Real Analysis continuity question

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trickslapper

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Given that f is a uniformly continuous function from the interval (0,1) into R, and given a sequence x[sub:5gkep5gd]n[/sub:5gkep5gd] in (0,1) that converges to 0, prove that the sequence f(x[sub:5gkep5gd]n[/sub:5gkep5gd]) is convergent.

if someone could push me in the right direction or show me how to solve this, i'd greatly appreciate it. thanks!
 
Assume xn -> x. By the continuous extension thm, there is an extension F of f such that F is continuous on [0,1]. (it agrees with f on (0,1)).

so lim f(xn) = limF(xn) since they agree on (0,1). since F is continuous and lim(xn) exists we can move the limit inside. So limf(xn)=limF(xn) = F(lim xn) = F(x).
 
is there another name for the continuous extension theorem? I don't think i've seen it in my class.

thanks!
 
Related Terms:

Continuous Linear Extension
Bounded Lienar Transformation or BLT
Hahn–Banach
Tietze Extension Theorem
 
What I typed works for an arbitrary limit and if you wished, an arbitrary interval (a,b). Replace x with 0.
 
You may also prove it using the fact that the reals are complete. That is, every cauchy sequence in R converges.

Note that f is uniformly continuous, so it satisfies the lipschitz condition, that is for some k>0, |f(x)-f(y)| < k|x-y|. Let epsilon > 0.

Then let N>0 such that n,m>N imply |xn-xm| < epsilon/k.

Got it?

Note this doesn't even use the fact that xn->0. As long as what is being approached is a limit point of the set, it doesn't matter.
 
Ok yes i am following somewhat now. I definitely need to use the definition of uniformly continous (its in my notes so i'm assuming thats what my professor wants me to use) I just don't see how to tie it all together to prove that f(xn) is convergent. Sorry theres a language barrier between the professor and the rest of the class so its hard to follow his examples.


Edit: nevermind i found the definition of a cauchy sequence in my book, should i be able to use the lipschitz definition of uniformly continuous function and the fact that xn is a cauchy sequence and tie those two results together to prove that f(xn) is convergent?
 
Sometimes it is seemingly impossible to prove "right from the definition".

Look at the implications of uniform continuity in your notes or book. It is standard to have proven almost immediately that 1) the derivative is bounded, 2) the Lipschitz condition, and 3) the continuous extension theorem and possible other things.

If you wanted to do it straight from the definition, you'll need to show that there exists some real number \(\displaystyle L\) such that for all \(\displaystyle \epsilon > 0\), there is an \(\displaystyle N > 0\) such that for every \(\displaystyle n > N\) we have \(\displaystyle |f(x_n) - L| < \epsilon\).

That looks a little more complicated, don't ya think? The continuous extension theorem guarantees a unique function \(\displaystyle F\) exists, and even gives you the answer, which is \(\displaystyle F(0)\).

That's why mathematicians develop a hierarchy of theorems, to make proofs simpler and (hope at least) make them easier to understand. At times you'll se obscure statements of theorems and definitions that won't make sense until much later.

edit: didnt see your edit
 
I understand the continuous extension theorem and it definitely makes it easier, and i'll ask my professor tomorrow if i can use it. But the Lipschitz condition is in my notes and a lot of the surrounding problems deal with the Lipschitz condition so i'm thinking thats the way to go (at least the way my professor intended). Do i need to show that f(xn) is a cauchy sequence which would then imply that it converges? Also will the interval (0,1) come in to play at all using the Lipschitz condition?
 
The fact that \(\displaystyle f(x_n)\) is Cauchy follows almost immediately from Lipschitz's condition and what I wrote above. That implies that \(\displaystyle f(x_n)\) converges.
 
k>0, |f(x)-f(y)| < k|x-y|. Let epsilon > 0.

Then let N>0 such that n,m>N imply |xn-xm| < epsilon/k.



so.. would it be |f(xn)-f(xm)|<k|xn-xm|, but |xn-xm|<epsilon/k which would imply that |f(xn)-f(xm)|<epsilon; which would prove that f(xn) is a cauchy sequence and therefore it converges ?

by the way thanks for the help, either way i'm gonna talk to my professor more about this (and the continous ext. theorem)
 
Yes, it looks fine. Don't be afraid to use real sentences though with, dare I say, simple language. I hate to see math being done entirely in symbols... it completely turns off an average interested reader. I like to treat my proofs like I would a short story :).
 
haha alright i usually pretty up my proofs also so that anyone could read them.

thanks for walking me through this problem!
 
So, i was reading my textbook and some of the other questions, and it seems like just because f is uniformly continuous doesn't mean that it satisfies the lipschitzian conditions. But the definition of uniformly continuous is awfully close: f is uniformly continuous on S if for every epsilon>0 there exists a delta>0 such that for all x,t in S the inequality |x-t| <delta implies |f(x)-f(t)|<epsilon.

Should i use this instead of this lipschitz condition? I also asked my professor and he said that if i was going to use it i would have to justify using it, but all i know about the function f is that it is uniformly continuous on (0,1) thats not enough to say that it satisfies the lipschitzian condition is it?

thanks!
 
Hmm, don't know how I let that slip by. The first solution I gave, I know is right. The generalized version of this question was on a qualifying exam for me ;)

\(\displaystyle f(x_n)\) must be Cauchy though. Since \(\displaystyle \{x_n\}\) converges, it is Cauchy. If \(\displaystyle \epsilon >0\) there is an \(\displaystyle N >0\) such that if \(\displaystyle n,m > N\) then \(\displaystyle |x_n-x_m| < \epsilon\).

\(\displaystyle f\) is uniformly continuous. So for all \(\displaystyle \epsilon'>0\), there is a \(\displaystyle \delta > 0\) such that \(\displaystyle |f(x)-f(y)|<\epsilon'\) whenever \(\displaystyle |x-y|<\delta\).

Try meshing these two together. It does work! Try using \(\displaystyle |x_n-x_m|\) along with \(\displaystyle \delta\).
 
Yea i asked my professor and he laughed when i mentioned the continuous ext theorem and said that i couldn't use it. I did end up doing what you said to do and yup it worked out.
 
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