# Why Banach spaces?

1. Jan 20, 2014

### Paalfaal

A fellow student of mine asked a question to our teacher in functional analysis, and the answer we got was not very satisfactory. In our discussion on Banach spaces the student asked "Why is it interesting/important for a normed space to be complete?". To my surprise the teacher said something along the lines; "well, we want sequences where the elements become very close to have a limit in the space".

What my fellow student and I was hoping for, was some deeper reason or explanation for why we are interested in complete normed spaces. Some deep theorems or results that require that a space is complete, for instance.

Does anyone have something to add to this discussion?

2. Jan 20, 2014

### jgens

This hits the nail on the head. Banach spaces should be nice generalizations of ordinary Euclidean space. Since all Euclidean spaces have this property it is a reasonable one to impose.

Another reason to make this requirement: There are some results essentially stating that infinite-dimensional linear algebra without topology is intractable. In addition to normability the completeness condition on the topology provides enough information to actually prove interesting theorems.

Most results in functional analysis have some reliance on completeness or convexity conditions. For example the bounded inverse theorem which says a continuous bijective linear operator has a continuous inverse.

Last edited: Jan 20, 2014
3. Jan 22, 2014

### Robert1986

We all know the the epsilon delta definition of a limit of a sequence. But, if I give you a sequence, to use this definition to determine if the sequence converges, you have to know (or have a good guess) about what the limit is converging to. If this is a sequene of numbers, it might not be so hard, but in a general normed vector space, it is quite difficult. However, it is usually easier to tell that the sequence is Cauchy, and so you know it converges to something in the space, if it is a Banach space.

Now, non Banach spaces also serve a purpose. Consider the normed vector space of polynomials on the interval [0,1] with L^1 norm. This is not a Banach Space, but is still important.

4. Jan 22, 2014

### WWGD

There is also the fact that a non-complete metric space M, in a sense, is as good as complete metric space, in that M has a metric completion, so you might as well use the advantages of completeness.

I think jgens was referring to the fact that having a topology allows you to talk of convergence when you deal with infinite sums ; in plain linear algebra, infinite linear combinations do not make sense without a notion of convergence, i.e., without Schauder bases. Hamel basis only take you so far, and do not allow you to work with infinite sums.

5. Jan 22, 2014

### jgens

This is not quite compelling (IMHO) since it can be difficult to determine properties of the original space just by looking at the completion. To each his own I suppose.

This is a good point although not the "no-go theorem" mentioned before. The result in mind essentially boils down to algebraic duals of infinite dimensional vector spaces not behaving especially well.

6. Jan 24, 2014

### Xiuh

7. Jan 25, 2014

### pwsnafu

8. Jan 29, 2014

### FactChecker

It is good to know that properties proven for a space will still hold for their limits. That is not true a lot of times. A property proved for every member of a complete space will still hold for limits of those members.

9. Jan 30, 2014

### mathwonk

In analysis we usually solve problems by finding a sequence of approximate solutions whose limit is an actual solution. Hence we need the limit to exist in our space. This is why we do calculus in the real numbers, a Banach space, rather than the rational numbers. Otherwise problems like: find x with x^2 = 2 would have no solution.

A trick often used is to take the formal completion of a given space, find a solution in that larger space, and then try to prove the solution actually lies in the original space. Examples of this are the use of "distributions" to solve differential equations, coupled with smoothness theorems to show the solutions found are actual smooth functions after all. This is not always easy, but math is not always an easy subject.

(consider the regularity theorem on p. 223 of warner's book:

https://www.amazon.com/Foundations-...8&qid=1391099094&sr=1-1&keywords=frank+warner)

specific examples of theorems which are true in the presence of the complete ness hypothesis include the open mapping theorem and the closed graph theorem, as in these notes:

http://www.math.ucsd.edu/~bdriver/240-01-02/Lecture_Notes/current_versions/chapter16.pdf

Last edited: Jan 30, 2014
10. Mar 15, 2014

### ndjokovic

Well, there's a lot of reasons why Banach spaces are so important. For example in the field of random and deterministic processes or also called Stochastics and Evolution Equations. Partial differential equations can be written in the form of abstract functional differential equations:
$$\frac{dx(t)}{dt}=Ax(t),$$ where $A$ is an operator operating in infinite dimensional spaces. As you see this equation can be solved using exponentials:
$$x(t)=e^{tA}x(0),$$
So we need usually to define the exponential of an operator which has the same properties as the real exponential function. But to extend this definition to operators we need to have a property about series well known in the real analysis: "Any absolutely convergent series in R is convergent",
But to have this property in an abstract space (infinite dimensional) we need it to be a Banach space: http://en.wikipedia.org/wiki/Absolute_convergence#Proof_that_any_absolutely_convergent_series_in_a_Banach_space_is_convergent
In this case, we can successfully define the exponential of an operator by the formula:
$$e^A=\sum_{n=0}^{\infty}\frac{A^n}{n!}$$
This is just an example, there's lot of other reasons why Banach spaces are important. I can give you more examples if you want.

Last edited: Mar 15, 2014
11. Mar 15, 2014

### economicsnerd

Some examples of things that work for Banach space, but stop being true if we replace "Banach space" with "normed linear space":
- Every absolutely summable series in a Banach space has a sum.
- If $X,Y$ are Banach spaces, and $T:X\to Y$ is a linear continuous bijection, then the inverse map $T^{-1}:Y\to X$ is also a linear continuous map. This really helps us get some nice structure on $\mathcal B(X)=\{T:X\to X: \enspace T \text{ is linear and continuous}\}$ for a Banach space $X$.
- If $X,Y$ are Banach spaces, and $(T_n)_{n=1}^\infty$ is a sequence of linear continuous maps $X\to Y$ which converges pointwise to $T:X\to Y$ (i.e. $T_n(x)\to T(x)$ as $n\to\infty$ for every $x\in X$), then $T$ is itself a linear continuous map.

Of course, there are important reasons to also concern ourselves with normed linear spaces in which the above may fail. That said, preserving the facts above helps make infinite-dimensional linear analysis look a lot more like finite-dimensional real analysis.

12. Mar 16, 2014

### mathwonk

"Every absolutely summable series in a Banach space has a sum." for a normed space, this is equivalent to being a banach space, i.e. to completeness. so this is a tautology.

13. Mar 16, 2014

### johnqwertyful

I think of "complete" as "everything that "should" have a limit DOES have a limit". Does it make sense why we like that? Would you like spaces where things that "should" have limits DON'T?

14. Mar 17, 2014

Agreed.