Gruber P. Convex and Discrete Geometry

14 Convex Functions
Proof (by affine support properties). By induction, x0 = λ1x1 +···+λnxn ∈ I .Let
a(x) = f (x0) + u(x − x0) be an affine support of f at x0. Then f (xi) ≥ a(xi) for
i = 1,...,n, and thus
λ1 f (x1) +···+λn f (xn) ≥ λ1a(x1) +···+λna(xn)
= (λ1 +···+λn) f (x0) + u � �
λ1x1 +···+λnxn − (λ1 +···+λn)x0
= f (x0) = f (λ1x1 +···+λnxn). ⊓⊔
Remark. If f is strictly convex and λ1,...,λn > 0, then equality holds in Jensen’s
inequality precisely in case when x1 =···= xn.
Mechanical Interpretation on Jensen’s Inequality
The center of gravity of the masses λ1,...,λn at the points � x1, f (x1) � ,...,
� xn, f (xn) � on the graph of f is the point
(xc, yc) = � λ1x1 +···+λnxn,λ1 f (x1) +···+λn f (xn) � .
It is contained in the convex polygon with vertices � x1, f (x1) � ,..., � xn, f (xn) �
which, in turn, is contained in the epigraph of f (as can be shown). Thus (xc, yc)
is also contained in the epigraph of f which is equivalent to Jensen’s inequality.
Inequality Between the Arithmetic and the Geometric Mean
As a direct consequence of Jensen’s inequality we have the following inequality.
Corollary 1.2. Let x1,...,xn ≥ 0 and λ1,...,λn ≥ 0 be such that λ1 +···+λn =
1. Then
In particular,
x λ1
1 ···xλn
n ≤ λ1x1 +···+λnxn.
(x1 ···xn) 1 n ≤ x1 +···+xn
.
n
Proof. We may suppose that x1,...,xn > 0. Since exp : R → R + is convex by
Theorem 1.8, an application of Jensen’s inequality to y1 = log x1,...,yn = log xn
then gives the desired inequality:
x λ1
1 ···xλn n = exp � log(x λ1
1 ···xλn n ) � = exp(λ1 log x1 +···+λn log xn)
≤ λ1 exp(log x1) +···+λn exp(log xn) = λ1x1 +···+λnxn. ⊓⊔
Actually, exp is strictly convex.