Room for t.b. and Srivatsan

3:54 PM

Sorry, it seems the ell vs. L distinction never registers in my mind, notationwise. Even though it's all clear in my head.

t.b.

So we're considering the polynomials of degree $\leq n$ on $[0,1]$ and we look at the $L^p$ norm $\|f\|_p = \left( \int_{0}^1 |f|^p \right)^{1/p}$

Srivatsan

@tb Right. Now I see. You said that finite dimensional spaces are always closed?

t.b.

Probably :)

All finite-dimensional normed spaces are isomorphic.

Srivatsan

@tb OK, I am not sure I completely understand the argument, so could we go over the small steps?

@tb So, let me try to summarise it:

t.b.

Sure

Srivatsan

3:58 PM

Restricted to the space of polynomials of degree $\leq n$, all the $L^p$'s are equivalent; in particular the topologies are the same.

t.b.

Yes.

Srivatsan

Separately, I do know that polynomials of degree $\leq n$ form a closed subspace of $L^p([0,1])$ for any $p \geq 2$. I am not sure if we want to use this fact here.

t.b.

okay

Not necessary.

Srivatsan

@tb I added the degree condition.

t.b.

okay :)

Srivatsan

4:01 PM

I know one proof of the second proposition, not sure if it is optimal. For $p=2$, we have a finite-dimensional Hilbert-space; it is necessarily closed.

t.b.

Just do it in the abstract. The polynomials of degree $n$ are an $(n+1)$-dimensional subspace of $L^p$. So let's look at a finite-dimensional subspace of a normed space instead.

Srivatsan

@tb Aw, it seems that is what I am missing.

I have a proof of this proposition for $p=2$ (and using the $p=2$ case and Holder inequality, I can handle the $p > 2$ case as well.)

t.b.

Okay. I see.

Srivatsan

(Because p=2 corresponds to a Hilbert space as I mentioned before.)

t.b.

Sure, I understand.

But ponder this: it is isomorphic to $\mathbb{R}^{n+1}$ and this isomorphism gives you a norm on $\mathbb{R}^{n+1}$. The norm is equivalent to the usual norm, hence the space is complete, hence it must be a closed subspace.

Srivatsan

4:06 PM

Isomorphic to $(\mathbb R^{n+1}, \| \cdot \|_p)$? Or the euclidean space? // OK, you mean the p-norm. // OK, you don't care about the norm yet... =)

t.b.

Just linear algebra isomorphic no topologies yet.

Srivatsan

@tb Are you exploting the fact that while being closed is a relative property, completeness is absolute: it depends on the space and not the ambient space?

t.b.

So you have an injective map $\varphi: \mathbb{R}^{n+1} \to L^p$ sending a vector to a polynomial in your preferred way. The expression $\|\varphi(x)\|_p$ defines a norm on $\mathbb{R}^{n+1}$.

@Srivatsan not really.

Srivatsan

Let me read what you said above.

@tb Sure.

This norm on $\mathbb R^{n+1}$ is equivalent to the standard one.

t.b.

This norm is equivalent to the usual norm $\|\cdot\|$, that is to say there's a constant $C \gt 0$ such that $C^{-1} \|x\| \leq \|\varphi(x)\|_p \leq C\|x\|$.

Srivatsan

4:11 PM

Right.

$\mathbb R^{n+1}$ with the standard norm is complete, so the $\mathbb R^{n+1}$ with this new norm is also complete. This means that the space of $\leq n$-degree polynomials is also complete.

t.b.

Now take a Cauchy sequence of degree $n$-polynomials $p_k = \varphi(x_k)$ in $L^p$. By the equivalence of these norms $x_k$ is Cauchy, hence it converges to some $x$. But by equivalence again $\varphi(x_k) \to \varphi(x)$.

@Srivatsan exactly.

Srivatsan

Er, that proof feels a little weird. You did not seem to do any work anywhere.

t.b.

Well, the work is hidden in the equivalence of norms on $\mathbb{R}^{n+1}$. This uses compactness of the unit ball among other things.

Srivatsan

@tb Right, that makes sense.

t.b.

You know the proof, right?

Srivatsan

4:15 PM

@tb Yes, I do.

(1) Show that any given norm is bounded (hence, continuous) w.r.t. the standard norm on $\mathbb R^{n+1}$. (2) Then invoke extreme-value theorem (my favorite!) to say the norm attains a minimum on the unit sphere.

t.b.

Exactly. And in $(1)$ you hide an easy case of Hölder or something like that.

Srivatsan

@tb Well, you need some bound, but any crummy one will do. You don't need anything as strong as Holder for this.

I guess that is why you mentioned "easy case of Holder".

t.b.

Well, usually you'll choose a base, get an $\ell^1$-estimate then bound the $\ell^1$-norm by the $\ell^2$-norm from above. This is AM-GM, Cauchy-Schwarz or whatever else you like to call it...

Srivatsan

Ah, right, OK.

Compactness is weird. ;)

t.b.

To be honest, I don't think there are easy explicit bounds for the norm from below.

Srivatsan

4:24 PM

@tb Yes. I thought about it a little, and decided that compactness is necessary.

t.b.

However, I'm pretty sure that in books on approximation theory you could find some good bounds but these tend to be tough to decipher.

Srivatsan

Suppose $S$ is an arbitrary finite set of unit vectors. Then there exists a norm $N$ such that $N(x) \geq 1$ for all $x \in S$, and yet it attains arbitrarily small $\delta$ (that is set beforehand) value on the unit sphere.

t.b.

Sure. But I was thinking of your explicit polynomials in $L^p$ problem.

Srivatsan

Oh, ok.

OK, I will chew on this a little bit. The proof is easy but not exactly trivial. Thanks.

t.b.

have fun, no problem :)

Srivatsan

4:32 PM

Have to go now, bye.

Srivatsan

10:41 PM

@t.b. I just wanted to tell you that I got a different-but-not-so-different proof.

It suffices to show that every $f \in \operatorname{C}([0,1])$ has a "best" projection on to the space of degree-$\leqslant n$ polynomials. This follows from:

Now (1.) is easy. But proving (2.) again requires me to use compactness. In fact, the proof for (2.) uses essentially the same idea as the previous one.

I like this approach only because it is kind of more transparent to me. :=)

t.b.

11:13 PM

@Srivatsan I'm sorry but I don't follow the argument: why does it suffice to show that you have a "best" projection?

(are you saying it is unique?)

Srivatsan

OK. Let me state it more clearly.

It suffices to show that the set $\{ \| h - f \|_p : h \text{ is a degree } \leqslant n \text{ polynomial} \}$ attains its minimum. The argmin of that set is the best approximation. (Sorry, projection is the wrong word, I guess.)

t.b.

But why does it suffice to show that?

Srivatsan

@tb Because if $f$ is not in $\mathcal P_n$ (the set of polynomials of degree $\leqslant n$), then this minimum will be strictly positive.

t.b.

@Srivatsan but doesn't this assume that $\mathcal{P}_n$ is closed already?

Srivatsan

@tb No. I am proving that it is closed. I want to prove that if $f$ is not in that set, then it is separated from it by a positive distance. To do this, I am claiming that it suffices to show that there exists $h \in \mathcal P_n$ that is the closest-approximation to $f$ in the $L^p$ sense.

t.b.

11:21 PM

Ah, you're saying this minimum is achieved by some $h \in \mathcal{P}_n$.

Srivatsan

@tb That is what I said right? "The set attains its minimum"

t.b.

yeah, I was instinctively thinking of infimum.

Srivatsan

Oh ok. Perhaps I could've stressed but I thought the word "attain" would be enough.

t.b.

No, no, I just wasn't reading carefully enough :)

Srivatsan

Now $h \mapsto \| h - f \|_p$ is continuous, so it attains a minimum over all compact sets. If $\mathcal P_n$ were compact, then we are done. It is not, but the fix is simple.

t.b.

11:24 PM

And actually the minimum is unique unless maybe when $p=1$.

The rest is fine with me. Nice :)

Srivatsan

@tb It's the same set of ideas rehashed =)

@tb How do we know that the minimum is unique?

t.b.

@Srivatsan Sure, it's closely related. But uniqueness gives you that you have a projection.

Srivatsan

Oh, does uniqueness follow from strict convexity?

t.b.

This is because the $p$-norm is strictly convex.

Srivatsan

Right, ok. Interesting.

t.b.

11:28 PM

Clarkson's inequalities

In mathematics, Clarkson's inequalities are results in the theory of Lp spaces. They give bounds for the Lp-norms of the sum and difference of two measurable functions in Lp in terms of the Lp-norms of those functions individually. Statement of the inequalities Let (X, Σ, μ) be a measure space; let f, g : X → R be measurable functions in Lp. Then, for 2 ≤ p < +∞, :\left\| \frac{f + g}{2} \right\|_{L^{p}}^{p} + \left\| \frac{f - g}{2} \right\|_{L^{p}}^{p} \le \frac{1}{2} \left( \| f \|_{L^{p}}^{p} + \| g \|_{L^{p}}^{p} \right)...

(overkill, but whatever)

Srivatsan

I see. Now I get a feel for why some people like (or hate, depending on who it is) the $L^1$ norm.

It behaves genuinely differently.

t.b.

I like the $L^1$ norm. My mathematical life depends on it :)

Srivatsan

@tb Dude, way to get your name attached to an inequality. That seems almost easy. =)

@tb Aha, nice.

t.b.

@Srivatsan The $p \geq 2$ case is easy, yes. The case $1 \lt p \lt 2$ is distinctly murky...

(at least I've never seen a very nice proof of that)

and it proves uniform convexity, not only strict convexity.

Srivatsan

@tb Hm, interesting.

@tb reading the Wikipedia page. No points for guessing who introduced the notion of "uniformly convex space"... =)

t.b.

11:33 PM

Speaking of $L^p$. Have you seen this answer?. This might be one of my favorites :)

Srivatsan

@tb I saw it a few minutes back. I didn't read it because I didn't understand the question =)

What does topologically "isomorphic" mean?

I guess I'm not in a position to really appreciate your favorite. =) (Sorry!)

t.b.

@Srivatsan There is a bijective bounded linear map (topologically isomorphic as opposed to isometrically isomorphic) between them. I don't like the terminology but it's well-established, unfortunately.

Srivatsan

I see. Should the inverse also be bounded? Or does that follow?

t.b.

It is usually quite hard to tell whether two Banach spaces are isomorphic or not.

@Srivatsan that follows from the open mapping theorem.

Srivatsan

@tb Oh ok.

t.b.

11:40 PM

(a Baire trickery showing that a surjective linear map between Banach spaces is an open map)

Srivatsan

Wait: how can $\ell_p$ and $L^p([0,1])$ be isomorphic? That's comparing apples and oranges.