Superadditivity

In mathematics, a sequence {a_n}, n ≥ 1, is called superadditive if it satisfies the inequality

${\displaystyle a_{n+m}\geq a_{n}+a_{m}}$

for all m and n. The major reason for the use of superadditive sequences is the following lemma due to Michael Fekete.[1]

Lemma: (Fekete) For every superadditive sequence {a_n}, n ≥ 1, the limit lim a_n /n exists and is equal to sup a_n /n. (The limit may be positive infinity, for instance, for the sequence a_n = log n!.)

Similarly, a function f is superadditive if

${\displaystyle f(x+y)\geq f(x)+f(y)}$

for all x and y in the domain of f.

For example, ${\displaystyle f(x)=x^{2}}$ is a superadditive function for nonnegative real numbers because the square of ${\displaystyle x+y}$ is always greater than or equal to the square of ${\displaystyle x}$ plus the square of ${\displaystyle y}$, for nonnegative real numbers ${\displaystyle x}$ and ${\displaystyle y}$ ((x + y)² = x² + y² + 2xy).

The analogue of Fekete's lemma holds for subadditive functions as well. There are extensions of Fekete's lemma that do not require the definition of superadditivity above to hold for all m and n. There are also results that allow one to deduce the rate of convergence to the limit whose existence is stated in Fekete's lemma if some kind of both superadditivity and subadditivity is present. A good exposition of this topic may be found in Steele (1997).[2][3]

The term "superadditive" is also applied to functions from a boolean algebra to the real numbers where ${\displaystyle P(X\lor Y)\geq P(X)+P(Y)}$, such as lower probabilities.

If f is a superadditive function, and if 0 is in its domain, then f(0) ≤ 0. To see this, take the inequality at the top: ${\displaystyle f(x)\leq f(x+y)-f(y)}$. Hence ${\displaystyle f(0)\leq f(0+y)-f(y)=0.}$

The negative of a superadditive function is subadditive.