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Problem:


Let \(f:\mathbb{R}\to \mathbb{R}\) be a continuous bounded function. Then

A. f has to be uniformly continuous

B. there exists an \(x\in \mathbb{R}\) such that \(f(x)=x\)

C. f cannot be increasing

D. \(\lim_{x\to \infty}f(x)\) exists.


Discussion:


Let us discard some answers first.

Define \(f(x)=sin(x^2)\). Then \(f\) is bounded, continuous.

Take \(x_1=\sqrt{n\pi+\pi/2}\), \(x_2=\sqrt{n\pi}\).

Then \(|x_1-x_2|=\frac{\pi/2}{\sqrt{n\pi+\pi/2}+\sqrt{n\pi}}<\frac{1}{\sqrt{n\pi}}\).

Since \(\frac{1}{\sqrt{n\pi}}\to 0\), given \(\delta >0\), we can have

\(|x_1-x_2|<\delta\) for large values of \(n\).

But \(|f(x_1)-f(x_2)|=1\). So given \(\epsilon =1/2\) we can never find a \(\delta >0\) for which \(|x_1-x_2|<\delta\) would imply that \(|f(x_1)-f(x_2)|<\epsilon\).

So this \(f\) is not uniformly continuous. So (A) is false.

Also from this example, since \(f\) does not have limit as \(x\to \infty\), we conclude that (D) is false.

Increasing does not mean strictly increasing. So, you are allowed to take a constant function as an example of increasing function. And it is bounded,continuous. This disproves (C).

Now we are left only with (C). Since we disproved all of the others, and one option is correct, (C) has to be true. We inspect the proof just to be sure.

Since \(f\) is bounded, there exists \(M>0\) such that \(|f(x)|<M\).

That means \(-M<f(x)<M\) for all \(x\in \mathbb{R}\).

Look at \(g(x)=f(x)-x\). Finding fixed point of \(f\) is same as finding a zero of \(g\).

We know that \(f\) is continuous, therefore \(g\) is continuous. We would like to find two points where \(g\) takes values with opposite signs. Now

\(g(M)=f(M)-M<0\) and \(g(-M)=f(-M)-(-M)=f(-M)+M>0\).

Therefore, \(g\) must cut the x-axis. Here we are using the intermediate value theorem for continuous functions.

This proves (B).


Helpdesk

  • What is this topic: Real Analysis
  • What are some of the associated concept: Intermediate value theorem, Bounded Function
  • Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert