32
votes
Accepted
Can we take a supremum over all Hilbert spaces?
It is true that we cannot use an arbitrary property$^1$ $P$ to define a set, in the sense that the collection of all things with property $P$ need not be a set. However, the axiom (scheme) of ...
- 19.3k
29
votes
What is the intuition for the trace norm (nuclear norm)?
One potential intuition for the trace norm is as a way of turning the rank of a matrix (which is very discontinuous) into a norm (which is continuous). Specifically, the trace norm is the unique norm ...
- 5,755
21
votes
Nonseparable Hilbert spaces
To the contrary, my feeling is that nonseparable Hilbert spaces are in some sense artifacts and can almost always be avoided. And more generally, to your comment about nonseparable Banach spaces being ...
- 42.1k
21
votes
Accepted
What is the intuition for the trace norm (nuclear norm)?
Another answer is that $M_n$, the space of $n\times n$ complex matrices, carries an operator norm where the norm of a matrix is its norm as a linear operator from $\mathbb{C}^n$ to itself (giving $\...
- 42.1k
17
votes
Rigged Hilbert spaces and the spectral theory in quantum mechanics
Why are rigged Hilbert spaces a paper subject, not usually treated in rigorous textbooks: this is a good question. If you learn QM the way I did, you start by understanding that the physicists' Dirac ...
- 42.1k
15
votes
$x f'$ bounded by $x^2f $ and $f''$?
By integration by parts,
$$\int (xf')^2 = \int (x^2f') f' = -\int 2xf'f - \int x^2 f'' f.$$
$$-\int 2xf'f = \int f^2=\|f\|^2.$$
By the Cauchy-Schwarz inequality
$$\bigg|\int x^2 f'' f\bigg|\le \|x^2f\|...
- 2,169
14
votes
Rigged Hilbert spaces and the spectral theory in quantum mechanics
@Nik Weaver: The answer one is looking for is some clearly important, basic topic in QM which can be treated more easily with rigged Hilbert spaces.
The foremost topic is the theory of resonances.
A ...
- 178k
14
votes
Accepted
Is the "space of physical quantities" a field of transcendence degree $6$ or $7$ over the rationals?
I don't think there are many meaningful situations where physical quantities of different dimension are added together (in fact, this is widely regarded as a taboo, and precisely the sort of mistake ...
- 30.2k
13
votes
Accepted
$x f'$ bounded by $x^2f $ and $f''$?
By a cutoff function argument, it suffices to assume $f$ is compactly supported, so we can integrate by parts without picking up boundary terms.
Thus
$$\int (xf')^2 = \int (x^2f') f' = -\int 2xf'f - \...
- 29.2k
13
votes
Rigged Hilbert spaces and the spectral theory in quantum mechanics
You ask 'why spectral theory instead of rigged Hilbert spaces?'.
There are some practical/pedagogical reasons: One is that you'll need basic spectral theory in quantum mechanics anyways, so might as ...
- 5,879
12
votes
Accepted
Criterion for compactness
$T$ is not necessarily compact. Let me produce a counterexample.
Let $H$ be any infinite dimensional real or complex separable Hilbert space. Let $(f_{j,k})_{1\leq k\leq j},(e_{j})_{j=1}^{\infty}$ be ...
- 27.8k
11
votes
Accepted
When $\lambda$-commutativity implies commutativity?
I don't see which kind of condition you are looking for, as there are a lot of pairs $T,S$ such that $TS=\lambda ST$ and $\lambda\ne1$, even in finite dimension. Such pairs are said to $\lambda$-...
- 51.6k
11
votes
Accepted
Subspaces isomorphic with quotients
Every separable Banach space is a quotient of $\ell_1$, so in particular every subspace of $\ell_1$ is a quotient of $\ell_1$.
- 11.3k
10
votes
Are dagger categories truly evil?
Here's a notion of undirected category, which is equivalent to the notion of dagger category. It's a bit cumbersome to work with, but the main point is to underscore Qiaochu Yuan and Manuel Bärenz's ...
- 347
10
votes
Accepted
Multiple of identity plus compact
OK, let me try too. It is going to be a somewhat long story. WLOG, $\|T\|\le 1$.
Step 1: It is enough to show that for every finite-dimensional subspace $E$ and every $\delta>0$, there exists a ...
- 59.8k
10
votes
Accepted
What is the difference between $L^2(\mathbb{R})$ and $L^2(\mathbb{R}^2)$?
We can tell $\mathbb{R}$ and $\mathbb{R}^2$ apart by topology, and we can tell $L^2(\mathbb{R})$ and $L^2(\mathbb{R}^2)$ apart by quantum topology. In the sense that a choice of C*-algebra contained ...
- 42.1k
10
votes
Accepted
Trace norm of operators obtained by restricting the matrix of a trace class operator
Here's an algorithm for testing an ad-hoc conjecture $C$ about Hilbert space operators. :-)
Set up the runtime environment correctly by loading the information "Most conjectures are false" ...
- 11.6k
9
votes
What is the intuition for the trace norm (nuclear norm)?
If you are interesting in geometrical intuition, here is the possible one:
Any matrix (operator) $A$ transforms a unit ball to an ellipsoid. Singular values of $A$ correspond to the lengths of ...
- 415
9
votes
Space of compact operators defined on separable Hilbert space
Let $X$ be a Banach space such that $X^*$ is separable. I claim that the space of compact operators $\mathcal K(X)$ on $X$ is separable.
It is enough to show that $nB_{\mathcal K(X)}$, the space of ...
- 11.3k
9
votes
Accepted
A homeomorphism between the unit interval $[0,1]$ and a linearly independent subset of a Hilbert space
It's not true.
Edit: thinking twice there's a simpler solution.
On $L^2([0,1])$, define $f(x)(t)=1+t^{1/2x}$ for $x\in\mathopen]0,1]$ and $f(0)=1$. It's free (because the $t\mapsto t^x$, $x\ge 0$, ...
- 60.4k
9
votes
Does eigenvalue exist in a Hilbert space?
As noted in comments, there are no bounded operators whose commutator with their adjoint is a non-zero scalar. The prototype of unbounded operators with the desired property is the 1-dimensional Dirac ...
- 22.6k
9
votes
Accepted
Example of linear functionals on $B(H)$
So, you are asking about non-normal functionals on $B(H)$. This is very similar to the question of what are the functionals on $\ell_\infty$ that are not in $\ell_1$?
Fix an ultrafilter $U$ on $\...
- 11.3k
9
votes
Accepted
Initial conditions in the Klein-Gordon equation
One must remark that derivatives in Sobolev spaces are usually taken in the sense of distributions: given $k\in\mathbb{N}_0=\{0,1,2,\ldots\}$, $H^k(\mathbb{R}^n)$ is the space of tempered ...
- 7,109
9
votes
Accepted
Name for certain property of equivalent norms on finite-dimensional subspaces of a Banach space
Your condition implies that $X$ is isomorphic to a Hilbert space with isomorphism constant at most $M^2$. The distance condition implies that both type 2 constant and cotype 2 constant of $X$ is ...
- 2,409
9
votes
Accepted
Is Solèr’s theorem true in constructive mathematics?
Suppose you have a classical classification theorem saying
Each structure (of a certain kind) is either an $A$ or a $B$.
Then you cannot exhibit constructively a $C$ which is neither $A$ nor $B$ ...
- 48k
9
votes
Accepted
Contact points for John's ellipsoid
Looks true. A necessary and sufficient condition for these points (let $E$ be a standard ball) is that the identity operator $I$ is a non-negative linear combination of projectors $P_i$ on lines ...
- 103k
8
votes
Matrices Representing Bounded Operators and Absolute Values
As requested, I've moved my comments to an answer.
The question is equivalent to the following:
Suppose that a doubly-infinite matrix $A=(A_{ij})_{i,j\in {\bf Z}}$ represents a bounded hermitian ...
- 25.5k
8
votes
A characterization of metric spaces admitting a bi-Lipschitz embedding into a Hilbert space?
Schoenberg's criterion can be extended to bi-Lipschtiz embeddings.
This is Corollary 3.5 in:
N. Linial, E. London, Y. Rabinovich, The geometry of graphs and some of its algorithmic applications. ...
- 361
8
votes
Accepted
Extending maps from dense $*$-algebras of $C^*$-algebras
I believe this is a counter-example.
Let $\newcommand{\mc}{\mathcal}\mc A$ be the algebra of complex polynomials restricted to $[0,1]$, with closure $A=C[0,1]$. Let $X\in\mc A$ be the coordinate ...
- 18.5k
8
votes
Accepted
What is the consistency strength of non-existence of outer automorphisms of Calkin algebra?
Perhaps surprisingly, the Open Coloring Axiom (even OCA + MA$_{\aleph_1}$) has no additional consistency strength. The situation is described in Velickovic's paper "Applications of the open coloring ...
- 19.3k
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