# A Holder space is a Banach space

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1. Dec 16, 2015

### Domenico94

Hi everyone. I was just reading Evans' book on PDE, and, at some point, it asked to prove that an holder space is a Banach space, and I tried to do that. I just want to ask you if my proof is correct (if you see dumb errors, just notice also that I study EE, so I'm not much into doing proofs .
A banach space is defined if :
i) it has a norm;
ii)Its cauchy series converge to any point of space;
We can start with proving i), by saying that an Holder space has a norm by definition.
To prove ii), we can make the statement that in a function u, belonging to a Holder space, |u(x) - u(y)|<= C|x-y|^t , with t defined as an arbitrary exponent and c a costant in R. Then, if a Cauchy sequence is convergent to a point, that means that the distance between any two elements u(x) and u(y) will always be smaller than a small value, let's say, e.
so
d(u(x), u(y))<e.
That means that, if we choose a costant c small enough, and eventually, let's make it tend to 0, we would obtain
|u(x) - u(y)| <= c|x - y|^t <e, which proves ii.

Last edited: Dec 16, 2015
2. Dec 17, 2015

### Domenico94

Anyone?

3. Dec 17, 2015

### Krylov

Homework?
"sequences", not "series".
Usually $0 < t \le 1$. Maybe it's better to use another symbol, though.
Just to make sure: Note that $C$ will depend on $u$.
I'm not convinced. Hint: It is a bit easier when you use the fact that $C^r(\overline{\Omega})$ with the usual norm (= the sum of the suprema of the derivatives of orders up to and including $r$, which is dominated by the Hölder norm) is already a Banach space. Here $\Omega$ is a bounded domain in $\mathbb{R}^n$ and $r \in \mathbb{N}$.

4. Dec 17, 2015

### Svein

Yes, but...
1. A Cauchy sequence is of the form x1, x2, ...
2. Of course you can have functional elements
3. A Cauchy sequence is characterized by d(xn, xm)<ε for all n, m greater than some N.
4. A Cauchy sequence converges if there exists a "point" X in the underlying space such that d(xn, X)<ε for all n greater than some N.
5. (Definition) A normed linear space is called complete if every Cauchy sequence in the space converges.

5. Dec 17, 2015

### Domenico94

First of all, thanks for answering, but I didn't understand why C can depend on u.
If we take the definition of a Liptizchian function, it just says that |f(x) - f(y)|<K|x-y| for some K>0. It doesn't talk about depending on other functions at all.
To Svein: I understand, but the fact that teh quantities |f(x) - f(y)| and |x-y| are always <e, doesn't it imply that the distance between any two points d(x1, x2)<e?

6. Dec 17, 2015

### Krylov

True, $K$ does not depend on other functions, but different functions may have different $K$'s. For example, when you have a Cauchy sequence $(u_n)_{n \in \mathbb{N}}$ in the Hölder space $C^{r,\gamma}(\overline{\Omega})$, then each $u_n$ may have a different constant $C_n$. (However, since Cauchy sequences are bounded, it follows that $\sup_n{C_n} < \infty$. You will need this fact in the proof.)

• Take a Cauchy sequence $(u_n)_{n \in \mathbb{N}}$ in $C^{r,\gamma}(\overline{\Omega})$.
• Argue that this is also Cauchy in the Banach space $C^r(\overline{\Omega})$, hence converges to a limit in this space, say $u$.
• Prove that $D^ru$ is Hölder continuous with exponent $\gamma$, hence $u \in C^{r,\gamma}(\overline{\Omega})$.
• Prove that $u_n \to u$ in the Hölder norm. (For the latter, it suffices to prove that for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that $\|D^r(u-u_n)(x) - D^r(u-u_n)(y)\| \le \varepsilon \|x - y\|^{\gamma}$ for every $x, y \in \Omega$.)
Please do not take this wrongly, but maybe it would be good if you would take the time to read up on some elementary (functional) analysis? You seem a little lost and might feel more confident afterwards. Perhaps it would also help if you would tell a little more about your background and what you are wishing to accomplish.

7. Dec 17, 2015

### Domenico94

Now I seem to understand my mistake. Amyway, I m just reading about PDEs because I just like the topic and became interested in it. My background is that of an electronics engineering graduate (so, knowledge of ODEs, and integral transforms (Fourier and Laplace), and vector calculus, although not so deeply(divergence, curl, laplacian, ecc.)

8. Dec 18, 2015

### Svein

Yes. possibly. It does not, however imply that the sequence has a limit in the space.

Example: Let our space U be the open interval (0, 1). Define $x_{n}=\frac{2^{n}-1}{2^{n}}$ Then {xn} is a Cauchy sequence, but it has no limit in U.

9. Dec 18, 2015

### Krylov

I think it is very nice that you became interested in it, since it is a beautiful topic, and I'm sure it will be of relevance to you. (Actually, I know of quite a few EEs that have made meaningful contributions to applied mathematics in general.)

If you allow me to offer some unwanted advice, I think it would be easier to read Evans' PDE book after learning a little more analysis.
• A title that I like quite a bit is Zeidler's Applied Functional Analysis, published by Springer as books 108 and 109 in the Applied Mathematical Sciences series. His style is very accessible. You could start with the first book, which concludes with a nice discussion of classical PDE in the context of functional analysis. Cauchy sequences are introduced in $\S 1.3$, so right at the start.
• Although not familiar with it myself, I learned that mathematically inclined engineers often like to read Kreyszig's Introductory Functional Analysis with Applications published by Wiley. Cauchy sequences appear in $\S 1.4$.
• I like Brezis' Functional Analysis, Sobolev Spaces and Partial Differential Equations, also published by Springer, but it may be better to read this after learning a bit more analysis.
On the other hand, if you are more interested in how to use PDEs mainly or exclusively for modelling, then I am not the right person to offer advice. Also, if you want some more recommendations on background reading of any sort, you could look in the textbook forum or ask @micromass.

Finally, if you fancy more details on the steps I gave in post #6, I can provide them. An alternative would be to read up a bit and try to fill in the gaps yourself. It is not difficult, and you can always ask questions here.

10. Dec 18, 2015

### Krylov

Erratum: In the fourth step in post #6 should of course be added at the end the phrase: "and for every $n \ge N$".

11. Dec 18, 2015

### Domenico94

Yes..I ve also read the proof on internet...I noticed that I focused too much on distance between two points,and I didn't care about a sequence at all, while proving that. Anyway, I just want to thank you for the advice regarding the books.
I think I ll give a glance on them, although there s an Italian one as well, that I think I should read( gilardi, functional analysis). I m more interested in the theory anyway, in the "analytical" rather than "numerical" methods for solving them :)

12. Dec 18, 2015

### Krylov

Is it from this author? I think so. As far as I can see, it has not been translated to English. I'm always curious about new books on the topic. Enjoy!

13. Dec 22, 2015

### Domenico94

I don t think it has been translated,but I m Italian, so wouldn t be a problem :)