ABSTRACT
Let f : [a, b]→ R be a convex mapping on [a, b]. Then
f
( a+ b
) ≤ 1 b− a
f (x) dx ≤ f (a) + f (b) 2
, (2.1)
which is known as the Hermite-Hadamard (H.-H.) integral inequality.
PROOF By the convexity of f on [a, b] , we have
f (ta+ (1− t) b) ≤ tf (a) + (1− t) f (b)
for all t ∈ [0, 1] . Integrating over t on [0, 1] , we obtain∫ 1
f (ta+ (1− t) b) dt ≤ f (a) ∫ 1
tdt+ f (b) ∫ 1
(1− t) dt. (2.2)
tdt = ∫ 1
(1− t) dt = 1 2
and, by the change of variable x = ta+ (1− t) b,∫ 1 0
f (ta+ (1− t) b) dt = 1 b− a
f (x) dx,
we get the second part of (2.1) from (2.2). By the convexity of f we also have
1 2
[f (ta+ (1− t) b) + f ((1− t) a+ tb)]
≥ f [ ta+ (1− t) b+ (1− t) a+ tb
] = f
( a+ b
) .
