another proof of Jensen’s inequality

First of all, it’s clear that defining

$${\lambda }_k=\frac{{\mu }_k}{{\sum }_{k=1}^n{\mu }_k}$$

we have

$$\sum _{k=1}^n{\lambda }_k=1$$

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

Let’s proceed by induction.

1) $n=2$; we have to show that, for any $x_1$and $x_2$ in $[a,b]$,

$$f\left({\lambda }_1{x}_1+{\lambda }_2{x}_2\right)\le {\lambda }_{1}f(x_1)+{\lambda }_{2}f(x_2).$$

But, since ${\lambda }_1+{\lambda }_2$ must be equal to 1, we can put ${\lambda }_2=1-{\lambda }_1$, so that the thesis becomes

$$f\left({\lambda }_1{x}_1+\left(1-{\lambda }_1\right)x_2\right)\le {\lambda }_{1}f(x_1)+\left(1-{\lambda }_1\right)f(x_2),$$

which is true by definition of a convex function.

2) Taking as true that $f\left({\sum }_{k=1}^{n-1}{\mu }_k{x}_k\right)\le {\sum }_{k=1}^{n-1}{\mu }_{k}f\left(x_k\right)$, where ${\sum }_{k=1}^{n-1}{\mu }_k=1$, we have to prove that

$$f\left(\sum _{k=1}^n{\lambda }_k{x}_k\right)\le \sum _{k=1}^n{\lambda }_{k}f\left(x_k\right),$$

where ${\sum }_{k=1}^n{\lambda }_k=1$.

First of all, let’s observe that

$$\sum _{k=1}^{n-1}\frac{{\lambda }_k}{1-{\lambda }_n}=\frac{\left({\sum }_{k=1}^n{\lambda }_k\right)-{\lambda }_n}{1-{\lambda }_n}=\frac{1-{\lambda }_n}{1-{\lambda }_n}=1$$

and that if all $x_k\in [a,b]$, ${\sum }_{k=1}^{n-1}\frac{{\lambda }_k}{1-{\lambda }_n}{x}_k$ belongs to $[a,b]$ as well. In fact, $\frac{{\lambda }_k}{1-{\lambda }_n}$ being non-negative,

$$a\le x_k\le b\Rightarrow \frac{{\lambda }_k}{1-{\lambda }_n}a\le \frac{{\lambda }_k}{1-{\lambda }_n}{x}_k\le \frac{{\lambda }_k}{1-{\lambda }_n}b,$$

and, summing over $k$,

$$a\sum _{k=1}^{n-1}\frac{{\lambda }_k}{1-{\lambda }_n}\le \sum _{k=1}^{n-1}\frac{{\lambda }_k}{1-{\lambda }_n}{x}_k\le b\sum _{k=1}^{n-1}\frac{{\lambda }_k}{1-{\lambda }_n},$$

that is

$$a\le \sum _{k=1}^{n-1}\frac{{\lambda }_k}{1-{\lambda }_n}{x}_k\le b.$$

We have, by definition of a convex function:

$f\left({\displaystyle \sum _{k=1}^n}{\lambda }_k{x}_k\right)$ $=$ $f\left({\displaystyle \sum _{k=1}^{n-1}}{\lambda }_k{x}_k+{\lambda }_n{x}_n\right)$

$=$ $f\left(\left(1-{\lambda }_n\right){\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \frac{{\lambda }_k}{1-{\lambda }_n}}{x}_k+{\lambda }_n{x}_n\right)$

$\le$ $\left(1-{\lambda }_n\right)f\left({\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \frac{{\lambda }_k}{1-{\lambda }_n}}{x}_k\right)+{\lambda }_{n}f\left(x_n\right).$

But, by inductive hypothesis, since ${\sum }_{k=1}^{n-1}\frac{{\lambda }_k}{1-{\lambda }_n}=1$, we have:

$$f\left(\sum _{k=1}^{n-1}\frac{{\lambda }_k}{1-{\lambda }_n}{x}_k\right)\le \sum _{k=1}^{n-1}\frac{{\lambda }_k}{1-{\lambda }_n}f\left(x_k\right),$$

so that

$f\left({\displaystyle \sum _{k=1}^n}{\lambda }_k{x}_k\right)$ $\le$ $\left(1-{\lambda }_n\right)f\left({\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \frac{{\lambda }_k}{1-{\lambda }_n}}{x}_k\right)+{\lambda }_{n}f\left(x_n\right)$

$\le$ $\left(1-{\lambda }_n\right){\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \frac{{\lambda }_k}{1-{\lambda }_n}}f\left(x_k\right)+{\lambda }_{n}f\left(x_n\right)$

$=$ ${\displaystyle \sum _{k=1}^{n-1}}{\lambda }_{k}f\left(x_k\right)+{\lambda }_{n}f\left(x_n\right)$

$=$ ${\displaystyle \sum _{k=1}^n}{\lambda }_{k}f\left(x_k\right)$

which is the thesis.

Title	another proof of Jensen’s inequality
Canonical name	AnotherProofOfJensensInequality
Date of creation	2013-03-22 15:52:53
Last modified on	2013-03-22 15:52:53
Owner	Andrea Ambrosio (7332)
Last modified by	Andrea Ambrosio (7332)
Numerical id	12
Author	Andrea Ambrosio (7332)
Entry type	Proof
Classification	msc 26D15
Classification	msc 39B62