which this condition holds.
f ? o(g) says, essentially
For every choice of a constant k > 0, you can find a constant a such that the inequality 0 <= f(x) < k g(x) holds for all x > a.
Once again, note that o(g) is a set.
In Big-O, it is only necessary that you find a particular multiplier k for which the inequality holds beyond some minimum x.
In Little-o, it must be that there is a minimum x after which the inequality holds no matter how small you make k, as long as it is not negative or zero.
These both describe upper bounds, although somewhat counter-intuitively, Little-o is the stronger statement. There is a much larger gap between the growth rates of f and g if f ? o(g) than if f ? O(g).
One illustration of the disparity is this: f ? O(f) is true, but f ? o(f) is false. Therefore, Big-O can be read as "f ? O(g) means that f's asymptotic growth is no faster than g's", whereas "f ? o(g) means that f's asymptotic growth is strictly slower than g's". It's like <= versus <.
More specifically, if the value of g(x) is a constant multiple of the value of f(x), then f ? O(g) is true. This is why you can drop constants when working with big-O notation.
However, for f ? o(g) to be true, then g must include a higher power of x in its formula, and so the relative separation between f(x) and g(x) must actually get larger as x gets larger.
To use purely math examples (rather than referring to algorithms):
The following are true for Big-O, but would not be true if you used little-o:
The following are true for little-o:
Note that if f ? o(g), this implies f ? O(g). e.g. x² ? o(x³) so it is also true that x² ? O(x³), (again, think of O as <= and o as <)