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Extreme Value Theorem

EVT + Functions

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,bRa,b\in\mathbb{R}
obey
a<ba<b
. Show if
f ⁣:[a,b]Rf\colon[a,b]\rightarrow\mathbb{R}
and
g ⁣:[a,b]Rg\colon[a,b]\rightarrow\mathbb{R}
are continuous with
f0f\geq 0
, then there is
c[a,b]c\in[a,b]
such that
abf(x)g(x) dx=g(c)abf(x) dx.\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=0f=0
or not), and use linearity of integration.
If
f=0f=0
, then the result follows immediately. Otherwise, the integral on the right hand side is positive, and so we can write
abf(x)g(x) dxabf(x) dx=g(c).\displaystyle\dfrac{\int_a^b f(x)g(x)\ \mathrm{d}x}{\int_a^b f(x)\ \mathrm{d}x}=g(c).
Since
gg
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)=0f(x)=0
for all
x[a,b]x\in[a,b]
, then
0=abf(x)g(x) dx=g(α)abf(x) dx,for all α[a,b].\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
gg
is not identically zero. Since
gg
is defined on the closed interval
[a,b][a, b]
, it follows from the extreme value theorem that
gg
attains a maximum
MM
and minimum
mm
on the interval
[a,b][a,b]
. Linearity of the integral yields
mabf(x) dx=abmf(x) dxabf(x)g(x) dxabMf(x) dx=Mabf(x) dx.\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
ff
is not identically zero, there is
x[a,b]x^\star \in [a,b]
such that
f(x)>0f(x^\star)>0
. Set
C[a,b](xδ,x+δ)\mathcal{C}\triangleq [a,b]\cap(x^\star-\delta,x^\star+\delta)
. Because
ff
is continuous, there is
δ>0\delta > 0
such that
f(x)>f(x)/2f(x) > f(x^\star)/2
for all
xCx\in\mathcal{C}
. This implies
|
abf(x) dx dxCf(x) dxCf(x)2 dx=Cf(x)2>0.\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
mabf(x)g(x) dxabf(x) dxM\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[a,b]c\in[a,b]
such that
g(c)=abf(x)g(x) dxabf(x) dx.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