Cauchy sequences in an inner product space

[Oxymoron]

Im in need of some guidance. No answers, just guidance.

Question.

Let $$(x_m)$$ be a Cauchy sequence in an inner product space, show that

$$\left\{\|x_n\|:n=1,\dots,\infty\right\}$$

is bounded.

$$proof$$

From the definition we know that all convergent sequences are Cauchy sequences. But it is not true that the converse always holds. However, in normed linear spaces the converse is always true. That is our Cauchy sequence is convergent to a point $$x$$.

Since $$(x_m)$$ is a convergent sequence we take an $$\epsilon$$ such that

$$\|x_m - x\| \leq \epsilon$$ if $$m > N$$.

Now we fix $$N$$ so that

$$\|x_m\| \leq \|x_m - x\| + \|x\| < \epsilon + \|x\|$$

which holds for all $$m > N$$. This is just a matter of adding $$\|x\|$$

Now we define

$$M = \mbox{max}\{\|x_1\|,\|x_2\|,\dots,\|x_N\|, (1+\|x\|)\}$$

Then $$\|x_m\| < M \, \forall \, m$$

That is, the sequence is bounded.

 

[Hurkyl]

From the definition we know that all convergent sequences are Cauchy sequences. But it is not true that the converse always holds. However, in normed linear spaces the converse is always true.

Are you sure? What about, say, Q, which forms a normed linear space over itself, and we all know that Q is full of Cauchy sequences that don't converge...


The first approach I can see is to review the proof that Cauchy sequences converge in R and adapt it to your purposes.

 

[Data]

You probably know what my first request is going to be already :P

What's the definition of a Cauchy sequence (and don't just give me a limit, because that has its own definition too! State it in terms of the definition of a limit)?

With the help of the "other" triangle inequality, $$\| y - z \| \geq \|y\| - \|z\|$$, see if you can get anywhere from that definition.

 

[Oxymoron]

You guys, Data and Hurkyl, do you just look at my questions and instantly know how to do them? I really appreciate your wisdom and effort.

So I guess my method of solving it is flawed because I used assumed the Cauchy sequence converges to some point then used the fact that it converges to show that it must be bounded above by some number K.

One question before I go on. In regards to this question, does $$\left\{\|x_n\| : n = 1, \dots, \infty\right\}$$ form a subset of an inner product space?

Ok, Data, starting from basics.

$$Definition$$

$$(x_n)$$ is said to converge to a point $$x_0 \in X$$ (has a limit $$x_0$$) if for all $$\epsilon > 0 \, \exists \, n_0 \in \mathbb{N}$$ such that

$$\|x_n - x_0\| < \epsilon$$

or equivalently

$$x_n \in B(x_0;\epsilon)$$

We want to show that
$$\left\{\|x_n\| : n = 1, \dots, \infty\right\}$$
is bounded. If we can show that $$(x_m)$$ is convergent then it is bounded. Am I correct in assuming that the sequence $$(x_m)$$ is formed by taking consecutive $$\|x_n\|$$ for $$n = 0, 1, \dots, \infty$$?

Let $$x_0 = \lim x_n$$, then $$\exists \, n_0 \in \mathbb{N}$$ such that $$\|x_n - x_0\| < \epsilon \, \forall \, n \geq n_0$$

The picture you should have in your mind is a closed ball centered on $$x_0$$ (the limit) with radius $$\epsilon$$. As you follow the sequence eventually all the $$x$$'s will stay inside the ball of radius $$\epsilon$$

So now we take the radius of the ball

$$r = \max\{\epsilon, \|x_1\|, \|x_2\|, \dots, \|x_{n-1}\|\}$$

then

$$\|x_n - x_0 \| \leq r$$

so it is bounded by $$r$$

 

[Data]

I don't instantly know how to do them, of course! If I'm lucky sometimes I can figure it out after a few minutes of looking at it though (for example, I had to look up what a Cauchy sequence was, although I did guess based on the Cauchy criterion for convergence of real sequences. I've never used them before!) :).

As to your first question, $$\{ \|x_n\| : n = 1, 2, ... \}$$ is a subset of an inner product space. Namely, of the real numbers. Remember that $$\| x \|$$ is just a real number! On the other hand, $$\{ x_n : n = 1, 2, ... \}$$ is a subset of the inner product space mentioned in the statement of your problem.

Anyways, your definition is a little off. Try this:

If $(x_n)$ is a sequence with elements in an inner product space, then we say that

$$( \ (x_n) \ \mbox{is a Cauchy sequence} \ ) \Longleftrightarrow (\ \exists N \ \forall \epsilon > 0 \ \mbox{s.t.} \ n, \ m \geq N \Longrightarrow \|x_n - x_m\| < \epsilon \ )$$

Your assumption is almost correct. According to your question, there is one $x_n$ for each $$n=1, 2, 3, ...$$.

Now, according to the definition I posted above, we can just choose $$m = N$$ for a particular $\epsilon$ (we can choose the $\epsilon$). By doing this, what can I say about $$\| x_n \|$$ for every $$n > m$$? Remember, all you can use are definitions, and known results (like the inequality I mentioned in my first post, which I think you will find in handy).

 

[Data]

Just a little clarification: From your last post, you seem to be a little bit confused about the difference between the notion of convergence and a Cauchy sequence. Unfortunately, for this problem, we can't prove that $$(x_n)$$ is convergent (because it isn't necessarily).

 

[Oxymoron]

$$\|x_n\| \leq \|x_n - x_m\| + \|x_m\|$$

 

[Hurkyl]

By the way, R is an inner product space, and the sequence of the ||x_i|| lives in R. (Can you show the ||x_i|| form a Cauchy sequence in R? If so, you know R is complete, and thus this sequence converges, even if the {x_i} don't)

 

[Data]

$$\|x_n\| \leq \|x_n - x_m\| + \|x_m\|$$

Correct. And what can you replace the $$\|x_n - x_m\|$$ by while keeping the inequality valid (something that you get to choose...)?

Alternatively, Hurkyl's suggestion also works out very nicely~

 

[Hurkyl]

You guys, Data and Hurkyl, do you just look at my questions and instantly know how to do them?

The more familiar you become with things, the easier they become.

For instance, I know how Cauchy sequences "behave" -- they don't head off towards infinity, and they can't have multiple limit points... a Cauchy sequence that doesn't converge behaves as if it's converging to something "missing".

More precisely, there's a bigger metric space (called the Cauchy completion) that contains your space, but every Cauchy sequence converges.

(As an example, R is the Cauchy completion of Q)

So, even if your Cauchy sequence doesn't converge, I know it should behave like it does, and in particular you should be able to find a bound for the sequence.

The easiest way I could imagine for finding that bound would be to appropriately modify the proof that Cauchy sequences converge in complete spaces like R^n -- one of the steps involved is bounding tails of the sequence.

 

[Oxymoron]

$$\|x_n\| \leq \epsilon + \|x_m\|$$

 

[Data]

Right, and like I said, you can just choose $$m = N$$. Since $$\| x_N \|$$ is just a real number, you should be missing just a couple of minor details to finish the proof. I'm sure you can figure those out~

 

[Oxymoron]

$$\|x_n\| \leq \epsilon + \|x_M\|$$
$$\|x_n\| - \|x_M\| \leq \epsilon$$
$$\|x_n - x_M\| \leq \epsilon$$

So by the definition the Cauchy sequence converges to $$x_M$$.

 

[Data]

Not quite. Remeber, we chose a particular epsilon, which is not enough to say that anything converges. The important point here is that $\| x_n \| \leq \epsilon + \| x_N \|$ holds $\forall n \geq N$, ie. you've bounded [itex] \| x_n \|[/tex] for $n \geq N$. The details I referred to simply have to do with the terms $x_n$ with $n < N$.

 

[Oxymoron]

So have we proved that $$\|x_n\|$$ is bounded?

 

[Data]

Here's what we have so far:

From the definition of a Cauchy sequence, we know that

$$\lim_{\mbox{min}(m, n) \rightarrow \infty} \| x_n - x_m\| = 0$$

or in other words

$$\exists N \forall \epsilon >0 \ \mbox{s.t.} \ n, m \geq N \Longrightarrow \|x_n - x_m\| < \epsilon$$

Thus, if we choose a particular $\epsilon >0$, we can find a particular N such that

$$\forall n, m \geq N, \ \|x_n - x_m \| < \epsilon$$

and thus

$$\forall n \geq N, \ \|x_n - x_N \| < \epsilon$$

and since, by the inequality I mentioned above, $\forall n \geq N$,

$$\|x_n\| - \|x_N\| \leq \| x_n - x_N \| < \epsilon \Longrightarrow \|x_n\| < \epsilon + \|x_N\| = M \in \mathbb{R}$$

so we have bounded [itex] \|x_n\| \forall n > N[/tex]. I'm sure you can fill in the rest (which is trivial!)

 

[Data]

On the other hand, here's a possible argument using Hurkyl's idea:

As in my last post, we know that

$$\exists N \ \forall \epsilon>0 \ \mbox{s.t.} \ n, m \geq N \Longrightarrow \|x_n - x_m \| < \epsilon$$

but by the inequality I mentioned,

$$\|x_n\| - \|x_m\| \leq \|x_n - x_m\|$$

and similarly, since $\|x_n - x_m\| = \|x_m - x_n\|$,

$$-(\|x_n\| - \|x_m\|) = \|x_m\| - \|x_n\| \leq \|x_n - x_m\|$$

so we can certainly conclude that

$$| \|x_n\| - \|x_m\| | \leq \| x_n - x_m \|$$

which gives us

$$\exists N \ \forall \epsilon >0, \ \mbox{s.t.} \ n, m \geq N \Longrightarrow | \|x_n\| - \|x_m\| | < \epsilon$$

and thus $(\|x_n\|)$ is also a Cauchy sequence, over the real numbers. But since Cauchy sequences over $\mathbb{R}$ always converge, we find that $(\|x_n\|)$ converges and is thus bounded.

 

[Oxymoron]

I'm sure you can fill in the rest (which is trivial!)

I hope I don't disappoint you too much but to me it looks like $$\|x_n\|$$ is bounded.

 

[Data]

it certainly is! Not getting a problem isn't any reason to be dissapointed... everyone has trouble with some problems!

 

[Oxymoron]

I have another question I'd like your help on.

$$\textsl{Question}$$

Let $$(x_n)$$ be a sequence of elements on a Hilbert space $$H$$ for which

$$\Sigma_{n=1}^{\infty}\|x_n\|<\infty$$.

Show that the sequence of partial sums

$$s_n = \Sigma_{k=1}^{n}x_n$$

converges to a point in $$H$$.

$$\textsl{Answer}$$

Since we know that $$H$$ is a Hilbert space, we know that every Cauchy sequence in $$H$$ is convergent. So to answer this question we have to prove that $$(x_n)$$ is Cauchy.

$$\|x_n - x_m\|^2 \rightarrow x$$ as $$n,m \rightarrow \infty$$

$$\|x_n - x_m\|^2 = \Sigma_{j=1}^{\infty}|x_{nj} - x_{mj}|^2 \geq |x_{nj} - x_{mj}|^2$$

So

$$|x_{nj} - x_{mj}| \rightarrow x$$ as $$j \rightarrow \infty$$

Therefore $$(x_{nj})$$ is a Cauchy sequence convergent to $$x$$

But is $$x \in H$$? Now we need to show that $$x \in H$$ and $$x_{n} \rightarrow x$$

The partial sum $$\Sigma_{j=1}^{n}|x_j|^2 = \Sigma_{j=1}^n|\lim_{k\rightarrow\infty}x_{j}|^2$$
$$= \lim_{k\rightarrow\infty}\Sigma_{j=1}^{n}|x_{nj}|^2$$

But

$$\Sigma_{j=1}^m|x_{nj}|^2 = \|x_n\|^2 \in H$$

So

$$\lim_{k\rightarrow\infty}\left(\Sigma_{j=1}^{\infty}|x_{nj}|^2\right)= \lim_{k\rightarrow\infty}\|x_n\|^2 \leq \lim_{k\rightarrow\infty}M$$

where $$\|x_n\|^2 \leq M \, \forall \, k$$

We know that such an $$M$$ exists because Cauchy sequences are bounded. So the partial sum

$$\Sigma_{j=1}^{\infty}|x_n|^2 = \lim_{k\rightarrow\infty}\|x_n\|^2 \leq M$$

This is an increasing sequence that is bounded above. So

$$\Sigma_{j=1}^{\infty}|x_n|^2 < \infty$$

ie $$x \in H$$

 

[Data]

Ok, let's review a few important things and then you can see if you can improve your proof.

We are dealing with a Hilbert space. What is a Hilbert space should be the first question you ask (and you did ask it!).

A Hilbert space is simply an inner product space where every Cauchy sequence happens to be convergent.

The next question (or perhaps multiquestion), as you might guess, is what are inner product spaces and what are Cauchy sequences, and what does it mean for a sequence to converge?

Well, I'm going to assume that you've seen enough of inner product spaces by now that I can safely skip answering that part explicitly. One thing to always remember is that the magnitude of a vector $x$ in our Hilbert space is always a real number, or in other words $\| x \| \in \mathbb{R} \ \forall x$ in the Hilbert space.

I'll move on to the other part of the question I just asked now.

A sequence $\{x_n\}$ is just an ordered sequence of elements of some set, indexed by positive integers $n$ (bored yet?). In this case, the elements are just vectors from our Hilbert space (call the space $H$ as in your question) when we talk about $\{x_k\}$, and just real numbers when we talk about $\{\|x_k\|\}$.

Now, remember that sequence $\{y_n\}$ (in any set) converges to $x$ if the terms get arbitrarily close to $x$, provided that we choose $n$ sufficiently large, and provided that $x$ is in the set in the first place. In other words,

$$\{ x_n \} \ \mbox{converges to} \ x \Longleftrightarrow \lim_{n \rightarrow \infty} x_n = x \Longleftrightarrow \left( \ \exists N \ \forall \epsilon > 0 \ \mbox{s.t.} \ n > N \Longrightarrow \| x_n - x \| < \epsilon \ \right)$$

Now, some sequences have a special property. These sequences are called Cauchy sequences. This special property is that we can make the magnitude of the difference between terms of the sequence arbitrarily small, provided that we choose their indices sufficiently large. In other words,

$$\{x_n\} \ \mbox{is a Cauchy sequence} \Longleftrightarrow \lim_{\mbox{min}(m, n) \rightarrow \infty} \| x_n - x_m \| = 0 \Longleftrightarrow \left( \ \exists N \ \forall \epsilon > 0 \ \mbox{s.t.} \ n, m > N \Longrightarrow \|x_n - x_m \| < \epsilon \right)$$

Now, we have a couple of important results relating to Cauchy sequences:

1) Every convergent sequence is a Cauchy sequence (in any set).

2) By the definition of a Hilbert space, every Cauchy sequence in a Hilbert space is also a convergent sequence.

In other words

If $\{ x_n \}$ is a sequence in a Hilbert space, then $\{x_n\}$ is convergent if and only if it is a Cauchy sequence.


Keep in mind that the real numbers $\mathbb{R}$ form a Hilbert space (and thus the statement above applies to a sequence of reals).

Be careful in your proof not to mix up elements of $H$ with elements of $\mathbb{R}$, because we do not know $\mathbb{R} = H$. In particular, keep in mind that $\sum_{k=0}^\infty \|x_k\| \in \mathbb{R}$ but $s_n = \sum_{k=0}^n x_k \in H$.



Good luck

 

[Data]

I will give you an extra hint too:

You will need to use induction with the triangle inequality (at least if you think about it the same way as me :tongue:).

Post your work here, and I will try to provide more, useful, hints, if you think you need them

These proofs are by no means simple, at least not until you see quite a few of them. You really need to understand what the definitions mean. Don't worry if you don't see how to do it right away~