# Jensen's inequality

### From Wikimization

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<math>\phi(ta + (1-t)b) \leq t \phi(a) + (1-t) \phi(b)</math> | <math>\phi(ta + (1-t)b) \leq t \phi(a) + (1-t) \phi(b)</math> | ||

- | whenever <math>\,0 \leq t \leq 1\,</math> and <math>\,a, b\,</math> are in the domain of <math>\,\phi\,</math>. | + | whenever <math>\,0 \leq t \leq 1\,</math> and <math>\,a\,, b\,</math> are in the domain of <math>\,\phi\,</math>. |

It follows by induction on | It follows by induction on | ||

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<br>'''Proof 1:''' By some limiting argument we can assume | <br>'''Proof 1:''' By some limiting argument we can assume | ||

- | that <math>\,f\,</math> is simple ( | + | that <math>\,f\,</math> is simple. (This limiting argument is a missing detail to this proof...) |

- | detail | + | <br>That is, <math>\,X\,</math> is the disjoint union of <math>\,X_1 \,\ldots\, X_n\,</math> |

and <math>\,f\,</math> is constant on each <math>\,X_j\,</math>. | and <math>\,f\,</math> is constant on each <math>\,X_j\,</math>. | ||

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<math>\,(f(a) - f(b)) / (a - b) \leq (f(a') - f(b')) / (a' - b')\quad\diamond</math> | <math>\,(f(a) - f(b)) / (a - b) \leq (f(a') - f(b')) / (a' - b')\quad\diamond</math> | ||

+ | The lemma shows: | ||

+ | *<math>\,\phi\,</math> has a right-hand derivative at every point, and | ||

+ | *the graph of <math>\,\phi\,</math> lies above the "tangent" line through any point on the graph with slope equal to the right derivative. | ||

- | + | Say <math>\,a = \int f d \mu\,</math> | |

- | + | ||

- | + | ||

- | + | ||

- | + | Let <math>\,m\,</math> be the right derivative of <math>\,\phi\,</math> | |

at <math>\,a\,</math>, and let | at <math>\,a\,</math>, and let | ||

## Revision as of 11:54, 26 July 2008

By definition is convex if and only if

whenever and are in the domain of .

It follows by induction on that if for then

(1)

Jensen's inequality says this:

If is a probability
measure on ,

is a real-valued function on ,

is integrable, and

is convex on the range
of then

(2)

**Proof 1:** By some limiting argument we can assume
that is simple. (This limiting argument is a missing detail to this proof...)

That is, is the disjoint union of
and is constant on each .

Say and is the value of on . Then (1) and (2) say exactly the same thing. QED.

**Proof 2:**

Lemma. If and then

The lemma shows:

- has a right-hand derivative at every point, and
- the graph of lies above the "tangent" line through any point on the graph with slope equal to the right derivative.

Say

Let be the right derivative of at , and let

The comment above says that for all in the domain of . So

D. Ullrich