Show that a space is a Banach space

[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?

 

[Ray Vickson]

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.

 

[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.

 

[member 428835]

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?

 

[fresh_42]

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.

 

[Ray Vickson]

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.

 

[member 428835]

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]

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.

 

[fresh_42]

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.

 

[Ray Vickson]

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?

 

[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.

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?

 

[member 428835]

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!

 

[fresh_42]

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.