Jensen’s inequality

If $f$ is a convex function on the interval $[a,b]$, for each $\left\{x_{k}\right\}_{k=1}^{n}\in[a,b]$ and each $\left\{\mu_{k}\right\}_{k=1}^{n}$ with $\mu_{k}\geq 0$ one has:

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

A common situation occurs when $\mu_{1}+\mu_{2}+\cdots+\mu_{n}=1$; in this case, the inequality simplifies to:

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

where $0\leq\mu_{k}\leq 1$.

If $f$ is a concave function, the inequality is reversed.

Example:
$f(x)=x^{2}$
is a convex function on $[0,10]$. Then

 $(0.2\cdot 4+0.5\cdot 3+0.3\cdot 7)^{2}\leq 0.2(4^{2})+0.5(3^{2})+0.3(7^{2}).$

A very special case of this inequality is when $\mu_{k}=\frac{1}{n}$ because then

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

that is, the value of the function at the mean of the $x_{k}$ is less or equal than the mean of the values of the function at each $x_{k}$.

There is another formulation of Jensen’s inequality used in probability:
Let $X$ be some random variable, and let $f(x)$ be a convex function (defined at least on a segment containing the range of $X$). Then the expected value of $f(X)$ is at least the value of $f$ at the mean of $X$:

 $\mathrm{E}[f(X)]\geq f(\mathrm{E}[X]).$

With this approach, the weights of the first form can be seen as probabilities.

 Title Jensen’s inequality Canonical name JensensInequality Date of creation 2013-03-22 11:46:30 Last modified on 2013-03-22 11:46:30 Owner Andrea Ambrosio (7332) Last modified by Andrea Ambrosio (7332) Numerical id 13 Author Andrea Ambrosio (7332) Entry type Theorem Classification msc 81Q30 Classification msc 26D15 Classification msc 39B62 Classification msc 18-00 Related topic ConvexFunction Related topic ConcaveFunction Related topic ArithmeticGeometricMeansInequality Related topic ProofOfGeneralMeansInequality