Date: Feb 28, 2013 7:56 PM Author: David Bernier Subject: Re: Invariant subspace problem solved? On 02/28/2013 01:24 AM, David Bernier wrote:

> On 01/28/2013 11:19 AM, José Carlos Santos wrote:

>> On 28-01-2013 15:35, David Bernier wrote:

>>

>>>> I found several texts across the Internet stating that the Invariant

>>>> Subspace problem for Hilbert spaces was solved by Carl Cowen (USA) and

>>>> Eva Gallardo (Spain). However, each of these texts is written either in

>>>> Spanish or in Portuguese. This is rather odd, given the importance of

>>>> the problem. Did anyone around here read something about this subject?

>>> [...]

>>>

>>> I found something similar to what you found.

>>>

>>> From the "Real Sociedad Matemática Española" web-site, a 2013 meeting:

>>>

>>> "del 21 al 25 de Enero" 21-25 January?

>>>

>>> http://www.rsme.es/content/view/1199/1/

>>

>> Interesting. This site is the website of the Royal Spanish Mathematical

>> Society. It looks serious.

>>

>>> I think they mention John von Neumann and the 1930s ...

> [...]

>

> This development seems to be of some interest, and me included,

> to some degree.

>

> The authors announced in early February that a gap in their proof

> of the invariant subspace problem had been found. They had

> submitted a paper to a journal containing that claimed proof,

> and now they have withdrawn their submission.

>

> They are working to bridge the gap, and there are results

> independent of the existence of the gap. Their statement

> was posted at cafe metematico blog here:

>

> http://cafematematico.com/2013/02/05/statement-from-cowen-and-gallardo/

>

> ---

>

> For complex Hilbert spaces H of dimension n, n finite and n>=2,

> the related question is if an linear operator T: H-> H

> that is continuous has a Hilbert sub-space W

> such that T(W) is contained within W, and

> W is non-trivial, meaning both

> dim(W) >= 1 and dim(W) < dim(H) = n.

>

> So the sought-for W has 1<= dim(W) <= n-1, if it exists.

>

> All n-dimensional Hilbert spaces are isomorphic in the strongest

> sense, essentially they are all like C^n with

> (z1, ... z_n).(w1, ... w_n) = sum_{i=1, .. n} z_i w_i^{bar}.

>

> Operators on H of finite dimension n>=2 can be thought of

> as nxn square matrices with entries in C.

>

> If A is an nxn matrix, the equation for eigenvalues is:

> det(A - lambda I) = 0. 'I' is the nxn identity matrix.

> lambda is the unknown scalar in C.

>

> det(A - lambda I) = 0 is equivalent (possibly after mutiplying by +/-1)

> to lambda^n + a_{n-1} lambda^(n-1) + ... + a_0 = 0,

> a degree 'n' polynomial equation with indeterminate

> lambda obtained by fully expanding the determinant:

> det(A - lambda I)

> by using the coefficients of A as things that don't

> depend on lambda (i.e. parameters, coefficients).

>

> Then, we apply the fundamental theorem of algebra for

> C to this degree n polynomial equation.

>

> Suppose lambda = r0 is one root of the polynomial.

> Then r0 is an eigenvalue of the matrix A, and

> there exists a non-zero vector v in C^n such that

> Av = r0 v. If k in C is any scalar,

> A (kv) = k (Av) = k (r0 v) = r0 (kv),

> so (A - r0 I) (kv) = 0.

>

> Or, one could say that for any scalar k in C,

> A maps kv to (r0 k) v, an element of the 1-dim.

> space C.v . So W = {kv, k in C the complex numbers},

> v is non-zero, dim(W) = 1 exactly , W is a non-trivial

> subspace of C^n, and W is mapped by A to W (although

> not always onto; this is shown in case A is the zero

> matrix).

>

> In countably infinite dimension, there's essentially one

> complex Hilbert space, say l^2, the Banach space

> of square-summable sequences of complex numbers.

>

> When the topic is Hilbert spaces, by a subspace one

> always means/assumes a complete (vector) subspace, unless

> it's stated otherwise. That's always been my working

> assumption.

>

> This also applies, if I remember correctly, to Banach spaces

> and subspaces, where the assumption would be that a

> subspace must be norm-complete, a complete metric space.

>

> So now if H is l^2 or any other essentially equivalent

> separable infinite dimensional complex Hilbert space,

> one supposes given a continuous complex-linear mapping

> T: H -> H .

>

> That's what an operator from H to H is. (However,

> in quantum mechanics they need mappings that aren't

> necessarily continuous, so-called unbounded operators).

>

> Just like there is a ring of nxn matrices

> in finite dimension, there is a ring with identity

> of all the continuous complex-linear mappings

> T: H -> H that can be. That huge ring is denoted

> B(H).

>

> B(H) is deceptively large.

> For example, if sigma: N^* -> N^* is a permutation

> on N^*, and e_i is the sequence

> (0, 0, ... 1, ... 0, ....)

> ^ i'th component ,

>

> then one can arrange provisionally to map

> each e_i to e_{sigma(i)} , for i = 1, 2, 3, ad infinitum

>

> This can be extended by linearity to all linear combinations

> of the e_i, i in N^*. Then, by continuity, the mapping

> can be extended to the closure in l^2 of these linear

> combinations.

>

> This works and will give a bounded operator U_sigma on

> l^2. This U_sigma has norm 1 in the operator norm.

> There are 2^\omega permutations on N^*, resulting in

> 2^\omega = continuum distinct operators U_sigma.

> If sigma and sigma' are distinct permutations, then

> for the operators U_sigma and U_sigma' , I get:

> || U_sigma - U_sigma' || >= 1 (hopefully ... ).

>

> { U_sigma, all permutations sigma: N^*->N^*} forms

> a set of the power of the continuum of points in

> B(H) any two of which are at an operator-norm distance

> of at least 1 from each other.

> It follows that B(H) has no countable dense subset, i.e.

> B(H) is a non-separable metric space.

>

> For any A: H-> H, there's something called the

> polar decomposition. One lets P = sqrt(A A^*) (I think)

> and then A = O P , where O is an isometry or a partial

> isometry and P is a so-called "positive" operator.

> This might seem like a promising divide-and-conquer

> approach to the invariant sub-space problem, but

> it doesn't seem to say much about what happens when,

> starting with a non-zero x in H, one iterates

> A, constructing Ax, AAx, A^3 x, etc or using

> the polar decomposition,

> (OPx), (OPOP)x, (OPOPOP)x, etc [i.e. which way are things

> going? we still don't know any better after applying

> the polar decomposition to A].

>

> Per Enflo first constructed an infinite-dim. Banach space

> and an operator on it that had no non-trivial closed

> sub-space. See e.g.:

> archive.numdam.org/article/SAF_1975-1976____A11_0.pdf

>

> and/or

>

> http://en.wikipedia.org/wiki/Per_Enflo

>

>

> For separable Hilbert spaces, the best known case is when

> the operator T is self-adjoint, where the so-called

> spectral theorem in functional analysis applies

> (Rudin: Functional Analysis).

[...]

I was reading on problems considered "modulo a compact

operator K".

If T is a bounded operator on a separable, infinite

dimensional Hilbert space H,

can one show that for some compact operator K

on H, T+K has a non-trivial invariant

subspace?

David Bernier

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