Regulated Functions

[Dave]

Since everyone's posting questions here, I thought I'd post a nice question I got on my Analysis III assignment last week. It's a bit involved (and somewhat pointless) but I thought it was pretty nice for some reason:

 

Let [math]f : [a,b] \to \mathbb{R}[/math] be a regulated function, and let [math](x_n)_{n=1}^{\infty}[/math] be a sequence in [math](a,b)[/math] such that [math](x_n) \to a[/math] as [math]n \to \infty[/math]. Show that [math]( f(x_n) )_{n=1}^{\infty}[/math] is a Cauchy sequence.

[Dave]

It just struck me that for some reason, when I looked up regulated functions the other day, nobody seemed to know about them so maybe they're known under a different name; if anyone doesn't know what the hell I'm on about, please let me know

[bloodhound]

i ve been doing some stuff on sequences and sub sequences, but havent come across regulated functions or cauchy sequences. could u explain what they are

[Dave]

A Cauchy sequence is defined as follows:

 

A sequence [math](a_n)[/math] is Cauchy if [math]\forall \epsilon > 0 \,\,\, \exists N \in \mathbb{N}[/math] such that [math]\forall n, m > N, \,\,\, | a_n - a_m | < \epsilon[/math].

 

A function [math]f : [a,b] \to \mathbb{R}[/math] is regulated if [math]\forall \epsilon > 0\,\,\, \exists \varphi[/math] where [math]\varphi[/math] is a step function such that [math]| \varphi(x) - f(x) | < \epsilon[/math].

[matt grime]

Actually, thinking about this (i've never heard of regulated functions so was interested in doing this):

 

[math]|f(x_n)-f(x_m)|=|f(x_n)-\varphi(x_n)+\varphi(x_n)-\varphi(x_m)+\varphi(x_m)-f(x_m)|[/math]

 

use the triangle inequality to make that into three terms, and that [math]\varphi[/math] is a step function to conclude that since x_n converges to a, that all those terms can be made either less than e, and the whole thing less than 3e, or each less than e/3, and the whole less than e, for all n,m greater than some N.

 

Is that the proof you found? I only write it out to check because that style of argument is very useful, and is what you'll see when you show that a continuous function on a closed bounded interval is uniformly continuo

[bloodhound]

If a function is regulated, does that automatically imply that its continuous?

[matt grime]

No, a step function is trivially a regulated function.

[Dave]

Actually' date=' thinking about this (i've never heard of regulated functions so was interested in doing this):

 

[math']|f(x_n)-f(x_m)|=|f(x_n)-\varphi(x_n)+\varphi(x_n)-\varphi(x_m)+\varphi(x_m)-f(x_m)|[/math]

 

use the triangle inequality to make that into three terms, and that [math]\varphi[/math] is a step function to conclude that since x_n converges to a, that all those terms can be made either less than e, and the whole thing less than 3e, or each less than e/3, and the whole less than e, for all n,m greater than some N.

 

Is that the proof you found? I only write it out to check because that style of argument is very useful, and is what you'll see when you show that a continuous function on a closed bounded interval is uniformly continuo

 

That's the one. It's quite a nice proof, or so I thought

 

The advantage of regulated functions is that you can define their integral as follows:

 

If f is regulated on [a,b], with a sequence of step functions [math](\psi_n)[/math] converging uniformly to f (i.e. [math]d(f, \psi_n) \to 0[/math] as [math]n\to\infty[/math]), then define [math]\int_a^b f = \lim_{n\to\infty} \int_a^b \psi_n[/math].

 

I'm not sure how this relates to Riemann sums (mainly because I haven't done them yet) so I'll leave that to someone else.

[Dave]

Actually' date=' thinking about this (i've never heard of regulated functions so was interested in doing this):

 

[math']|f(x_n)-f(x_m)|=|f(x_n)-\varphi(x_n)+\varphi(x_n)-\varphi(x_m)+\varphi(x_m)-f(x_m)|[/math]

 

use the triangle inequality to make that into three terms, and that [math]\varphi[/math] is a step function to conclude that since x_n converges to a, that all those terms can be made either less than e, and the whole thing less than 3e, or each less than e/3, and the whole less than e, for all n,m greater than some N.

 

Is that the proof you found? I only write it out to check because that style of argument is very useful, and is what you'll see when you show that a continuous function on a closed bounded interval is uniformly continuo

 

That's the one. It's quite a nice proof, or so I thought

 

The advantage of regulated functions is that you can define their integral as follows:

 

If f is regulated on [a,b], with a sequence of step functions [math](\psi_n)[/math] converging uniformly to f (i.e. [math]d(f, \psi_n) \to 0[/math] as [math]n\to\infty[/math]), then define [math]\int_a^b f = \lim_{n\to\infty} \int_a^b \psi_n[/math].

 

I'm not sure how this relates to Riemann sums (mainly because I haven't done them yet) so I'll leave that to someone else.

[bloodhound]

We just started sequences of functions in our analysis module. We should be doing this stuff in a short while

[bloodhound]

We just started sequences of functions in our analysis module. We should be doing this stuff in a short while

[Dave]

Yeah, it's all pretty interesting stuff. We've just moved onto Fourier series, so expect the same kind of thing.

[Dave]

Yeah, it's all pretty interesting stuff. We've just moved onto Fourier series, so expect the same kind of thing.

[bloodhound]

ahhhh, ur one step ahead of me in everything.. have a module on that next term

[bloodhound]

ahhhh, ur one step ahead of me in everything.. have a module on that next term

[Dave]

Yeah, they seem to be piling the work on atm, completely snowed under with all the assignments that are due in

[Dave]

Yeah, they seem to be piling the work on atm, completely snowed under with all the assignments that are due in