unbdd linear functionals

masfenix

Oct 15, 2013 02:09

HEY GUYS

sorry caps.

I was wondering about a problem, does there exist an unbounded linear map $F: l^2(\mathbb{N}) \rightarrow \mathbb{C}$

Kevin Driscoll

@Pedro Sorry I was away. Yeah at 150 mph (240 kph) I buy it. Its very similar to high speed footage of water balloons at lowish speeds and golf balls are surprisingly elastic

Ted Shifrin

@masfenix: Why do you think there should be one?

Fernando Martin

@KarlKronenfeld: let's see

@PedroTamaroff: No, just took a break

masfenix

@TedShifrin I think there is one because $l^2(\mathbb{N})$ is an infinite-dimensional Hilbert space.

so I remember my professor saying that there can be unbounded maps

Ted Shifrin

I know of discontinuous linear maps on Banach spaces, but not on Hilbert spaces.

masfenix

Oct 15, 2013 02:20

So the answer is no?

well then I would have to prove that.

Ted Shifrin

Inner products are powerful things.

Hmm, I take it back.

masfenix

@TedShifrin en.wikipedia.org/wiki/Discontinuous_linear_map says that it may be hard to find a concrete example for a complete space (so Hilbert spaces?)

Ted Shifrin

I am just very rusty ... It's been about 35 years since I've thought about this :P

No, I untake it back. I can think of examples $H\to H$, but not for a linear functional on $H$.

Kevin Driscoll

@Ted @masfenix I dont think the inner product helps you because an unbounded map is precisely one for which $\left| F( \phi ) \right|$ is unbounded even though $\left| \phi \right|$ is finite

but what do I know

Ted Shifrin

Right @Kevin. Can you give a map to $\Bbb C$ ?

Maybe so. What if we take $H=L^2([0,1])$ and something like evaluation at $0$?

No, that's not defined.

Kevin Driscoll

Oct 15, 2013 02:27

@Ted No, I can't give an example. @Masfenix talked about this the other day and my instinct was that no such map exists, but I'm totally unsure

Ted Shifrin

I remember loving learning about densely defined discontinuous linear maps in the context of PDEs when I was in grad school. But I'm old and I never use this stuff at all :P

masfenix

Discontinuous linear map

In mathematics, linear maps form an important class of "simple" functions which preserve the algebraic structure of linear spaces and are often used as approximations to more general functions (see linear approximation). If the spaces involved are also topological spaces (that is, topological vector spaces), then it makes sense to ask whether all linear maps are continuous. It turns out that for maps defined on infinite-dimensional topological vector spaces (e.g., infinite-dimensional normed spaces), the answer is generally no: there exist discontinuous linear maps. If the domain of defin...

look at the section under General existence theorem

Anthony Carapetis

The "general existence theorem" section seems to be exactly what you want

masfenix

I think this may help us

yes @AnthonyCarapetis exactly.

Kevin Driscoll

@Ted I see. So it may be that the function is integrable, but its value at a point may still be infinite

Ted Shifrin

Oct 15, 2013 02:29

yeah, but values of functions don't make sense, because elements of $L^2$ are equivalence classes.

Hmm, @masfenix @Anthony, I was thinking about an example like that but convinced myself it wouldn't work in a Hilbert space.

masfenix

Now, I understand the example but I could use a little intuition on it.

Ted Shifrin

@Anthony is our resident fancy analyst ... I defer.

Anthony Carapetis

hey don't be calling me things like that

Ted Shifrin

@masfenix, $\ell^2$ has an obvious orthonormal basis :P

Anthony Carapetis

I think the key thing to notice is that it's an algebraic basis, not a Schauder basis

Kevin Driscoll

Oct 15, 2013 02:32

@Ted equivalence classes......... yes, of course............ quite right

Ted Shifrin

well, I'm thinking Hilbert space basis ... not algebraic

masfenix

@TedShifrin, right which is $(1, 0, 0 \ldots), (0, 1, 0 \ldots), (0, 0, 1, \ldots)$

Ted Shifrin

I've been spending all night thinking about Chern classes and degeneracy classes. I should go back to that ... :P

yes @masfenix

Anthony Carapetis

If you try to define a map like that using only a Hilbert space basis

Ted Shifrin

So, mapping the $n$th one to $n$ and extending by linearity is an example.

Anthony Carapetis

Oct 15, 2013 02:33

Then you're only defining it on a dense subset

Ted Shifrin

Well, unbounded operators are usually densely defined, anyhow :P

Anthony Carapetis

Usually we extend by continuity to get a bounded linear map, but obviously in this example we don't want to

So you need a full algebraic basis to "complement" it I guess

Ted Shifrin

UGH at algebraic basis.

I don't do that stinking uncountable stuff.

Anthony Carapetis

Hahaha

Karl Kronenfeld

@FernandoMartin I think we may lose the vector space structure present in the hom-set, but if there is a way to represent addition and scalar multiplication "categorically", or to otherwise show that functors are linear we'd be making good progress.

masfenix

Oct 15, 2013 02:36

So if I understand this correctly, we take the usual orthonormal set of $l^2$ and define a linear functional by $F(e_n) = n\cdot |e_n|$

so how do I show this map is discontinuous?

Kevin Driscoll

@Ted what happened to your geometrical answer for the 'building a bridge' question? Did you find an error?

masfenix

because the distance between two points is always one?

Fernando Martin

@KarlKronenfeld: sorry, I got distracted and just finished reading

Anthony Carapetis

@masfenix: that's only defined for the subspace spanned by the $e_n$

Fernando Martin

let's see

Ted Shifrin

Oct 15, 2013 02:37

Just show it's unbounded, @masfenix. Continuity is equivalent to boundedness.

masfenix

right, but I am not really sure how to show its unbounded. Isn't it easier to show discontinuity?

Ted Shifrin

Well, @Kevin, they convinced me it was wrong. And I had convinced myself it was right based on symmetry, and there was a 7th grade algebra flaw in that. So I think my "obvious" triangle inequality step was wrong.

masfenix

@AnthonyCarapetis, the orthonormal basis $e_n = \delta_n$ where $\delta_n$ is the Kronicker delta function spans the entire space, no?

Fernando Martin

@KarlKronenfeld: I don't know much about this, but wouldn't this have to do with enriched categories?

Ted Shifrin

You have elements of length $1$ whose images have arbitrarily large lengths.

masfenix

Oct 15, 2013 02:43

@TedShifrin, thank you. So I've formulated something like the following: Define $F(e_n) = n |e_n| $. Recall that a unbounded map is exactly one for which, $\forall x \in H$ where H is a Hilbert space, $|F(x)|$ is unbounded whereas $|x|$ is bounded, and so since the map defined above takes elements (that are bounded) to images with increasingly larger lengths, it follows that the image is not bounded and hence the map is unbounded.

Fernando Martin

@KarlKronenfeld: How would you represent, for instance, the trivial endofunctor?

Kevin Driscoll

@Ted ah okay. That's unfortunate because I didn't understand your answer, and maybe there was something there I need to understand

Ted Shifrin

I still think the path should consist of parallel beginning and end, @Kevin, but my proof was flawed (since dilating the inner ball distorts distance).

@masfenix: First, $|e_n|=1$, so it's just $F(e_n)=n$. But I guess you should address @Anthony's point of whether this extends to give a well-defined linear map on all of $\ell^2$.

Anthony Carapetis

@masfenix: Hilbert space orthonormal bases only span the whole space only if you allow infinite linear combinations. This normally works fine to extend the definition of a linear functional to the whole space (by continuity); but since this one is unbounded that won't work. For example $\sum_k e_k / k^2 \in \ell_2$ but $F$ would send it out of $\ell_2$.

Ted Shifrin

@Anthony: We're defining a linear functional! So what do you mean?

Anthony Carapetis

Oct 15, 2013 02:48

@Ted: what's the issue? We've only defined it on a dense subspace, and we clearly can't extend by continuity (since it's not even bounded on the subspace)

Ted Shifrin

Right. But my experience (years ago) with unbounded operators is that they're essentially always only densely defined (e.g., differentiation on $L^2$). I dunno.

Anthony Carapetis

Oh right, forgot about that

Yeah if that's the convention being used then you don't need to worry about extending it

masfenix

@AnthonyCarapetis could you explain "extending" a little bit more. I want to understand it a bit more

Anthony Carapetis

@masfenix: the example we have is defined only for those $x \in \ell_2$ with finitely many non-zero components, so it is defined only on a subspace of $\ell_2$. It is conventional that an "unbounded linear operator $X \to Y$" actually means a linear map $Z \to Y$ for some subspace $Z \subset Y$; so if you are using this convention then the example suffices.

If not (i.e. if you want the map to have domain $X$) then you need to define it for sequences with infinitely many non-zero components; i.e. "extend it to all of $\ell_2$".

I always found that terminology confusing... "unbounded operator" really means "partially-defined operator" rather than "operator that is not bounded"

Pedro Tamaroff

@KevinDriscoll Physics approved!.

masfenix

Oct 15, 2013 03:01

@AnthonyCarapetis wow, thanks.

$Mathematics$ Mathematics

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