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单词 AnotherProofOfJensensInequality
释义

another proof of Jensen’s inequality


First of all, it’s clear that defining

λk=μkk=1nμk

we have

k=1nλk=1

so it will we enough to prove only the simplified version.

Let’s proceed by inductionMathworldPlanetmath.

1) n=2; we have to show that, for any x1and x2 in [a,b],

f(λ1x1+λ2x2)λ1f(x1)+λ2f(x2).

But, since λ1+λ2 must be equal to 1, we can put λ2=1-λ1, so that the thesis becomes

f(λ1x1+(1-λ1)x2)λ1f(x1)+(1-λ1)f(x2),

which is true by definition of a convex function.

2) Taking as true that f(k=1n-1μkxk)k=1n-1μkf(xk), where k=1n-1μk=1, we have to prove that

f(k=1nλkxk)k=1nλkf(xk),

where k=1nλk=1.

First of all, let’s observe that

k=1n-1λk1-λn=(k=1nλk)-λn1-λn=1-λn1-λn=1

and that if all xk[a,b], k=1n-1λk1-λnxk belongs to [a,b] as well. In fact, λk1-λn being non-negative,

axkbλk1-λnaλk1-λnxkλk1-λnb,

and, summing over k,

ak=1n-1λk1-λnk=1n-1λk1-λnxkbk=1n-1λk1-λn,

that is

ak=1n-1λk1-λnxkb.

We have, by definition of a convex function:

f(k=1nλkxk)=f(k=1n-1λkxk+λnxn)
=f((1-λn)k=1n-1λk1-λnxk+λnxn)
(1-λn)f(k=1n-1λk1-λnxk)+λnf(xn).

But, by inductive hypothesis, since k=1n-1λk1-λn=1, we have:

f(k=1n-1λk1-λnxk)k=1n-1λk1-λnf(xk),

so that

f(k=1nλkxk)(1-λn)f(k=1n-1λk1-λnxk)+λnf(xn)
(1-λn)k=1n-1λk1-λnf(xk)+λnf(xn)
=k=1n-1λkf(xk)+λnf(xn)
=k=1nλkf(xk)

which is the thesis.

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更新时间:2025/5/5 9:56:57