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.

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

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

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].

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

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

No, a step function is trivially a regulated function.

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.

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.

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

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

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

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

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

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

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

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