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

__Bounded Functions__

In this lecture we will consider

*real functions*

**of**

*f(x)**real argument*

**.**

*x*The

*domain*of these functions will be a contiguous interval, finite or infinite, including or not including the endpoints.

A finite contiguous interval with endpoints included we will call

*segment*.

We will prove the following theorem.

*Boundedness Theorem*

A

*continuous*function defined on a

*segment*(finite interval with endpoints) is bounded from above and from below.

*Proof*

The proof is based on two main properties introduced in prior lectures:

(a) Bolzano - Weierstrass Theorem that states that from any bounded sequence we can extract a convergent subsequence.

(b) Continuity property.

We will only prove the boundedness from above, the one from below is completely analogous.

Let's assume the opposite, that our function

**is defined and continuous on segment**

*f(x)***[**, and is not bounded from above.

*a,b*]Then for any, however large, number

**we will be able to find an argument**

*N***∈**

*x*_{N}**[**such that

*a,b*]

*f(x*_{N}) ≥ NThe sequence

**{**consists from points in segment

*x*}_{N}**[**and, therefore, is

*a,b*]*bounded*by the endpoints of this segment.

According to Bolzano - Weierstrass Theorem, we can extract from it a subsequence

**{**of points in this segment that converges to some point

*y*}_{n}**∈**

*Y***[**.

*a,b*]Since a set of values of our function is unbounded on sequence

**{**, it is also unbounded on subsequence

*x*}_{N}**{**and, therefore, there no limit of

*y*}_{n}**as**

*f(y*_{n})**→∞.**

*n*But function

**is**

*f(x)**continuous*on segment

**[**, which means that, if

*a,b*]**{**→

*y*}_{n}**∈**

*Y***[**, then

*a,b*]**{**→

*f(y*}_{n})**. So, the limit of**

*f(Y)***does exist.**

*f(y*_{n})Came to a contradiction. Hence,

**is bounded from above.**

*f(x)*
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