If a sequence is Uniformly Cauchy, that implies uniform convergence for the sequence?

  • #1
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Homework Statement


Let ##X \subset \mathbb{C}##, and let ##f_n : X \rightarrow \mathbb{C}## be a sequence of functions. Show if ##f_n## is uniformly Cauchy, then ##f_n## converges uniformly to some ##f: X \rightarrow \mathbb{C}##.

Homework Equations


Uniform convergence: for all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that for any ##x \epsilon X## if ##n > N(\varepsilon)## then ##\vert f_n(x) - f(x) \vert < \varepsilon##.

Uniformly Cauchy: for all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that for any ##x \epsilon X##: if ##n, k > N(\varepsilon)## then ##\vert f_n(x) - f_k(x) \vert < \varepsilon##.

The Attempt at a Solution


Proof: Suppose ##f_n## is uniformly Cauchy. First we show ##f_n## converges point wise to some ##f : X \rightarrow \mathbb{C}##.

Fix some ##x \epsilon X##. Then ##f_n(x)## is a Cauchy sequence of complex values. By completeness, ##f_n(x)## converges to some ##f(x): X \rightarrow \mathbb{C}##. Thus, ##f_n## converges point wise to some ##f## on ##X##.

Let ##\varepsilon/2 > 0##. By definition of uniformly Cauchy, there exists some ##N_1 = N_1(\varepsilon/2)## such that for any ##x \epsilon X##: if ##l, k > N_1## then ##\vert f_l(x) - f_k(x) \vert < \varepsilon / 2##. By point wise convergence, there exists ##N_2 = N_2(\varepsilon, x)## such that if ##n > N_2## then ##\vert f_n(x) - f(x) \vert < \varepsilon / 2##.

..
Not sure, I think I'm on the wrong track, I was going to try to pick max(N_1, N_2) and use triangle inequality but I don't see how to do this/how this would help to show uniform conv.

Any hints?, please
 
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Answers and Replies

  • #2
wrobel
Science Advisor
Insights Author
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Try to assume the converse: there exist a constant ##c>0## , a sequence ##x_n## and a subsequence ##f_{i_n}## such that
$$|f(x_n)-f_{i_n}(x_n)|\ge c$$
............................
Contradiction.
 
Last edited:
  • #3
326
41
Try to assume the converse: there exist a constant ##c>0## , a sequence ##x_n## and a subsequence ##f_{i_n}## such that
$$|f(x_n)-f_{i_n}(x_n)|\ge c$$
............................
Contradiction.
Thanks for the reply,

The converse would be if ##f_n## converges uniformly to some ##f: X \rightarrow \mathbb{C}##, then ##f_n## is uniformly Cauchy. The text proved this case so I can use it, but I'm not sure how this statement implies, or why we assume ##\vert f(x_n) - f_{i_{n}}(x_n) \vert \ge c## for some constant ##c \epsilon \mathbb{C}##.

From Wikipedia(https://proofwiki.org/wiki/Limit_of_Subsequence_equals_Limit_of_Sequence): if ##a_n \rightarrow L## then for any subsequence of ##a_n## we have ##a_{n_k} \rightarrow L##.

From there though, we can say ##\lim_{n\rightarrow\infty} f_{i_{n}}(x_n) \neq f(x_n)##. By contrapositive of wikipedia statement, we have ##\lim_{n\rightarrow \infty} f(x_n) \neq f(x_n)##, a contradiction. We can conclude...
 
  • #5
326
41
I meant proof by contradiction
Thanks,

I have: Assume (the converse) if ##f_n## converges uniformly to some ##f: X \rightarrow \mathbb{C}##, then ##f_n## is uniformly Cauchy. Furthermore, assume ##f_n## is uniformly Cauchy.

Assume by contradiction, there exists some ##c \epsilon \mathbb{C}## and subsequence ##f_{i_n}## such that for all ##N(c)## if ##n > N(c)## then ##\vert f(x_n) - f_{i_n}(x_n) \vert \ge c##. But this implies ##\lim_{n\rightarrow\infty} f(x_n)## does not exist, a contradiction?. We can conclude for all ##\varepsilon > 0## there exists ##N(\varepsilon, x_n)##? such that if ##n > N(\varepsilon, x_n)## then ##\vert f(x_n) - f_{i_n}(x_n) \vert < \varepsilon##.

But I don't know where to get a contradiction because we only know ##f_n## converges point wise. Since we have a sequence of ##x##'s and not a fixed ##x## i'm not sure if we can use point wise convergence as a contradiction?
 
  • #6
wrobel
Science Advisor
Insights Author
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we can write
$$|f(x_n)-f_{i_n}(x_n)|\le |f(x_n)-f_k(x_n)|+|f_k(x_n)-f_{i_n}(x_n)|$$
Take ##\varepsilon=c/2## and choose ##n## big such that if ##j>i_n## then ##|f_j(x)-f_{i_n}(x)|<\varepsilon## for any ##x\in X.##
Choose ##k>i_n## such that for this fixed ##n## (and fixed ##x_n##) it follows that ##|f(x_n)-f_k(x_n)|<\varepsilon##
we have got ##|f(x_n)-f_{i_n}(x_n)|<c##
 
  • #7
wrobel
Science Advisor
Insights Author
739
451
Actually the proof can be simplified a lot.

You have: for any ##\varepsilon>0## there is a number ##N## such that if ##i,j>N## then
$$|f_i(x)-f_j(x)|<\varepsilon\qquad (*)$$ for each ##x\in X##
now fix ##i## and for each ##x## pass to the limit as ##j\to \infty## in formula (*)
you will have ##|f_i(x)-f(x)|\le\varepsilon##
 

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