# 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 $[a,b]$,

 $f\left(\lambda_{1}x_{1}+\lambda_{2}x_{2}\right)\leq\lambda_{1}f(x_{1})+\lambda% _{2}f(x_{2}).$

But, since $\lambda_{1}+\lambda_{2}$ must be equal to 1, we can put $\lambda_{2}=1-\lambda_{1}$, so that the thesis becomes

 $f\left(\lambda_{1}x_{1}+\left(1-\lambda_{1}\right)x_{2}\right)\leq\lambda_{1}f% (x_{1})+\left(1-\lambda_{1}\right)f(x_{2}),$

which is true by definition of a convex function.

2) Taking as true that $f\left(\sum_{k=1}^{n-1}\mu_{k}x_{k}\right)\leq\sum_{k=1}^{n-1}\mu_{k}f\left(x_% {k}\right)$, where $\sum_{k=1}^{n-1}\mu_{k}=1$, we have to prove that

 $f\left(\sum_{k=1}^{n}\lambda_{k}x_{k}\right)\leq\sum_{k=1}^{n}\lambda_{k}f% \left(x_{k}\right),$

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,

 $a\leq x_{k}\leq b\Rightarrow\frac{\lambda_{k}}{1-\lambda_{n}}a\leq\frac{% \lambda_{k}}{1-\lambda_{n}}x_{k}\leq\frac{\lambda_{k}}{1-\lambda_{n}}b,$

and, summing over $k$,

 $a\sum_{k=1}^{n-1}\frac{\lambda_{k}}{1-\lambda_{n}}\leq\sum_{k=1}^{n-1}\frac{% \lambda_{k}}{1-\lambda_{n}}x_{k}\leq b\sum_{k=1}^{n-1}\frac{\lambda_{k}}{1-% \lambda_{n}},$

that is

 $a\leq\sum_{k=1}^{n-1}\frac{\lambda_{k}}{1-\lambda_{n}}x_{k}\leq b.$

We have, by definition of a convex function:

 $\displaystyle f\left(\sum_{k=1}^{n}\lambda_{k}x_{k}\right)$ $\displaystyle=$ $\displaystyle f\left(\sum_{k=1}^{n-1}\lambda_{k}x_{k}+\lambda_{n}x_{n}\right)$ $\displaystyle=$ $\displaystyle f\left(\left(1-\lambda_{n}\right)\sum_{k=1}^{n-1}\frac{\lambda_{% k}}{1-\lambda_{n}}x_{k}+\lambda_{n}x_{n}\right)$ $\displaystyle\leq$ $\displaystyle\left(1-\lambda_{n}\right)f\left(\sum_{k=1}^{n-1}\frac{\lambda_{k% }}{1-\lambda_{n}}x_{k}\right)+\lambda_{n}f\left(x_{n}\right).$

But, by inductive hypothesis, since $\sum_{k=1}^{n-1}\frac{\lambda_{k}}{1-\lambda_{n}}=1$, we have:

 $f\left(\sum_{k=1}^{n-1}\frac{\lambda_{k}}{1-\lambda_{n}}x_{k}\right)\leq\sum_{% k=1}^{n-1}\frac{\lambda_{k}}{1-\lambda_{n}}f\left(x_{k}\right),$

so that

 $\displaystyle f\left(\sum_{k=1}^{n}\lambda_{k}x_{k}\right)$ $\displaystyle\leq$ $\displaystyle\left(1-\lambda_{n}\right)f\left(\sum_{k=1}^{n-1}\frac{\lambda_{k% }}{1-\lambda_{n}}x_{k}\right)+\lambda_{n}f\left(x_{n}\right)$ $\displaystyle\leq$ $\displaystyle\left(1-\lambda_{n}\right)\sum_{k=1}^{n-1}\frac{\lambda_{k}}{1-% \lambda_{n}}f\left(x_{k}\right)+\lambda_{n}f\left(x_{n}\right)$ $\displaystyle=$ $\displaystyle\sum_{k=1}^{n-1}\lambda_{k}f\left(x_{k}\right)+\lambda_{n}f\left(% x_{n}\right)$ $\displaystyle=$ $\displaystyle\sum_{k=1}^{n}\lambda_{k}f\left(x_{k}\right)$

which is the thesis.

Title another proof of Jensen’s inequality AnotherProofOfJensensInequality 2013-03-22 15:52:53 2013-03-22 15:52:53 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 12 Andrea Ambrosio (7332) Proof msc 26D15 msc 39B62