Convex function

santosguzella

Convex function

**Posted:** Sun Jul 08, 2007 12:10 am

zgorkster

This is a very commonly known theorem called Jensen's inequality. Here is a good link:

[url]http://planetmath.org/encyclopedia/JensensInequality.html [/url]

**Posted:** Sun Jul 08, 2007 5:01 am

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santosguzella

#3

**Posted:** Sun Jul 08, 2007 5:38 am

BaBaK Ghalebi

#4

I didnt find any proof in the above link,so here is a proof for jensen inequality:

first we know that $f$ is convex iff for every $\lambda \in [0,1]$ we have:

$f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y)$

i.e. $f(w_{1}x+w_{2}y)\leq w_{1}f(x)+w_{2}f(y)$ where $w_{1},w_{2}$ are psoitive numbers s.t $w_{1}+w_{2}=1$ ...

now we want to prove:

jensen inequality:for convex function $f$ we have:

$f(\sum_{i=1}^{n}w_{i}x_{i})\geq \sum_{i=1}^{n}w_{i}f(x_{i})$ where $\sum_{i=1}^{n}w_{i}=1$

proof:we prove the above inequality with induction on $n$ ...

first of all for $n=2$ it follows immediately from the definition of convex functions mentiond above...

now assume that its true for $n-1$ i.e. for every $n-1$ numbers $w_{1},...,w_{n-1}$ with sum equal to $1$ we have:

$f(\sum_{i=1}^{n-1}w_{i}x_{i})\leq \sum_{i=1}^{n-1}w_{i}f(x_{i})$

now we want to prove that:

$f(\sum_{i=1}^{n}w_{i}x_{i})\leq \sum_{i=1}^{n}w_{i}f(x_{i})$

we have:

$RHS=w_{1}f(x_{1})+\sum_{i=2}^{n}w_{i}f(x_{i})$

now put $w'_{i}=\frac{w_{i}}{1-w_{1}}$ for every $2\leq i\leq n$ so we get that $\sum_{i=2}^{n}w'_{i}=1$

now we ge that:

$RHS=w_{1}f(x_{1})+(1-w_{1})\sum_{i=2}^{n}w'_{i}f(x_{i})$

now according to the hypothesis of the induction we get that $\sum_{i=2}^{n}w'_{i}f(x_{i})\geq f(\sum_{i=2}^{n}w'_{i}x_{i})$ so the
last inequality becomes:

$RHS\geq w_{1}f(x_{1})+(1-w_{1})f(\sum_{i=2}^{n}w'_{i}x_{i})$

now according to jensen for $n=2$ the above inequality becomes:

$RHS\geq f(x_{1}w_{1}+(1-w_{1})\sum_{i=1}^{n}w'_{i}x_{i})=f(\sum_{i=1}^{n}x_{i}w_{i})$

and we are done...

**Posted:** Sun Jul 08, 2007 8:29 pm

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