# Show that a space is a Banach space

• member 428835
Read moreIn summary, the given conversation is discussing the proof that a specific space equipped with a given norm is a Banach space. The space in question is ##C^k[a,b]##, defined as the set of all continuous functions ##u:[a,b]\to \mathbb R## that have continuous derivatives up to order ##k## on the interval ##[a,b]##. The norm for this space is defined as $$||u||:= \sum_{j=0}^k \max_{a\leq x\leq b}|u^{(j)}(x)|$$ where ##u^{(j)}## is the ##j##th derivative. The conversation focuses on how to prove that
member 428835

## Homework Statement

Show the following space equipped with given norm is a Banach space.

Let ##C^k[a,b]## with ##a<b## finite and ##k \in \mathbb{N}## denote the set of all continuous functions ##u:[a,b]\to \mathbb R## that have continuous derivatives on ##[a,b]## to order ##k##. Define the norm $$||u||:= \sum_{j=0}^k \max_{a\leq x\leq b}|u^{(j)}(x)|$$
where ##u^{(j)}## is the ##j##th derivative.

## Homework Equations

Nothing comes to mind.

## The Attempt at a Solution

No idea how to start this. I believe a Banach space is when every Cauchy sequence converges. So then I need to show

$$||u_n - u_m||= \sum_{j=0}^k \max_{a\leq x\leq b}|u_n^{(j)}(x)-u_m^{(j)}(x)| < \epsilon$$

But how?

Delta2
joshmccraney said:

## Homework Statement

Show the following space equipped with given norm is a Banach space.

Let ##C^k[a,b]## with ##a<b## finite and ##k \in \mathbb{N}## denote the set of all continuous functions ##u:[a,b]\to \mathbb R## that have continuous derivatives on ##[a,b]## to order ##k##. Define the norm $$||u||:= \sum_{j=0}^k \max_{a\leq x\leq b}|u^{(j)}(x)|$$
where ##u^{(j)}## is the ##j##th derivative.

## Homework Equations

Nothing comes to mind.

## The Attempt at a Solution

No idea how to start this. I believe a Banach space is when every Cauchy sequence converges. So then I need to show

$$||u_n - u_m||= \sum_{j=0}^k \max_{a\leq x\leq b}|u_n^{(j)}(x)-u_m^{(j)}(x)| < \epsilon$$

But how?

No, you do not need to show this, because it is your starting assumption; that is, you are assuming that ##\{ u_n \}## is a Cauchy sequence, so you are assuming that, given any ##\epsilon > 0## you have ##\| u_n - u_m \| < \epsilon## for all ##m,n > N(\epsilon)##. What you DO need to show is that this implies ##u_n \to u## for some ##u \in C^k[a,b]##. In other words, you need to show that the limit function ##u## is ##k##-times continuously differentiable.

Last edited:
Delta2
Another criterion for whether a normed space is Banach, that might be of use here is that

Every absolutely convergent series of the space V, converges in space V that is

if ## \sum ||u_n||## converges (in R) then ##\sum u_n ## converges in V then V is Banach space.

Last edited:
Ray Vickson said:
No, you do not need to show this, because it is your starting assumption; that is, you are assuming that ##\{ u_n \}## is a Cauchy sequence, so you are assuming that, given any ##\epsilon > 0## you have ##\| u_n - u_m \| < \epsilon## for all ##m,n > N(\epsilon)##. What you DO need to show is that this implies ##u_n \to u## for some ##u \in C^k[a,b]##. In other words, you need to show that the limit function ##u## is ##k##-times continuously differentiable.
I follow you, but I have no clue how to start. Any hints?

Delta2, I'm really not sure how this helps. Maybe I'm just blind?

joshmccraney said:
I follow you, but I have no clue how to start. Any hints?
I would start with a proof for ##k=0## and then ##k=1## to see, whether an induction will work.

Delta2
joshmccraney said:
I follow you, but I have no clue how to start. Any hints?

Delta2, I'm really not sure how this helps. Maybe I'm just blind?

How to start: consult calculus books about limit properties for convergent function sequences. There are tons of theorems available, and most good calculus III books will have all you need.

Ray Vickson said:
How to start: consult calculus books about limit properties for convergent function sequences. There are tons of theorems available, and most good calculus III books will have all you need.
Is that real advice: read a book? I read the chapter of our book, I googled examples, I did not come here hoping for you to give me all the answers. I wanted a place to start. Perhaps you're being genuine, but it comes across a little hurtful: "read a book".

Delta2
joshmccraney said:
Is that real advice: read a book? I read the chapter of our book, I googled examples, I did not come here hoping for you to give me all the answers. I wanted a place to start. Perhaps you're being genuine, but it comes across a little hurtful: "read a book".
You got a point, I can speak for myself and my hint wasn't good enough, maybe @Ray Vickson wasn't too hintful either. You caught me off guard in this , I need to go back and read my textbook and notes on complete normed spaces (dated back to my undergraduate years @ 1998!

But maybe @fresh42 gave a good hint, if you "study " how the norm behaves for k=0, and try for this case of k=0 to prove that every Cauchy sequence of continuous functions converges to a continuous function under this norm.

member 428835
A question ahead: Don't you also have to show that the defined norm is actually one, or is this given?

You have to construct the limit function ##u(x):=\lim_{n \to \infty} u_n(x) ## somehow, and pointwise is all we have. The proof for ##k=0## shows that a) ##u(x)## is continuous and b) the convergence ##u_n \longrightarrow u##, mainly by application of the triangle inequality.
In the next step I would try to extend this on ##k=1##. Once you've seen how this is done, the rest will probably be easy.

joshmccraney said:
Is that real advice: read a book? I read the chapter of our book, I googled examples, I did not come here hoping for you to give me all the answers. I wanted a place to start. Perhaps you're being genuine, but it comes across a little hurtful: "read a book".

OK, let me expand a bit. Do you think that convergence in the norm ##\| \cdot \|## implies uniform convergence? If you did have uniform convergence, what function properties would be conserved when taking the limit?

fresh_42 said:
A question ahead: Don't you also have to show that the defined norm is actually one, or is this given?

You have to construct the limit function ##u(x):=\lim_{n \to \infty} u_n(x) ## somehow, and pointwise is all we have. The proof for ##k=0## shows that a) ##u(x)## is continuous and b) the convergence ##u_n \longrightarrow u##, mainly by application of the triangle inequality.
In the next step I would try to extend this on ##k=1##. Once you've seen how this is done, the rest will probably be easy.
Okay, thanks for elaborating, I'll look into this. Regarding the norm question, I'm not sure. The examples in the book first show that the proposed norm is a norm, but after that they say they still need to show the proposed space is a Banach space. I don't think I have to show it's a norm, but I believe I can do this without help from PF.

I should also say all the examples in the book invoke the Cauchy Criterion, but the book never formally defines it (it's an applied functional analysis course, and I'm an engineer so this is a bit foreign). I seem to recall from an undergrad analysis course that all Cauchy sequences converge. Then it seems trivial: what am I missing?

Ray Vickson said:
OK, let me expand a bit. Do you think that convergence in the norm ##\| \cdot \|## implies uniform convergence? If you did have uniform convergence, what function properties would be conserved when taking the limit?
Thanks for elaborating. I have to leave for a bit, but I'll think about this and get back to you.

Thanks all for responses. Very helpful!

joshmccraney said:
I seem to recall from an undergrad analysis course that all Cauchy sequences converge.
They converge iff the space where they live in is complete. So to show completeness, which is a crucial property in the definition of a Banach space, you have to show that they actually do converge. E.g. you can have a sequence of rational numbers which approaches ##\sqrt{2}##, but as ##\sqrt{2} \notin \mathbb{Q}## it does not converge there. It converges in ##\mathbb{R}##, and this makes the difference between rationals and reals. Thus the crucial point here is to show, that the function ##u(x):= \lim_{n \to \infty}u_n(x)## is in ##C^k##, i.e. that it a) has the differentiation properties, and b) that ##||u_n - u||_\infty \longrightarrow 0## with the given norm. Then the limit we get by a pointwise convergence of ##u_n(x)##, which exists, since ##\mathbb{R}## is complete, is also a function in ##C^k## plus it converges in the given norm.

## 1. What is a Banach space?

A Banach space is a complete normed vector space. This means that it is a vector space equipped with a norm (a way to measure the size of a vector) that satisfies the completeness property. Completeness means that every Cauchy sequence (a sequence of vectors that gets arbitrarily close to each other) in the space converges to a limit that is also in the space.

## 2. How can I show that a space is a Banach space?

To show that a space is a Banach space, you need to prove two things: first, that it is a normed vector space, meaning it has a norm that satisfies certain properties. Second, you need to show that it is complete, meaning that every Cauchy sequence in the space converges to a limit that is also in the space. This can be done by using the definition of a Cauchy sequence and the completeness property of the space.

## 3. What is the importance of Banach spaces in mathematics?

Banach spaces are important in mathematics because they provide a framework for studying and understanding vector spaces in a more general and abstract way. They also have many applications in areas such as functional analysis, partial differential equations, and harmonic analysis.

## 4. Are all normed vector spaces Banach spaces?

No, not all normed vector spaces are Banach spaces. In order for a normed vector space to be a Banach space, it must also satisfy the completeness property. This means that not only does it have a norm, but every Cauchy sequence in the space must converge to a limit that is also in the space.

## 5. Can a Banach space be finite-dimensional?

Yes, a Banach space can be finite-dimensional. In fact, all finite-dimensional normed vector spaces are Banach spaces. This is because in a finite-dimensional space, all Cauchy sequences converge, so the completeness property is automatically satisfied.

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