Jensen’s inequality for non-convex functions
Holger STEPHAN 1
Jensen’s inequality is well known: Let P
f (x) be a convex function on x ∈ I ⊂ R,
x1 , ..., xn ∈ I and qi ≥ 0 are weights with i qi = 1. Then, Jensen’s inequality
n
X
i=1
qi f (xi ) ≥ f
n
³X
qi xi
i=1
´
holds.
This inequality is in some sense equivalent to the definition of convexity. That’s
why, it is not well known that Jensen’s inequality holds even for non-convex functions
(moreover, this is believed to be false).
√
Let’s start with an easy contest problem (from the journal Die wurzel, Jena,
05/2005, problem µ19):
Let x1 ,. . . , xn be positive real numbers with 0.075 h(x)
0.05
x1 +· · ·+xn = n. Prove the following inequality.
n
X
i=1
0.025
n
X 2
1
≥
1 + xi
3 + xi
i=1
(1)
We use Jensen’s inequality with the function
h(x) =
x
10
20
30
40
-0.025
-0.05
-0.075
2
1−x
1
−
=
.
1+x 3+x
(1 + x)(3 + x)
and the weights qi = n1 . We get
¶ X
n µ
n
n
³1 X
´
2
1
1
1X
−
xi = h(1) = 0
=
h(xi ) ≥ h
n i=1 1 + xi
3 + xi
n
n i=1
i=1
and therefore √
(1). The
√ result is true but the proof is false, because h(x) is not convex
for x > 1 + 3 2 + 3 4 = 6.69464 . . . (see the picture). Why does the inequality
holds, nevertheless? To investigate this question, we try to prove Jensen’s inequality
n
P
deriving an identity for it. We fix x0 =
qi xi and define a function
i=1
g(x) =
f (x) − f (x0 )
x − x0
(for x = x0 we define g(x0 ) = f 0 (x0 )). Now, we set
n
n
n
³X
´ X
X
X=
qi f (xi ) − f
qi xi =
qi f (xi ) − f (x0 )
i=1
i=1
i=1
and check, when does X ≥ 0 hold. Using easy calculations, we obtain
1
Cercetător dr., Institutul Weierstrass, Berlin (e-mail: [email protected])
109
(2)
n
X
n
n
X
¢ X
¡
f (xi ) − f (x0 )
X=
qi f (xi ) − f (x0 ) =
qi f (xi ) − f (x0 ) =
qi
(xi − x0 ) =
xi − x0
i=1
i=1
i=1
=
=
n
X
qi g(xi ) (xi − x0 ) =
i=1
n
X
i,j=1
=
qi qj xi g(xi ) −
X
1≤j<i≤n
n
X
n
X
i=1
n
n
³ X
´
X
g(xi )qi xi
qj −
qj xj =
qi qj xj g(xi ) =
i,j=1
¡
¢
qi qj (xi − xj ) g(xi ) − g(xj ) .
Thus, we get the identity
n
n
³X
´
X
X=
qi f (xi ) − f
qi xi =
i=1
j=1
n
X
i,j=1
X
i=1
j=1
¢
¡
qi qj xi g(xi ) − xj g(xi ) =
1≤j<i≤n
¡
¢
qi qj (xi − xj ) g(xi ) − g(xj )
(3)
We see that X ≥ 0 holds, if the function g(x) is monotone increasing. The function
g(x) is called slope function. g(x) is the slope of the secant through the points in
x0 and x. If f (x) in convex, g(x) is increasing for any point x0 . But this is not the
only case. Function g(x) is increasing, if looking from the point (x0 , f (x0 )) on the
graph of f (x), no point of f (x) "lies in the shadow" of the graph. This can happen
for some points x0 even if the function is not convex. The function h(x) above is an
example (here is x0 = 1).
An other amazing example are polynomials of
fourth order — typical non-convex functions. We
e1
f(x)
consider such a function f (x) and its inflection
points (the x-coordinates are x1 and x2 ). The
e2
inflection point tangents e1 and e2 intersect the
graph of f (x) in points with x-coordinates x3 and
x4 . Now, Jensen’s inequality holds for x0 ≥ x3 or
x
x3
x0 ≤ x4 .
x4
x2
x1
The typical proof of Jensen’s inequality starts from the inequality for convex
functions
(xi − x0 )f 0 (x0 ) ≤ f (xi ) − f (x0 ) ≤ (xi − x0 )f 0 (xi )
Multiplying by qi and adding up over i we get
n
n
n
n
X
X
X
X
qi (xi − x0 )f 0 (x0 ) ≤
qi f (xi ) − f (x0 )
qi ≤
qi (xi − x0 )f 0 (xi )
i=1
i=1
The left hand side in zero because of
i=1
n
P
qi = 1 and
i=1
i=1
n
P
qi xi = x0 . The middle is X
i=1
and transforming the right hand sind in a similar way as above (3) we obtain.
n
n
³X
´
X
X
¡
¢
0≤
qi f (xi ) − f
qi xi ≤
qi qj (xi − xj ) f 0 (xi ) − f 0 (xj )
i=1
i=1
1≤j<i≤n
This is similar to identity (3), but holds only for convex functions.
Moreover, identity (3) is very useful, if we are not only interested in Jensen’s
inequality, but if we want to estimate the difference on the left hand side of (3).
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