Showing Uniform Convergence of Cauchy Sequence of Functions

In summary, the text proved that if ##f_n## converges uniformly to some ##f: X \rightarrow \mathbb{C}## then ##f_n## is uniformly Cauchy. The text also assumed the converse, which is that there exist a constant ##c>0## and a sequence ##x_n## and subsequence ##f_{i_n}## such that$$|f(x_n)-f_{i_n}(x_n)|\ge c$$
  • #1
fishturtle1
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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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  • #2
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.
 
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  • #3
wrobel said:
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
wrobel said:
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
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
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##
 

FAQ: Showing Uniform Convergence of Cauchy Sequence of Functions

What is a Cauchy sequence of functions?

A Cauchy sequence of functions is a sequence of functions where the distance between any two functions in the sequence approaches zero as the sequence approaches infinity. In other words, the functions in the sequence become increasingly similar to each other as the sequence progresses.

How is uniform convergence of a Cauchy sequence of functions defined?

A Cauchy sequence of functions is said to converge uniformly if the distance between the functions in the sequence and the limit function approaches zero uniformly, meaning that the rate of convergence is the same at every point in the domain of the functions.

What is the difference between pointwise convergence and uniform convergence?

Pointwise convergence only requires that the functions in the sequence approach the limit function at each individual point in the domain, while uniform convergence requires that the convergence happens at the same rate at every point in the domain. Uniform convergence is a stronger condition than pointwise convergence.

How is uniform convergence of a Cauchy sequence of functions proven?

To prove uniform convergence of a Cauchy sequence of functions, one must show that the distance between the functions in the sequence and the limit function can be made arbitrarily small by choosing a sufficiently large index in the sequence. This can be done using the Cauchy criterion for uniform convergence.

What are some applications of uniform convergence of Cauchy sequence of functions?

Uniform convergence of Cauchy sequence of functions is a fundamental concept in real analysis and has many applications in mathematics and other fields. It is used in the proof of the Weierstrass approximation theorem, which states that any continuous function on a closed interval can be approximated by a sequence of polynomials. It is also used in the theory of Fourier series and in the study of infinite series. In addition, the concept of uniform convergence is important in the development of numerical analysis methods for solving mathematical problems.

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