Difference between revisions of "Jensen's Inequality"

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<math>F(a_1x_1+\dots+a_n x_n)\le a_1F(x_1)+\dots+a_n F(x_n)</math>
 
<math>F(a_1x_1+\dots+a_n x_n)\le a_1F(x_1)+\dots+a_n F(x_n)</math>
 
</center><br>
 
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The proof of Jensen's inequality is very simple. Since the graph of every convex function lies above its tangent line at every point, we can compare the function <math>\displaystyle{F}</math> with the linear function <math>\displaystyle{L}</math> whose graph is tangent to the graph of <math>\displaystyle{F}</math> at the point <math>a_1x_1+\dots+a_n x_n</math>. Then the left hand side of the inequality is the same for <math>\displaystyle{F}</math> and <math>\displaystyle{L}</math> while the right hand side is smaller for <math>\displaystyle{L}</math>. But the inequality for <math>\displaystyle{L}</math> is an identity!
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The proof of Jensen's inequality is very simple. Since the graph of every convex function lies above its tangent line at every point, we can compare the function <math>\displaystyle{F}</math> with the linear function <math>\displaystyle{L}</math>, whose graph is tangent to the graph of <math>\displaystyle{F}</math> at the point <math>a_1x_1+\dots+a_n x_n</math>. Then the left hand side of the inequality is the same for <math>\displaystyle{F}</math> and <math>\displaystyle{L}</math>, while the right hand side is smaller for <math>\displaystyle{L}</math>. But the inequality for <math>\displaystyle{L}</math> is an identity!
  
 
The simplest example of the use of Jensen's inequality is the [[quadratic mean]] - [[arithmetic mean]] inequality. Take <math>F(x)=x^2</math> and <math>a_1=\dots=a_n=\frac 1n</math>. You'll get <math>\left(\frac{x_1+\dots+x_n}{n}\right)^2\le \frac{x_1^2+\dots+ x_n^2}{n} </math>. Similarly, [[arithmetic mean]]-[[geometric mean]] inequality can be obtained from Jensen's inequality by considering <math>F(x)=-\log x</math>.
 
The simplest example of the use of Jensen's inequality is the [[quadratic mean]] - [[arithmetic mean]] inequality. Take <math>F(x)=x^2</math> and <math>a_1=\dots=a_n=\frac 1n</math>. You'll get <math>\left(\frac{x_1+\dots+x_n}{n}\right)^2\le \frac{x_1^2+\dots+ x_n^2}{n} </math>. Similarly, [[arithmetic mean]]-[[geometric mean]] inequality can be obtained from Jensen's inequality by considering <math>F(x)=-\log x</math>.

Revision as of 12:01, 21 June 2006

Let $\displaystyle{F}$ be a convex function of one real variable. Let $\displaystyle x_1,\dots,x_n\in\mathbb R$ and let $a_1,\dots, a_n\ge 0$ satisfy $a_1+\dots+a_n=1$. Then


$F(a_1x_1+\dots+a_n x_n)\le a_1F(x_1)+\dots+a_n F(x_n)$


The proof of Jensen's inequality is very simple. Since the graph of every convex function lies above its tangent line at every point, we can compare the function $\displaystyle{F}$ with the linear function $\displaystyle{L}$, whose graph is tangent to the graph of $\displaystyle{F}$ at the point $a_1x_1+\dots+a_n x_n$. Then the left hand side of the inequality is the same for $\displaystyle{F}$ and $\displaystyle{L}$, while the right hand side is smaller for $\displaystyle{L}$. But the inequality for $\displaystyle{L}$ is an identity!

The simplest example of the use of Jensen's inequality is the quadratic mean - arithmetic mean inequality. Take $F(x)=x^2$ and $a_1=\dots=a_n=\frac 1n$. You'll get $\left(\frac{x_1+\dots+x_n}{n}\right)^2\le \frac{x_1^2+\dots+ x_n^2}{n}$. Similarly, arithmetic mean-geometric mean inequality can be obtained from Jensen's inequality by considering $F(x)=-\log x$.