3

I am aiming to show the following inequality: $$\exp\left[\frac{1}{b-a}\int_{a}^bf(x)dx\right] \leq \frac{1}{b-a}\int_a^b \exp[f(x)]dx$$

where $f(x) \in C([a,b])$.

Intuitively, this makes sense since for some real numbers $r$ and $s$ $$\exp\left[\frac{r}{2}+\frac{s}{2}\right] = \exp\left[\frac{r}{2}\right]\exp\left[\frac{s}{2}\right] = \sqrt{\exp[r]\exp[s]} \leq \frac{\exp[r]}{2} + \frac{\exp[s]}{2}$$

i.e. the arithmetic mean of two numbers is at least as much as the geometric mean (e.g. geometric mean of 1 and 9 is 3 while the arithmetic mean of 1 and 9 is 5). But how do I deal with a more generalized version of this statement dealing with integrals as expressed above?

Zhanxiong
  • 14,040
Charles
  • 33
  • 5
    https://en.wikipedia.org/wiki/Jensen%27s_inequality – Robert Israel Aug 11 '15 at 23:21
  • IIRC, Hardy, Littlewood, and Polya often derive integral inequalities from the limit of summation inequalities. They comment that often a strict inequality (e.g., "<") gets converted to a non-strict inequality (e.g., "$\le$") when this happens. – marty cohen Aug 15 '15 at 03:51

2 Answers2

1

Arithmetic Geometric mean inequality states

$\dfrac 1 {b-a} \cdot \int_{a}^{b} g(x)dx\ge e^{\frac 1 {b-a} \cdot \int_{a}^b ln(g(x)) dx}$

for $g(x)$ being positive integrable function on $[a,b]$

To obtain your problem just replace $g(x)=e^{f(x)}$

Booldy
  • 1,070
  • 1
    Although your inequality is true, as Robert Israel commented, this is Jensen's Inequality applied with the convex function $e^x$. The AM-GM Inequality usually applies to finite sums. – robjohn Aug 14 '15 at 22:20
  • 1
    wel actually it is an aplication of Jensens inequality,but this inequality is known as integral AmGm inequality,for example see here https://www.google.ba/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&cad=rja&uact=8&ved=0CDYQFjAEahUKEwjR07azyqnHAhXFthoKHf_cBFg&url=http%3A%2F%2Frgmia.org%2Fpapers%2Fv2n1%2Fv2n1-10.pdf&ei=KmbOVdGWHcXtav-5k8AF&usg=AFQjCNHjh7ulX82N60tPydK_13CogZ3EWA – Booldy Aug 14 '15 at 22:25
1

I think you are on the right track. Consider $$ \sum_{i=1}^nf(x_i)\frac{b-a}{n}$$ where $x_i \in[a+(b-a)(i-1)/n,a+(b-a)i/n]$, is Riemann sum of $f$ continuous in $[a,b]$. Then for positive $n$, and because of the geometric mean being less than the arithmetic mean,

\begin{align} \exp\left(\frac{1}{b-a}\sum_{i=1}^nf(x_i)\frac{b-a}{n}\right)&=\exp\left(\frac{1}{n}\sum_{i=1}^nf(x_i)\right)\\ &=\sqrt[n]{\prod_{i=1}^n\exp\left(f(x_i)\right)}\\ &\le \frac{1}{n}\sum_{i=1}^n\exp(f(x_i))\\ &=\frac{1}{b-a}\sum_{i=1}^n\exp(f(x_i))\frac{b-a}{n} \end{align}

Take limits on both sides $n\rightarrow\infty$ and you get the result since the right side is Riemann sum of the function $\exp(f(\cdot))$. All continuous functions in a bounded closed interval are Riemann integrable.

Enredanrestos
  • 858