*Notes to a video lecture on http://www.unizor.com*

__Extreme Value Theorem__

In this lecture we will consider

*real functions*

**of**

*f(x)**real argument*

**defined on a**

*x**closed segment*.

We will prove the following theorem.

*Extreme Value Theorem*

A

*continuous*function defined on a

*segment*(finite interval with endpoints) attains its extreme values.

In other words, there is at least one point within its domain where it reaches its maximum and there is at least one point where it reaches its minimum.

*Proof*

We will prove this theorem for a case of attaining the

*maximum*value. The corresponding proof for

*minimum*is completely analogous.

The proof is based on the

*boundedness theorem*that states that any continuous function

**defined on segment**

*f(x)***[**is bounded from above and below. It was proven in the previous lecture.

*a,b*]That means that a set of real values of our function

**{**is bounded from above. According to the

*f(x)*}*axiom of completeness*, there must be the

*least upper bound*for this set of values.

Assume,

**is this**

*M**least upper bound*for function

**on segment**

*f(x)***=**

*S*_{1}**[**:

*a,b*]∀

**∈**

*x***[**

*a,b*]**≤**

*f(x)*

*M*Consider now a set

**of all points**

*S*_{2}**in the domain of our function**

*x***, where the function value is greater than**

*f(x)***.**

*M−1/2*This set cannot be empty because in this case

**would be an**

*M−1/2**upper bound*that is smaller than the

*least upper bound*

**.**

*M*Next, consider set

**of all points**

*S*_{3}**in the domain of our function**

*x***, where the function value is greater than**

*f(x)***.**

*M−1/3*This set cannot be empty because in this case

**would be an**

*M−1/3**upper bound*that is smaller than the

*least upper bound*

**.**

*M*In addition, set

**is a subset of set**

*S*_{3}**since the restrictions on the values of function**

*S*_{2}**are more strict within this set.**

*f(x)*Continue this process with set

**of all points**

*S*_{4}**in the domain of our function**

*x***, where the function value is greater than**

*f(x)***,**

*M−1/4***of all points**

*S*_{5}**in the domain of our function**

*x***, where the function value is greater than**

*f(x)*

*M−1/5*Generally, set

**contains all points**

*S*_{n}**in the domain of our function**

*x***, where the function value is greater than**

*f(x)***.**

*M−1/n*Each of these sets cannot be empty because in this case

*M−1/n**upper bound*that is smaller than the

*least upper bound*

**.**

*M*Every subsequent set

**is a subset of the previous set**

*S*_{n+1}**since the restrictions on the values of function**

*S*_{n }**are more strict within this set:**

*f(x)***⊃**

*S*_{1}**⊃**

*S*_{2}**⊃...⊃**

*S*_{3}**⊃**

*S*_{n}**⊃...**

*S*_{n+1}Now we can pick a single point from each of these sets to obtain a

*bounded*sequence of points

**{**within segment

*x*}_{n}**[**, which, according to Bolzano - Weierstrass theorem should have a convergent subsequence

*a,b*]**{**with the limit also lying within our closed segment

*y*}_{k}**[**. Assume this limit is

*a,b*]**.**

*L*This limit point

**must be the point of maximum for function**

*L***, that is**

*f(x)***.**

*f(L)=M*To prove this, consider the following:

Since

**as**

*y*_{k}→L**and**

*k→∞***is continuous,**

*f(x)***.**

*f(y*_{k})→f(L)The distance between

**and the**

*f(y*_{k})*least upper bound*

**is an infinitesimal as index**

*M***increases to infinity because, if**

*k***∈**

*y*_{k}**,**

*S*_{k}**|**≤

*f(y*|_{k})−M**.**

*1/k*Therefore,

**must be the limit of**

*M***, that is**

*f(y*_{k})

*M =**lim*

_{k→∞}**f(y**_{k}) = f(L)As we have proven, the

*maximum*

**of function**

*M***is attained at point**

*f(x)***∈**

*L***[**.

*a,b*]End of proof.

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