In mathematics, a function f {\displaystyle f} is superadditive if
f ( x + y ) ≥ f ( x ) + f ( y ) {\displaystyle f(x+y)\geq f(x)+f(y)}
for all x {\displaystyle x} and y {\displaystyle y} in the domain of f . {\displaystyle f.} Similarly, a sequence a 1 , a 2 , … {\displaystyle a_{1},a_{2},\ldots } is called superadditive if it satisfies the inequality
a n + m ≥ a n + a m {\displaystyle a_{n+m}\geq a_{n}+a_{m}}
for all m {\displaystyle m} and n . {\displaystyle n.}
The term "superadditive" is also applied to functions from a boolean algebra to the real numbers where P ( X ∨ Y ) ≥ P ( X ) + P ( Y ) , {\displaystyle P(X\lor Y)\geq P(X)+P(Y),} such as lower probabilities.
Examples of superadditive functions The map f ( x ) = x 2 {\displaystyle f(x)=x^{2}} is a superadditive function for nonnegative real numbers because f ( x + y ) = ( x + y ) 2 = x 2 + y 2 + 2 x y = f ( x ) + f ( y ) + 2 x y ≥ f ( x ) + f ( y ) . {\displaystyle f(x+y)=(x+y)^{2}=x^{2}+y^{2}+2xy=f(x)+f(y)+2xy\geq f(x)+f(y).}
The determinant is superadditive for nonnegative Hermitian matrix, that is, if A , B ∈ Mat n ( C ) {\displaystyle A,B\in {\text{Mat}}_{n}(\mathbb {C} )} are nonnegative Hermitian then det ( A + B ) ≥ det ( A ) + det ( B ) . {\displaystyle \det(A+B)\geq \det(A)+\det(B).} This follows from the Minkowski determinant theorem, which more generally states that det ( ⋅ ) 1 / n {\displaystyle \det(\cdot )^{1/n}} is superadditive (equivalently, concave) for nonnegative Hermitian matrices of size n {\displaystyle n} : If A , B ∈ Mat n ( C ) {\displaystyle A,B\in {\text{Mat}}_{n}(\mathbb {C} )} are nonnegative Hermitian then det ( A + B ) 1 / n ≥ det ( A ) 1 / n + det ( B ) 1 / n . {\displaystyle \det(A+B)^{1/n}\geq \det(A)^{1/n}+\det(B)^{1/n}.}
Horst Alzer proved that Hadamard's gamma function H ( x ) {\displaystyle H(x)} is superadditive for all real numbers x , y {\displaystyle x,y} with x , y ≥ 1.5031. {\displaystyle x,y\geq 1.5031.}
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Properties If f {\displaystyle f} is a superadditive function whose domain contains 0 , {\displaystyle 0,} then f ( 0 ) ≤ 0. {\displaystyle f(0)\leq 0.} To see this, simply set x = 0 {\displaystyle x=0} and y = 0 {\displaystyle y=0} in the defining inequality. The negative of a superadditive function is subadditive.
Fekete's lemma The major reason for the use of superadditive sequences is the following lemma due to Michael Fekete.
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