The reason I like the problem is that the reflex is: they can't possibly be isomorphic! But then, when thinking about it, it's getting hairy :)

@tb I understand that feeling.

Similar things happened in the MO thread Jonas linked to...

Okay since you like compactness: a normed space is finite-dimensional if and only if its unit ball is compact.

Do you know that statement?

I can anticipate it. One second; let me try a proof.

Try it for Hilbert spaces first.

Finite dimensional $\implies$ Unit ball compact is clear.

Hopefully :)

Try finding a sequence with no Cauchy subsequence for the other direction.

Other direction: Here's an attempt. Let me assume countable dimensional first. Fix a basis $B = \{ v_1, v_2, \ldots, v_n, \ldots \}$ of unit vectors. Then the map $\sum_{n=1}^\infty \lambda_n v_n \mapsto \sum_{n=1}^\infty \frac1n |\lambda_n|$ does not attain a minimum on the unit sphere.

nice :)

I really like to squeeze the extreme value theorem. That to me is the essence of compactness. =) OK, so we proved that fact now.

Another way of seeing it would be to say $\|v_i - v_j\| = \delta_{ij} \sqrt{2}$.

So $(v_n)$ doesn't have a convergent subsequence.

Ah, that's cool.

For normed spaces a similar trick works, but for the moment that's good enough.

Wait: one second. I might have erred. I haven't checked that the map I wrote is continuous.

make it $1/n^2$ then it surely is.

@tb Right. That's nice. Good that it can be fixed.

Now, let us consider a bounded linear map $T$ with finite-dimensional range.

Ok.

Compactness of balls gives us the following nice property: given any bounded sequence $x_n$ there is a subsequence $x_{n_i}$ such that $Tx_{n_i}$ converges.

Note that $x_{n}$ itself need not have a convergent subsequence since the domain is infinite dimensional. But we can always arrange that a subsequence in the range converges.

A projection on a finite-dimensional subspace, for example.

One minute, I'm checking with a concrete example.

I am thinking about the very same example. Suppose we are considering the projection on $\{ v_1, ..., v_n\}$ where $\{ v_i \}$ is an orthonormal basis. What happens to the sequence $v_1, ..., v_n, ...$?

Oops, that's eventually zero. :( Nevermind!

We can go ahead (unless you have a better example).

Now given any bounded sequence, pick a subsequence such that the first coordinate converges. Then pick another subsequence such that the second coordinate converges, etc.

Do this up to coordinate $n$, and there you go.

Cool.

Another essence of compactness: it allows diagonalization :)

(and comes from it)

@tb Hm. Yes, that's true. But I haven't used it many times myself. I have seen it being used...

[I usually try to find an alternate proof using the EVT. :D]

Well, that's how you prove that a countable product of compact metric spaces is compact, for example.

I am not allowed to say "Tychonoff's theorem", right? I see what you mean.

Why not: metric Tychonoff. Very useful.

Or think of Arzelà-Ascoli: pure diagonalization.

Right.

So, back to operators: An operator is called compact if the image of every bounded sequence has a convergent subsequence.

We've just convinced ourselves that operators with finite-dimensional range are compact.

And we already noted that any operator with finite dimensional range is compact.

:-)

But I suppose that these are not the only ones.

So, in order for this to be an interesting concept, we need to convince ourselves that there are operators with infinite-dimensional range that are compact, right?

@tb Yes. Now I am slightly confused.

The infinite-dimensional range itself is not compact. So there exists some sequence with no convergent subsequence. We can try taking the pre-image of this sequence, hoping to get a contradiction.

Well, yes but it only tells us that every bounded set is mapped to a compact set.

@Srivatsan So we will fail extracting a bounded pre-image sequence.

@tb Yes, that is true.

Basic examples are integral operators.

A priori, I don't get a feeling that such an operator should exist. Trying to think of some examples.

Are you saying something like $p \mapsto \int p$?

Let's start with the Volterra operator

I do not know it.

Map $f$ to $Vf(t) = \int_{0}^t f(x)\,dx$

Right. Wow, this has a name? =) Anyway, let me check that it is compact.

Well, we need to say where it is defined, first.

You mean the domain of $V$? I was thinking of polynomials as a simple start.

Okay, fine. Maybe even better the continuous functions (to get rid of notational ballast involved in polynomials)

OK, if continuous functions work then we can take that as well.

let's work on $C[0,1]$, then

with the sup-norm?

Yes. And I claim $V: C[0,1] \to C[0,1]$ is compact

To see that it is an interesting example note first that a monomial is mapped to a monomial of one degree higher, so its range is certainly infinite-dimensional.

Certainly interesting. =)

Now $Vf$ is certainly continuous and $V$ is linear, so we're set.

Yes.

Should we apply Arzela-Ascoli for showing compactness?

you're good :)

First convince yourself that $V$ is bounded.

=) Thanks. It looks like it certainly, the details are vague still.

One second: $V$ is bounded would mean (Don't reply; I'm just thinking aloud)

I see. $\| V f \| = \sup_t \left| \int_0^t f \right| \leqslant \sup_t \int_0^t \| f \| \leqslant \| f \|$.

Right. So given a bounded sequence the image sequence is certainly bounded.

What's missing for applying Arzelà-Ascoli?

Equicontinuity of the image sequence.

I claim that even more is true: the entire unit ball is mapped to an equicontinuous family.

Let me think about that.

The point is that $\displaystyle Vf(t) - Vf(s) = \int_{s}^t f$

Aha, of course. If $f$ is in the unit ball, then this is at most $|t-s|$.

Done.

Yes.

One side question: shouldn't this property be called equi-uniform-continuity?

instead of equicontinuity?

Yes.

Drop it. I'm not sure why I said that or if it makes any sense.

Some people define equicontinuity pointwise: For every $\varepsilon \gt 0$ and every $x$ there is $\delta_x$ such that for every $f \in \mathscr{F}$ we have $d(x,y) \lt \delta_x$ implies $d(f(x),f(y)) \lt \varepsilon$.

and they speak of uniform equicontinuity if $\delta$ can be chosen independently of $x$.

I see. So it does make some sense then.

Yes.

OK, but it is not terribly important here. Just that the terminology felt a little odd.

But I think calling uniform equicontinuity simply equicontinuity is (at least when working on compact spaces) more common.

Anyway....

One (basic) property of compact operators is that they map bounded sets to relatively compact sets.

Why is this...

well, by definition

take a sequence in the image, it has a bounded pre-image sequence, hence there's a subsequence such that the image sequence (so a subsequence of the sequence we started with) converges.

No, I don't see it. If I give you a sequence in the image, why is there a bounded preimage sequence?

the image is the image of a bounded set

I said: compact operators map bounded sets to relatively compact sets.

Ah, I missed that, yes.

@tb But relatively compact means thhat closure is compact, no? Where did the closure come into picture?

Are you asking why the closure is compact?

One second. Let me take a minute and do this properly.

Sure.

Closure is surely important: the image of the set need not be compact (even closed).

Yes.

E.g., take the image of $\{\frac{1}{n}\}$ under the identity map.

as a map from $\mathbb{R} \to \mathbb{R}$?

Yes, that is correct.

By the way: when is the identity of a normed space compact?

When the space is finite dimensional, by the very first proposition of tonight.

yep.

Is your confusion about closure gone?

Sort of. Fixing the details of the proof.

You haven't really proved it, no?

well, if you've got a sequence $x_n$ in the closure of the image of a bounded set, take $y_n$ in the image with $\|x_n - y_n\| \leq 1/n$. Then pass to a subsequence such that $y_{n_i}$ converges and then $x_{n_i}$ converges to the same limit which is in the closure of the image.

(and no, I haven't proved it before)

Ok, now it looks good. My main confusion was whether that was a proof or a sketch/idea. =)

Now: if the range of a compact operator is closed then it is finite dimensional

(hint: open mapping theorem)

Range is closed in some ambient space, right? What do you mean by simply closed?

Well, we've got a compact operator $T: X \to Y$ from one Banach space to another one. If $T(X) \subset Y$ is a closed subspace then it is finite-dimensional.

Why can't I take $Y = T(X)$ and claim that the range is closed (in itself)?

For instance: we considered the Volterra operator $V: C[0,1] \to C[0,1]$. A continuous function is in its image if and only if it is $C^1$.

And your proposition -- does it imply that $C^1$ is not closed in $C$?

Well, consider the sequence $x \mapsto \sqrt{x^2 +1/n}$. On $[-1,1]$ it converges uniformly to $x \mapsto |x|$.

(or did I just goof? the idea is that you can squeeze parabolas to the absolute value function)

No it works.

@Srivatsan ?

Okay, I guess you either fell asleep or are gone otherwise. I have to go now. See you later!

@tb Ah, sorry. I dozed off you presumed.

Sorry! :/

I guess we'll just drop this for now and start off from here later.

Thanks.

Sure, no problem :) Let's take this up when you're fresh again :) See you later