proving the theorems of limits of functions

[Guest chordbook]

CAN ANYONE PLEASE PROVIDE ME WITH THE PROOF OF THEOREMS ON LIMITS OF FUNCTIONS?

[Dave]

I'm not familiar with that particular theorem. Can you at least state it so I can attempt to prove it?

[MandrakeRoot]

There are many statements about limits of functions. Please state the statement you would like to proof !

 

Mandrake

[matt grime]

"Theorems on the Limits of Functions" - sounds like there could be a book of 'em, and old one written by Hadamard or something.

[bloodhound]

so is he asking us to prove EVERTHING relating to limits of functions... good luck

[premjan]

epsilon-delta?

[bloodhound]

isnt that just a definition of a limit

[Dave]

Epsilon-delta is used to define things, they don't really constitute a theorem.

[matt grime]

It's not even as simple as epsilon-delta. Firstly convergence of functions on what space in what norm? Oh, and yes, I think it more than likely that this is going beyond what the poster is thinking of, but it is supposed to highlight the difficulty of answering that open ended question.

 

Given any interval of the real line, say [0,1] the space of continuous real valued functions have an infinite number of norms on it: the L^n norms. none of which are complete. And there's L^{infinity}, which is complete (compact so continuous is uniformly continuous, and hence uniform limit of uniformly continuous functions is continuous. Not to mention the other innerproducts on it.

 

Or do we mean R as the domain?

 

Or are we talking about results independent of the norm, ie general metric space results.

 

Or is it more along the lines of: if f is continuous and g is continuous show f+g is continuous (at a point)....

[MandrakeRoot]

Maybe he wants to know the following. If f is a uniformly continuous from some set A, a subset of a metric space (lets say a subset of R) into some metric space (lets say R again), then f has a continuous extension on the closure of A into R (or the appropriate space).

 

There are quite a lot of different theorems on limits though.

I doubt he wants you to check that the space of continours functions from some vectorspace X into some vectorspace Y is a vectorspace.

 

Mandrake

[bloodhound]

mandrake. can u explain the bit. "then f has a continuous extension on the closure of A into R"

[MandrakeRoot]

I will give an example. If f :(a,b) -> R is some uniform continuous function, then it is possible to find a continuous function g: [a,b] (the closure in R of (a,b)) -> R such that f(x) = g(x) for all x in (a,b), this g would be the extension of f to the closure of (a,b)....

 

Do you see what i mean ?

 

Mandrake

[bloodhound]

i never came across "uniform continuous" . I know "uniform convergence of sequence of functions" .. but uniform continuous function?

 

I get the extension of f to the closure of (a,b)

 

so all you have to take is if f:(a,b) -> R then

g[a,b] -> R

 

defined by

 

g(x)=f(x) if x in (a,b)

 

[math]g(a)=\lim_{n\to\infty}f(a_n)[/math] where [math](a_n)\subseteq(a,b)[/math] and [math]a_n\to a[/math]

 

g(b) defined similarly. is that correct?

[matt grime]

Sort of but it's important that f is uniformly continuous.

 

Uniform continuity means that when you do the formal e-d proof of continuity oat x, the d you choose can be independent of the point x.

 

 

For instance the function 1/x does not have a continuation onto [0,1] even though it is continuous on (0,1).

 

You see, the limit as x tends to 0 of 1/x doesn't exist, but it would if it were a uniformly continuous function.

[bloodhound]

could you give me some example of functions which are uniformly continuous in a open domain.

 

does uniform continuity of a function imply other properties of that function as well?? like injectivity, linearlity etc...

[matt grime]

Absolutely nothing of that kind is inplied:

 

Any continuous function on a closed interval (or compact if you know what that is) is uniformly continuous.

[bloodhound]

well, the definition of compact i have been taught is that (in R^d)

 

a subset C is said to be compact if every sequence [math](x_n) \subseteq C[/math] has a subsequence converging to a point to C

 

A more useful equivalent definition is given as well

 

C is compact iff C is closed and bounded.

 

I know a set can be bounded set doesnt have to be closed,

 

but I dont understand why a closed set doesnt always have to be bounded. Can some one give me a counterexample

[bloodhound]

could you also give me an example of a uniformly continuous function on an open interval.

 

As there is no point of that statement Mandrakeroot wrote if we assume that f is already assumed to be continuous on an closed interval.

[MandrakeRoot]

Well the real topological definition of a compact set is a little more complicated. The equivalence closed and bounded and compact only holds in specific topological spaces (like the R^d's). In topological vector spaces compact implies bounded and closed but the inverse is not perse true.

 

A closed set that is not bounded in R would be [0,infty)

 

If i would give you a uniformly continuous function on some open interval, with the above argument it could be immediately extended to a continuous function on the closed interval that is the closure of this open interval, so i dont see how we could give non-trivial exemples ?

 

In fact the whole strenght of the theorem is when we consider functions on some metric space X (not even supposed of finite dimension), then the closure of a set is less trivial and also the extension to go with. Or functions on some general open sets of R (that could be the union of many open intervals)

 

Mandrake

[matt grime]

Or, to put it in slightly different terms. If f is unformly continuous on some set S, it is unformly continuous on every subset of it. So x^2 is continuous on (0,1).

 

R is closed, when thought of as a subset of R, and that certainly isn't bounded. The integers are a closed subset of R and aren't bounded.

[MandrakeRoot]

Let me write down the entire theorem

Let [math](X,d)[/math] be a metric space and [math](Y,\rho)[/math] a complete metric space. Let [math]f : U \rightarrow Y[/math] be a uniformly continuous function on the subset [math]U \subseteq X[/math] of X. There exists a unique continuous function [math]g : \bar{U} \rightarrow Y[/math] such that [math]g|_{U} =f[/math].

 

The proof is quite simple really, i am sure you can do it bloodhound.

 

Mandrake

[bloodhound]

this is a bit silly, but what is a difference between a metric space and a complete metric space.

[MandrakeRoot]

A complete metric space is a metric space, but in this metric space all Cauchy-sequences are convergent. Note that in any metric space all convergent sequences are Cauchy !

 

(A Cauchy sequence is a sequences that sticks together in the tail :

Let {x_n}_n be a sequence in the metric space (X,d), then {x_n}_n is called a Cauchy sequence if for every epsilon > 0 there exists a N (integer) such that

d(x_n,x_m) < epsilon for all n,m >= N)

 

Often it is easily shown that a sequences is Cauchy and so using the fact that your space is complete you know that there is some limit for your sequence !

 

If you would like some hints for the proof i could write them down ?

 

Mandrake

 

Ps: it is not a silly question

[bloodhound]

yes please. some hints would be good.

 

could u give me some cauchy sequence which does not converge.

 

[edit] oops x_n=a for all n is a cauchy sequence that doesnt converge rite[/edit] where d is the usual metric in R

 

[edit] dOH.,,,, that ovbisouly converges

 

it should be x_n = n

[MandrakeRoot]

Yes i could. Consider the space (Q,d), where Q is the set of fractions and d is absolute distance (like the metric on R in fact) (note that we will pretend that R does not exist and therefore in the above replace epsilon by 1/k)

 

Take the following sequence defined by [math]q_{n+1} = q_n - \frac{q_n^2 - 2}{2q_n}[/math] with q_0 say 2;

 

This is clearly a sequence in Q that should converge to sqrt(2), but well that is not an element of Q.

 

Mandrake