Hermite–Hadamard inequality - Alchetron, the free social encyclopedia

In mathematics, the Hermite–Hadamard inequality, named after Charles Hermite and Jacques Hadamard and sometimes also called Hadamard's inequality, states that if a function ƒ : [a, b] → R is convex, then the following chain of inequalities hold:

Generalisations - The concept of a sequence of iterated integrals

Suppose that −∞ < a < b < ∞, and let f:[a, b] → ℝ be an integrable real function. Under the above conditions the following sequence of functions is called the sequence of iterated integrals of f,where a ≤ s ≤ b.:

F ( 0 ) ( s ) := f ( s ) , F ( 1 ) ( s ) := ∫ a s F ( 0 ) ( u ) d u = ∫ a s f ( u ) d u , F ( 2 ) ( s ) := ∫ a s F ( 1 ) ( u ) d u = ∫ a s ( ∫ a t f ( u ) d u ) d t , ⋮ F ( n ) ( s ) := ∫ a s F ( n − 1 ) ( u ) d u , ⋮

Example 1

Let [a, b] = [0, 1] and f(s) ≡ 1. Then the sequence of iterated integrals of 1 is defined on [0, 1], and

F ( 0 ) ( s ) = 1 , F ( 1 ) ( s ) = ∫ 0 s F ( 0 ) ( u ) d u = ∫ 0 s 1 d u = s , F ( 2 ) ( s ) = ∫ 0 s F ( 1 ) ( u ) d u = ∫ 0 s u d u = s 2 2 , ⋮ F ( n ) ( s ) := ∫ 0 s u n − 1 ( n − 1 ) ! d u = s n n ! , ⋮

Example 2

Let [a,b] = [−1,1] and f(s) ≡ 1. Then the sequence of iterated integrals of 1 is defined on [−1, 1], and

F ( 0 ) ( s ) = 1 , F ( 1 ) ( s ) = ∫ − 1 s F ( 0 ) ( u ) d u = ∫ − 1 s 1 d u = s + 1 , F ( 2 ) ( s ) = ∫ − 1 s F ( 1 ) ( u ) d u = ∫ − 1 s ( u + 1 ) d u = s 2 2 ! + s 1 ! + 1 2 ! = ( s + 1 ) 2 2 ! , ⋮ F ( n ) ( s ) = s n n ! + s n − 1 ( n − 1 ) ! 1 ! + s n − 2 ( n − 2 ) ! 2 ! + ⋯ + 1 n ! = ( s + 1 ) n n ! , ⋮

Example 3

Let [a, b] = [0, 1] and f(s) = e^s. Then the sequence of iterated integrals of f is defined on [0, 1], and

F ( 0 ) ( s ) = e s , F ( 1 ) ( s ) = ∫ 0 s F ( 0 ) ( u ) d u = ∫ 0 s e u d u = e s − 1 , F ( 2 ) ( s ) = ∫ 0 s F ( 1 ) ( u ) d u = ∫ 0 s ( e u − 1 ) d u = e s − s − 1 , ⋮ F ( n ) ( s ) = e s − ∑ i = 0 n − 1 s i i ! ⋮

Theorem

Suppose that −∞ < a < b < ∞, and let f:[a,b]→R be a convex function, a < x_i < b, i = 1, ..., n, such that x_i ≠ x_j, if i ≠ j. Then the following holds:

∑ i = 1 n F ( n − 1 ) ( x i ) Π i ( x 1 , … , x n ) ≤ 1 n ! ∑ i = 1 n f ( x i )

where

Π i ( x 1 , … , x n ) := ( x i − x 1 ) ( x i − x 2 ) ⋯ ( x i − x i − 1 ) ( x i − x i + 1 ) ⋯ ( x i − x n ) , i = 1 , … , n .

In the concave case ≤ is changed to ≥.

Remark 1. If f is convex in the strict sense then ≤ is changed to < and equality holds iff f is linear function.

Remark 2. The inequality is sharp in the following limit sense: let x _ = ( x 1 , … , x n ) , α _ = ( α , … , α ) and a < α < b .
Then the limit of the left side exists and

lim x _ → α _ ∑ i = 1 n F ( n − 1 ) ( x i ) Π i ( x 1 , … , x n ) = lim x _ → α _ 1 n ! ∑ i = 1 n f ( x i ) = f ( α ) ( n − 1 ) !

References

Hermite–Hadamard inequality Wikipedia

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