Inverse Functions and Uniform Convergence

(1)
Let g:[a,b]->R be a discontinuous function. Suppose that g is 1-1 and has image g([a,b])=[c,d]. Can the inverse function g^-1 [c,d]->[a,b] be continuous?

My intuition says no, and my argument goes as follows.
I can show that g continuous => g^-1 continuous. Thus because both functions are inverses of one another, the converse is also true.

Can anyone spot the flaw in my reasoning? (There must be a flaw; questions are never this easy... :rolleyes:

)

(2)
Can anyone explain, in plain English, the meaning of uniform convergence (in the context of sequences of functions)? I know how to test for it, and I understand the whole "N doesn't depend on x" thing, but I just don't really see the point and I'm finding it hard to visualise what's going on.

Please help :smile:

Reply 1

DFranklin

18

James Gurung

(1)
Let g:[a,b]->R be a discontinuous function. Suppose that g is 1-1 and has image g([a,b])=[c,d]. Can the inverse function g^-1 [c,d]->[a,b] be continuous?

My intuition says no, and my argument goes as follows.
I can show that g continuous => g^-1 continuous. Thus because both functions are inverses of one another, the converse is also true.I agree with your logic, but I don't see that it's obvious that g cts => g^-1 cts. You need to prove it, and I can't say I see an easy proof.

(2) Can anyone explain, in plain English, the meaning of uniform convergence (in the context of sequences of functions)? I know how to test for it, and I understand the whole "N doesn't depend on x" thing, but I just don't really see the point and I'm finding it hard to visualise what's going on.

Basically, it means the functions that as N->infinity, the functions get "close" over the entire interval at once, rather than at individual points. If you have f_n->f uniformly, then when you do the epsilon>0 thing, you're basically drawing a "worm" lying slightly above and slightly below f, and saying for n>N, the f_n lie inside the "worm". Dunno if that helps.

The big thing about uniform convergence is that if f_n->f uniformly, lots of things we'd want to be true "just work". In particular, if the f_n are all continuous, then so is f.

Reply 2

James

OP

14

Thank you. I'm glad my intuition was right for once :smile:

. We have seen the proof that g cts => g^-1 cts in lectures; it's here if you're interested:

http://www.maths.ox.ac.uk/current-students/undergraduates/lecture-material/Mods/analysis2/pdf/analysis2-notes.pdf

Thanks also for explaining about uniform convergence.

Generally, this term of analysis seems easier than last, but things are making less sense. Sequences and series were very straightforward (although I wouldn't have said that at the time!), whereas some of this functions stuff seems a bit obscure. Why do we care about sequences of functions, uniform continuity and convergence, etc?? Also some of the proofs like the IVT are pretty horrendous. The worst one we had to do last term was AOL Products, and that wasn't exactly hard!!

Anyway, I'm not sure why I went into that. :biggrin:

Reply 3

DFranklin

18

James Gurung

Thank you. I'm glad my intuition was right for once :smile:

. We have seen the proof that g cts => g^-1 cts in lectures; it's here if you're interested:

http://www.maths.ox.ac.uk/current-students/undergraduates/lecture-material/Mods/analysis2/pdf/analysis2-notes.pdfFunny - I think that proof is a lot less obvious than the IVT.

Why do we care about sequences of functions, uniform continuity and convergence, etc??

So you can prove that lots of things that you'd like to be true actually are true. It's not all abstract stuff either - you'll end up proving that Taylor series actually do what they're supposed to and (later on) that Fourier series do as well.

But possibly it's not for you. At university level, most people will only really enjoy quite a small subset of their course.

Reply 4

James

OP

14

Lol, the IVT was just an example. I don't find either proof straightforward!

The funny thing is that I actually do quite enjoy Analysis. It takes me a long time to get things, but eventually they do click and then it's fun to tackle some of the more challenging problems (this contrasts with calculus, which so far has all been very mundane). The topic I'm currently finding a nightmare is probability - I just don't get it!

I suppose that everything becomes more enjoyable when you understand the basics. It's like with power series last term; initially I couldn't get my head around the ratio test (my notes were wrong), but when someone explained it on TSR everything suddenly became very simple!

I love maths :smile:

Reply 5

dvs

10

James Gurung

(2)
Can anyone explain, in plain English, the meaning of uniform convergence (in the context of sequences of functions)? I know how to test for it, and I understand the whole "N doesn't depend on x" thing, but I just don't really see the point and I'm finding it hard to visualise what's going on.

The way I like to think about uniform convergence is that the functions f_n converge as a whole to f, as opposed to pointwise convergence, where values of f_n(x) converge to f(x). Symbolically, I find
f_n -> f (uniform)
f_n(x) -> f(x) (pointwise)
to be very descriptive.

This type of convergence is of course more demanding that mere pointwise convergence, but as a consequence, it yields rich results (such as the retention of continuity and integrability).

Also, in regards to the proof of the IVT, you should realise that it's basically one (deep) idea: the least upper bound property. The rest is just mechanical work. You will learn to appreciate this property later on when you study topology.