analysis – eigenten

Throughout an undergrad math or engineering education, students like myself typically think of “spaces” as $\mathbb{R}, \mathbb{R}^n, \mathbb{C}$ . These are all finite-dimensional and constant spaces. Finite and constant meaning the actual element in the space is a finite list/vector of constants (reals/complex). Space of fixed size matrices, though at first may seem strange, are really just space of finite dim vectors. All of such spaces are typically also vector spaces and hence we can perform linear algebra on them. These are all the setting for a majority of undergraduate level problems in math and engineering.

However, one soon should get adjusted to thinking about infinite-dim spaces. And not just working on them as well but getting an intuition for it (which is the difficult part). My professor also mentioned that this is a good bridge to start going to less intuitive spaces. This is my first actual post, so I apologize for being overly pedantic (in both explanation and proof writing).

The Problem

A unique (at least for an undergrad) problem came up while working on a chapter on metrics and normed spaces. It involves a few definitions and is a good challenge if you’ve already done analysis at the level of Abbott’s and have a basic grasp of metric spaces. It’s short and concise but brings up quite a bit to comment on. It’s stated as follows:

Key words: Sequences, normed linear spaces, metric spaces, completion of a metric space, dense subset, convergence.

The space $X$ is just the space constituting of all real sequences that are finite (i.e., abruptly end at some point). Notice how an element $x \in X$ is thus defined a bit strangely.

Side tangent: Getting used to defining basic things (like spaces) rigorously can be tedious, but doing so allows you to later perform analysis on it much more simply. Also, understanding conventions for defining certain structures (such as this) allows one to more quickly read advanced math without having to get stuck on “what does this subscript mean here, why is there an N for every sequence” or “how is this even defined” itself later on. This, IMO, is a major part of the hidden curriculum of undergraduate math.

The problem uses a standard max norm function which simply takes the largest absolute element. Notice we use a max norm, in what case (what similar space X) would it be better to use a sup norm instead? Anyways, let’s begin with the solution for this.

Preliminaries

But wait, before calling it a norm, we need to verify that our space $X$ is a vector/linear space (we’ll stick to calling it linear, since someone at this level is soon to be studying functional analysis). After that, we need to check that our function $||\cdot||$ is a valid norm on our linear space $X$ . This is part a. All of this is fairly straight forward through verifying axioms, we’ll skip the details. So, we now know $(X, || \cdot ||)$ is a normed linear space.

And as hopefully you know, a normed linear space induces a metric space with metric $||x - y||, \; x,y \in X$ . Note that this may not be the only metric space involving space $X$ . Let’s move on to part b.

X is not Complete

In part b, our goal now is to show that the metric space induced from the normed linear space is not complete (i.e., that it is not a complete metric space). The notion of completion is something I’ve just recently learned, so bear with me while we work on this problem. Here’s one way to define the completeness property of a metric space.

A metric space is complete if all Cauchy sequences are convergent to a point in the space.

Let’s show our space is not complete. This means we want to show (WTS) that there exists some Cauchy sequence that converges to a limit outside of our space. These problems can be a blessing or a curse since one way to prove it is to find a specific example which can be either easy or annoyingly constructive.

This one is right in the middle. The plan is as follows: We have to first construct a sequence in our space (1), show it’s Cauchy (2), and then show that it converges to a point outside the space (3). Devising a basic plan to solve a problem is a useful and difficult skill to develop in doing proofs (I myself used to have a habit of “blindly” trying to solve it and praying that the steps I’m taking are leading me closer to the solution). It requires you to know how a problem statement can be divided into steps, and seeing how a certain step can lead to the next.

(1): This is a bit annoying notation wise, but we construct a sequence $(x^{(k)})$ with elements from our space of sequences $X$ , and define each element (a sequence) as:

$x^{(k)} = (1, 1/2, 1/3, ..., 1/k, 0, 0, ...)$

So our (main) sequence looks like

$(x^{(k)}) = (x^{(1)}, x^{(2)}, ...) = ((1, 0, 0, ...), (1, 1/2, 0, 0, ...), ...)$

(2): We now WTS that $(x^{(k)})$ is Cauchy. Intuitively, it should be somewhat clear that it is indeed Cauchy. For example, fix an epsilon, after some point N (which we want to find), the distance between any two elements $x^{(n)}, x^{(m)}$ should be sufficiently small (smaller than epsilon). This is because the points before that should cancel out to zero. So, we know it is indeed Cauchy.

Let $\epsilon > 0$ . Let $N = N(\epsilon)$ which we’ll define later. Then for all $n > m \geq N$ ,

$\| x^{(n)} - x^{(m)} \| = \| (1, 1/2, ..., 1/n, 0, 0, ...) - (1, 1/2, ..., 1/n, ..., 1/m, 0, 0, ...)\|$

$= \|(0, 0, 0, ..., 0, 1/(n+1), 1/(n+2), ..., 1/m, 0, 0, ... \|$

$= 1/(n+1) < 1/(N+1) < 1/N$

Observe that we can control how small this value is with $N$ . So for every $\epsilon$ , if we let $N(\epsilon) = 1/\epsilon = N$ , we have a Cauchy sequence since then $\| x^{(n)} - x^{(m)} \| = ... < 1/N = \epsilon$ .

(3): Now comes the main part — Showing our Cauchy sequence converges to a point outside the space. Let fixed $(x) = (x_1, x_2, ..., x_N, 0, 0, ...) \in X$ be any sequence in our space. We want to show that our Cauchy sequence of sequences $(x^{(k)})$ in fact does not converge to $x$ meaning it doesn’t converge to a point in our space. Assuming that our fixed $N$ from our $x$ ends earlier than some $k$ ,

$\|x^{(k)} - x\| = \|(1 - x_1, 1/2 - x_2, ..., 1/N - x_N, 1/(N+1), ..., 1/k, 0, 0, ...) \|$

Observe that as we take $k \to \infty$ , the value of above stays constant since our norm takes the absolute max. Thus, explicitly, $\forall k > N,$

$\|x^{(k)} - x\| = max \{1-x_1, 1/2-x_2, ..., 1/N - x_N, 1/(N+1)\} \geq 1/(N+1)$

Clearly $1/(N+1) \geq \epsilon$ for some $\epsilon$ . So for that $\epsilon$ , we know that the distance between our Cauchy sequence and $x$ never gets below a fixed constant. Therefore, our Cauchy sequence does not converge to a limit in our space. Hence our space is not complete.

X is dense in C_0

In part c), we define a new space $c_0$ to be the space of all real sequences $(x_0)$ that converge to 0, i.e., $x_n \to 0$ . In fact, $X$ is a subset of $c_0$ since each element in X eventually are all just zeros hence have elements that converge to 0. Thus, this part considers the normed linear space $(c_0, || \cdot ||)$ (which again induces a metric space).

The problem asks us to show that unlike $X$ , our new larger space $c_0$ is complete. We also want to show that $X$ is dense in $c_0$ . A common example of such is the rationals which are dense in the reals (which are complete under the typical metrics).

A subset $X$ of a metric space $c_0$ is said to be dense in $c_0$ if the closure of $X$ is the metric space.

Alternatively, subset $X$ of a metric space $c_0$ is dense in $c_0$ if $\forall c \in c_0, \exists (x_n) \in X$ such that $x_n \to c$ .

(1): First, let’s show that $X$ is dense in $c_0$ through the second definition. Let $x = (x_1, x_2, x_3, ...) \in c_0$ . Since $x \to 0$ , we know that $\forall \epsilon > 0, \exists N$ such that if $n > N$ , then

$|x_n - 0| = |x_n| < \epsilon$

Now we want to construct a sequence $(x^{(k)})$ in $X$ such that it converges to $x$ . Let $x^{(k)} = (x_1, x_2, ..., x_k, 0, 0, ...) \in X$ where element values are taken from $x$ . Expanded,

$(x^{(k)}) = (x^{(1)}, x^{(2)}, ...) = ((x_1, 0, ...), (x_1, x_2, 0, ...), ....)$

Therefore, $\forall \epsilon > 0, \exists N$ such that if $n > N$ , then

$\|x - x^{(n)}\| = \|(x_1, ..., x_N, ..., x_n, x_{n+1}, ...) - (x_1, ..., x_N, ..., x_n, 0, 0, ...)\|$

$= \|(0, 0, ..., x_{n+1}, x_{n+2}, ...)\| < \epsilon$

since $|x_n| < \epsilon, \; \forall n > N$ . Hence, we’ve shown $X$ is dense in $c_0$ .

(2): Now we want to show that $c_0$ in fact is complete. This part is very involved and incorporates a lot of nested indices, so we’ll skip this part for now. The goal nonetheless is as expected, to show that every Cauchy sequence converges to a point in the same space.

Ending

This wraps up the solution to a simple but somewhat involved problem introducing the space of sequences and other notions from metric spaces. Th proofs themselves are not atypical for an analysis problem. I suspect getting troubled by the bookkeeping of indices from all these constructions will be less of an issue as one advances in math. Getting lost in construction and manipulation is never the main focus. And building intuition for a problem through drawings, examples, and laying out a plan will always help clear things out.

This post is my first math related post and was written up in 3-4 hours as I was learning the material. These types of posts involving solving a specific problem shouldn’t be too common in the future.

eigenten

Tag: analysis

A problem on space of finite sequences