# Extreme Value Theorem

Example 1. TBD.

### EVT + Integrals

The remaining content on this page utilizes integration, which is covered later in the course. Feel free to skip these examples are return after learning basic properties of integration.
Example 1. Let
$a,b\in\mathbb{R}$
obey
$a
. Show if
$f\colon[a,b]\rightarrow\mathbb{R}$
and
$g\colon[a,b]\rightarrow\mathbb{R}$
are continuous with
$f\geq 0$
, then there is
$c\in[a,b]$
such that
 ​$\displaystyle\int_a^b f(x)g(x)\ \mathrm{d}x=g(c) \int_a^b f(x)\ \mathrm{d}x.$​ ​
Hint
Proof Sketch
Formal Write-Up
Break the problem into two cases (whether
$f=0$
or not), and use linearity of integration.
If
$f=0$
, then the result follows immediately. Otherwise, the integral on the right hand side is positive, and so we can write
 ​$\displaystyle\dfrac{\int_a^b f(x)g(x)\ \mathrm{d}x}{\int_a^b f(x)\ \mathrm{d}x}=g(c).$​ ​
Since
$g$
is continuous on a closed interval, we can bound it from above and below and find a range of values and use linearity of integration to get the denominator on the left side to cancel out. At that point, we can use the intermediate value theorem.
If
$f(x)=0$
for all
$x\in[a,b]$
, then
 ​$\displaystyle 0=\int_a^bf(x)g(x)\ \mathrm{d}x=g(\alpha)\int_a^b f(x)\ \mathrm{d}x,\quad \text{for all}\ \alpha\in[a,b].$​ ​
In what remains, assume
$g$
is not identically zero. Since
$g$
is defined on the closed interval
$[a, b]$
, it follows from the extreme value theorem that
$g$
attains a maximum
$M$
and minimum
$m$
on the interval
$[a,b]$
. Linearity of the integral yields
 ​\begin{aligned}m\int_a^b f(x)\ \mathrm{d}x&=\int_a^bm\cdot f(x)\ \mathrm{d}x\\&\leq \int_a^b f(x)g(x)\ \mathrm{d}x\\&\leq \int_a^b M\cdot f(x)\ \mathrm{d}x\\&=M\int_a^b f(x)\ \mathrm{d}x.\end{aligned}​ ​
Since
$f$
is not identically zero, there is
$x^\star \in [a,b]$
such that
$f(x^\star)>0$
. Set
$\mathcal{C}\triangleq [a,b]\cap(x^\star-\delta,x^\star+\delta)$
. Because
$f$
is continuous, there is
$\delta > 0$
such that
$f(x) > f(x^\star)/2$
for all
$x\in\mathcal{C}$
. This implies
|
\begin{aligned}\int_a^b f(x)\ \mathrm{d}x\ \mathrm{d}x&\geq \int_{\mathcal{C}} f(x)\ \mathrm{d}x\\&\geq \int_{\mathcal{C}} \dfrac{f(x^\star)}{2}\ \mathrm{d}x\\ & = |\mathcal{C}|\cdot \dfrac{f(x^\star)}{2}\\&>0.\end{aligned}
| | | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | :-: |
Dividing the first inequality in this proof by this integral reveals
 ​$\displaystyle m \leq \dfrac{\int_a^b f(x)g(x)\ \mathrm{d}x}{\int_a^b f(x)\ \mathrm{d}x}\leq M$​ ​
where the denominator is nonzero by the prior inequality. By the intermediate value theorem, there is
$c\in[a,b]$
such that
 ​$g(c)=\dfrac{\int_a^bf(x)g(x)\ \mathrm{d}x}{\int_a^b f(x) \ \mathrm{d}x}.$​ ​
Rearranging gives the desired result, and the proof is complete.
 ​$\blacksquare$​ ​