User jonas meyer - MathOverflow

Answer by Jonas Meyer for The reals as continuous image of the irrationals

2012-11-12T16:46:47Z

Count the rational numbers as $(a_1,a_2,a_3,\ldots)$. Let $f:\mathbb R\to \mathbb R$ be defined at the nonpositive integers by $f(n)=(-1)^nn$, and on the positive integer multiples of $\sqrt 2$ by $f(n\sqrt 2)=a_n$. Let $f$ be piecewise linear between the points just defined. Then the restriction of $f$ to $\mathbb R\setminus \mathbb Q$ is a continuous surjection from the irrationals to $\mathbb R$.

Answer by Jonas Meyer for Unbounded representations of Banach algebras

2011-08-13T08:45:26Z

As Dima Shlyakhtenko indicated, there are examples, obtained for example by modifying your zero product example to $x\mapsto\begin{bmatrix}0&f(x)\\ 0&0\end{bmatrix}$ for some unbounded linear functional $f$.

A generalization of the result you mentioned is found in Theorem 4.1.20 of Rickart's General theory of Banach algebras. A special case of that theorem says that if the range of the representation is a $*$-subalgebra of $B(H)$, then the representation is automatically continuous.

What looks like a good reference is Dales's Banach algebras and automatic continuity, but I don't currently have a copy. It is referenced in an Encyclopaedia of mathematics article by the same author called "Automatic continuity for Banach algebras," which also gives some other references that might help. According to the article, there are some Banach spaces $E$ such that every homomorphism from $B(E)$ to another Banach algebra is continuous. (It doesn't say which ones.)

Does equality of the operator norm and the cb norm for every bimodule map over a C*-subalgebra imply that the subalgebra is matricially norming?

2010-02-19T02:09:02Z

In this post, without further mention all C*-algebras are assumed to have an identity element and subalgebras inherit the identity.

Question: Let $\mathcal{C}$ be a C*-subalgebra of $\mathcal{B}$. Suppose that for every C*-algebra $\mathcal{A}$ containing $\mathcal{C}$ as a C*-subalgebra and every $\mathcal{C}$-bimodule map $\phi:\mathcal{A}\to\mathcal{B}$, $\|\phi\| = \|\phi\|_{cb}$. Does it follow that $\mathcal{C}$ is matricially norming for $\mathcal{B}$?

Definitions:

Let $\mathcal{A}$ and $\mathcal{B}$ be C*-algebras and let $\phi:\mathcal{A}\to\mathcal{B}$ be a linear map. Define, for each $n\in\mathbb{N}$, $\phi_n:M_n(\mathcal{A})\to M_n(\mathcal{B})$ by $\phi_n((a_{ij}))=(\phi(a_{ij}))$. The cb norm of $\phi$ is defined by $\|\phi\|_{cb}=\sup_{n\in\mathbb{N}}\|\phi_n\|$ . (The "cb" stands for completely bounded, and $\phi$ is called completely bounded when its cb norm is finite.)
A C*-subalgebra $\mathcal{C}$ of a C*-algebra $\mathcal{B}$ is called matricially norming for $\mathcal{B}$ if for every $n\in\mathbb{N}$ and every $n$-by-$n$ matrix $(b_{ij})$ in the C*-algebra $M_n(\mathcal{B})$, $$\|(b_{ij})\|=\sup\left\{ \left\| \sum_{i,j=1}^n x_i b_{ij} y_j \right\|: x_i,y_j\in\mathcal{C}, \left\| \sum_{i=1}^n x_i x_i^* \right\|\leq1, \left\| \sum_{j=1}^n y_j^* y_j \right\|\leq1\right\}.$$

Remarks:

The matricially norming condition says that matrices over $\mathcal{B}$ are "normed" by contractive row and column matrices over $\mathcal{C}$. That is, for $B\in M_n(\mathcal{B})$, $\|B\|=\sup\|RBC\|$ , where the sup is taken over all rows $R$ and columns $C$ of length $n$ of elements of $\mathcal{C}$ such that $\|RR^*\|\leq1$ and $\|C^*C\|\leq1$ . (So that, for instance, it is readily seen that $\mathbb{C}$ is matricially norming for $\mathbb{C}$.) Sometimes just "norming" is used instead of "matricially norming", but I am using the more descriptive term found in Paulsen's book.
The converse to the question holds. That is, if $\mathcal{C}$ is a matricialy norming C*-subalgebra of $\mathcal{B}$, then for every C*-algebra $\mathcal{A}$ containing $\mathcal{C}$ as a C*-subalgebra and every $\mathcal{C}$-bimodule map $\phi:\mathcal{A}\to\mathcal{B}$, $\|\phi\| = \|\phi\|_{cb}$. This is what inspired the question, brought to me by a fellow graduate student after we had studied chapter 8 in Paulsen's book.
The subalgebra of scalars is matricially norming for a commutative C*-algebra, so if there is a counterexample it is with a noncommutative $\mathcal{B}$.
I believe that results in this paper by Smith can be used to show that there is no counterexample when $\mathcal{C}=\mathbb{C}$. It is shown there that if all bounded linear maps from all C*-algebras into $\mathcal{B}$ are completely bounded, then $\mathcal{B}$ must be isomorphic to a subalgebra of $M_n(C(X))$ for some $n\in\mathbb{N}$ and some compact Hausdorff space $X$. But then if $\mathcal{B}$ is noncommutative, the transpose map should give an example of a map whose cb norm is bigger than its operator norm.
For more on matricially norming C*-subalgebras, the paper by Pop, Sinclair, and Smith is a good place to start.

Answer by Jonas Meyer for Must a surjective isometry on a dual space have a pre-adjoint?

2011-03-03T20:54:37Z

Let $X$ be the space of sequences, indexed by the nonzero integers, that tend to $0$ at $-\infty$ and to an arbitrary finite limit at $\infty$, with sup norm (a direct product of $c_0$ and $c$). Then $X^*$ can be identified with $\ell^1$ of $\mathbb{Z}$. If $f$ is in $\ell^1$, then the corresponding functional on $X$ sends $x$ to $\sum_{n\neq0}f_nx_{n} + f_0\cdot\lim_{n\to\infty}x_n$ . The map on $\ell^1\cong X^*$ defined by $(f_n)_{n\in\mathbb Z}\mapsto (f_{-n})_{n\in\mathbb{Z}}$ is a linear surjective isometry with no pre-adjoint. (If there were a preadjoint, it would have to send $(x_n)_{n\in \mathbb Z\setminus\{0\}}$ to $(x_{-n})_{n\in\mathbb Z\setminus\{0\}}$ .)

Answer by Jonas Meyer for Does pointwise convergence imply uniform convergence on a large subset?

2010-11-12T04:27:27Z

I did some Googling and came up with something that looks relevant, Theorem 10 quoted below from Morgan's Point set theory. It cites works of Sierpiński from the late 1930s, but I can't tell what works are cited because the preview won't let me see that page in the references.

The existence of a linear set having the power of the continuum that is concentrated on a denumerable set is equivalent to the existence of a pointwise convergent sequence of functions of a real variable that does not converge uniformly on any uncountable set.

Answer by Jonas Meyer for Most memorable titles

2010-10-31T20:56:00Z

The book Free rings and their relations by P.M. Cohn.

Answer by Jonas Meyer for Most memorable titles

2010-10-31T20:44:59Z

"On $O_n$" by D.E. Evans. ($\mathcal{O_n}$ is notation Cuntz gave for the algebras he introduced in "Simple $C^*$ -algebras generated by isometries".)

Answer by Jonas Meyer for Topological spaces whose continuous image is always closed

2010-10-21T00:14:16Z

If $Z$ is not compact, and $X=\{p\}\cup Z$ is the space whose nonempty open sets are of the form $\{p\}\cup V$ with $V$ open in $Z$, then $X$ is not compact, but every continuous function from $X$ to a Hausdorff space is constant.

Answer by Jonas Meyer for Converse of Picard's Big Theorem?

2010-09-02T05:16:11Z

If the boundary point is not an isolated singularity, then you can't say it is an essential singularity, so the answer to your final question is no. It is quite possible that $f$ cannot be extended to a larger domain that contains a punctured neighborhood of $x$. To be quasi-explicit, take a holomorphic function whose domain of holomorphy is $\Omega$ and multiply by a holomorphic function defined everywhere in the plane except with an essential singularity at $x$.

If $f$ has an isolated singularity at $x$, then yes, this characterizes essential singularities. $f$ goes to $\infty$ at a pole, so in particular is nonzero in a neighborhood of the pole. If the singularity at $x$ is removable, then either $f$ goes to a nonzero value, in which case it is nonzero in a neighborhood of $x$, or $f$ goes to 0. In the latter case, $f$ could be extended to a holomorphic function on a domain containing $x$, so if $x$ were a limit point of the zero set of $f$, then $f$ would be identically zero. And you could easily adjust this to nonzero cases.

You don't need anything near the strength of Picard's theorem. In fact, that is what is so amazing about Picard's theorem: Either $f$ has really nice behavior near an isolated singularity, or it maps everywhere except possibly one point in each neighborhood of the singularity.

Answer by Jonas Meyer for totally disconnected and zero-dimensional spaces

2010-09-01T16:13:48Z

Here's just a reference showing that a Hausdorff locally compact totally disconnected space is zero-dimensional: Proposition 3.1.7 of Arhangel'skii and Tkachenko.

Answer by Jonas Meyer for Representation of $*$-automorphism on finite dimensional matrix algebras

2010-08-30T19:23:50Z

Here is one generalization:

Every $*$-automorphism of the algebra of compact operators on a Hilbert space is conjugation by a unitary operator on that space.

Using the fact that the algebra of compact operators is irreducible, this can be seen as a special case of:

Every irreducible $*$-representation of the algebra of compact operators on a Hilbert space is unitarily equivalent to the identity representation.

A proof can be found for instance in Section 1.4 of Arveson's An invitation to C* algebras. Another proof of the first assertion that gives more information can be found in Proposition 1.6 of Raeburn and Williams's Morita equivalence and continuous trace C*-algebras.

The first part is still true if you take all bounded operators instead of only the compact ones. (And these are the same thing in the finite dimensional case.)

Answer by Jonas Meyer for infinitely many linear equations in infinitely many variables

2010-08-25T06:41:17Z

Take a look at Section 6 of Chapter III of Banach's book, which gives a result in the theory of $F$-spaces. The title of the section in the English translation is "Systems of linear equations in infinitely many unknowns".

(By coincidence I was reading this recently, and I admit that that is part of the reason I voted to reopen.)

Answer by Jonas Meyer for range projection of an unbounded idempotent affiliated to a finite von Neumann algebra

2010-08-25T05:58:38Z

Disclaimer: I fear I may be missing some subtlety here, as is often the case when I think about unbounded operators. This is an attempt to generalize the result.

A closed densely defined operator $T$ on $H$ has a unique polar decomposition $T=V|T|$ with $|T|=\sqrt{T^*T}$ and $V$ a partial isometry whose initial space is the closure of the range of $|T|$ and whose final space is the closure of the range of $T$. If $T$ is affiliated with a von Neumann algebra $M$, then $V$ is in $M$ (as stated e.g. in Nelson's paper on the bottom of page 111). Thus $VV^*$, the projection onto the closure of the range of $T$, is in $M$.

Answer by Jonas Meyer for How come nowhere dense subsets implies discrete?

2010-08-22T17:04:55Z

Because a nowhere dense set minus a point is nowhere dense, the hypothesis implies that points are relatively open in nowhere dense subsets. Perhaps more to the point, every subset of a nowhere dense set is nowhere dense, hence closed by hypothesis.

As for your question about cofinite topologies: Finite subspaces of cofinite topological spaces are discrete, because finite subsets are closed. (The same is true in any $T_1$ space.) This doesn't mean that subsets of finite sets are open, but rather that they are relatively open.

Answer by Jonas Meyer for Hausdorff dimension: subset of $\mathbb{R}^n$ vs. boundary of this subset

2010-08-09T19:20:30Z

The set of rational numbers has Hausdorff dimension 0, while its boundary is the set of real numbers, with Hausdorff dimension 1.

Answer by Jonas Meyer for Is the set of exponentials open?

2010-08-05T18:40:44Z

The exponential map is not open on the von Neumann algebra of all bounded linear operators on an infinite dimensional separable complex Hilbert space. A 1987 article by Conway and Morrel shows that the spectrum of an element of the interior of the image of the exponential map does not separate 0 from infinity. On the other hand, a bilateral shift U has spectrum equal to the unit circle centered at the origin, and every Borel logarithm on the circle applied to the unitary operator $U$ yields a preimage point for U under the exponential map. Hence U is in the image but not in the interior of the image.

I learned about Conway and Morrel's article from this answer by David Speyer.

Answer by Jonas Meyer for Question about von Neumann algebra generated by a complete algebra of projections

2010-07-23T22:21:55Z

The answer "yes" follows from Theorem 2.8 of Bade's "On Boolean algebras of projections and algebras of operators," 1955, which is in the more general context of algebras of operators on a Banach space.

Bade had previously proven a less general result that still covers your case, dealing with algebras of operators on reflexive spaces, in Theorem 3.4 of "Weak and strong limits of spectral operators," 1954.

Answer by Jonas Meyer for Examples of common false beliefs in mathematics.

2010-07-23T18:39:05Z

"Compact implies sequentially compact."

Answer by Jonas Meyer for 'Eigenvectors' of evolute operation

2010-07-18T22:13:49Z

Light studied the problem, but didn't solve it in his 1917 dissertation. A 1920 note of Light was followed by a 1921 note of Franklin showing the existence of infinitely many classes of examples not covered by Light, and pointing out that Puiseux had shown this in 1844. However, the question of determining all of them isn't addressed there, and I don't know what else is known.

Regarding your question about formalizing the notion of curves as eigenvectors, couldn't you just take the real vector space whose basis consists of representatives from each similarity class of curves? Scalar multiplication of a basis element by a real number corresponds to dilating or shrinking, and reflecting if negative. The evolute operator is formally extended to this vector space linearly, and the curves similar to their evolutes will be precisely (scalar multiples of) the basis vectors that are eigenvectors.

Answer by Jonas Meyer for Do names given to math concepts have a role in common mistakes by students?

2010-07-18T20:18:36Z

The article "Surprises from mathematics education research: student (mis)use of mathematical definitions" by Edwards and Ward addresses some of your concerns in the context of U.S. undergraduates.

From the introduction:

...[T]asks involving the definitions of "limit" and "continuity," for example, were problematic for some of the students. Ward's intuitive reaction was that those words were "loaded" with connotations from their nonmathematical use and from their less than completely rigorous use in elementary calculus. He said, "I'll bet students have less difficulty or, at least, different difficulties with definitions in abstract algebra. The words there, like 'group' and 'coset,' are not so loaded."

...

He was surprised to see his algebra students having difficulties very similar to those of Edwards's analysis students. (So he lost his bet.) In particular, he was surprised to see difficulties arising from the students' understanding of the very nature of mathematical definitions, not just from the content of the definitions.

Answer by Jonas Meyer for Standard name for basis-independent submatrices?

2010-01-05T22:10:56Z

The standard name in operator theory is "compression", and its partner in crime is "dilation". I.e., A is a compression of B if and only if B is a dilation of A (although sometimes "dilation" is reserved for cases where the compression respects powers). The Wikipedia entry is not proof, but here it is anyway. As for searches, you'll get some relevant hits from "compression of an operator" with quotes.

Here are some examples.

Some further remarks:

Sz.-Nagy and Foiaș in Harmonic analysis of operators on Hilbert space (1970) use the notation $\text{pr }T$ for the compression of $T$ onto $K$ (see page 10), but apparently without ever giving it a name. The notation is suggestive of "projection", and that is the terminology used by Sarason in "Generalized interpolation in $H^\infty$" (1967).

Lebow goes into more detail on terminology in "A note on normal dilations" (1965), saying in particular that Sz.-Nagy used "projection". In fact, this is the terminology used by Sz.-Nagy in the celebrated appendix to Riesz and Sz.-Nagy's Functional analysis (1955), which in turn refers to Halmos's paper "Normal dilations and extensions of operators" (1950) as the first place where "compression" and "dilation" were used. The terminology "strong compression" may be used when the compression respects powers, and this is the same as saying that $K$ is semi-invariant for $T$ (see Sarason's "On spectral sets having connected complement" (1965)). If $K$ is reducing for $T$, i.e., if both $K$ and $K^\perp$ are invariant subspaces for $T$, then Lebow calls the compression a "reduction".

Dixmier gives some terminology in von Neumann algebras (translated 1981 printing) for the case when the compression is applied to an entire von Neumann algebra of operators, which clashes somewhat with the terminology of Lebow. A von Neumann algebra compressed to the space of a projection in the algebra is called a "reduced" von Neumann algebra (page 19), even though the space is reducing only if the projection is in the center. The compression of a von Neumann algebra onto the space of a projection in the commutant (in which case the compression is a normal $*$-homomorphism) is called an "induction". If $P$ denotes the orthogonal projection you called $\pi_K$, then Dixmier uses the notation $T_K$ or $T_P$ for the compression, but without ever giving a name to the construction for single operators. On the other hand, Jones and Sunder use "reduction" for what Dixmier calls "induction", more in tune with Lebow, on page 21 of Introduction to subfactors (1997).

I stand by my answer that by now "compression" is most standard for single operators, and it is satisfying to find out that we have Halmos to thank for this.

Answer by Jonas Meyer for Relevance of the complex structure of a function algebra for capturing the topology on a space.

2010-05-25T21:55:48Z

Here is a slightly different, perhaps simpler take on showing that $C(X,\mathbb{R})$ determines $X$ if $X$ is compact Hausdorff. For each closed subset $K$ of $X$, define $\mathcal{I}_K$ to be the set of elements of $C(X,\mathbb{R})$ that vanish on $K$. The map $K\mapsto\mathcal{I}_K$ is a bijection from the set of closed subsets of $X$ to the set of closed ideals of $C(X,\mathbb{R})$. Urysohn's lemma and partitions of unity are enough to see this, with no complexification, Gelfand-Neumark, or (explicitly) topologized ideal spaces required. I remember doing this as an exercise in Douglas's Banach algebra techniques in operator theory in the complex setting, but the same proof works in the real setting.

Here are some details in response to a prompt in the comments. (Added later: See Theorem 3.4.1 in Kadison and Ringrose for another proof. Again, the functions are assumed complex-valued there, but you can just ignore that, read $\overline z$ as $z$ and $|z|^2$ as $z^2$, to get the real case.)

I will take it for granted that each $\mathcal{I}_K$ is a closed ideal. This doesn't require that the space is Hausdorff (nor that $K$ is closed). Suppose that $K_1$ and $K_2$ are unequal closed subsets of $X$, and without loss of generality let $x\in K_2\setminus K_1$. Because $X$ is compact Hausdorff and thus normal, Urysohn's lemma yields an $f\in C(X,\mathbb{R})$ such that $f$ vanishes on $K_1$ but $f(x)=1.$ Thus, $f$ is in $\mathcal{I}_{K_1}\setminus\mathcal{I}_{K_2}$, and this shows that $K\mapsto \mathcal{I}_K$ is injective. The work is in showing that it is surjective.

Let $\mathcal{I}$ be a closed ideal in $C(X,\mathbb{R})$, and define $K_\mathcal{I}=\cap_{f\in\mathcal{I}}f^{-1}(0)$, so that $K_\mathcal{I}$ is a closed subset of $X$. Claim: $\mathcal{I}=\mathcal{I}_{K_\mathcal{I}}$ .

It is immediate from the definition of $K_\mathcal{I}$ that each element of $\mathcal{I}$ vanishes on $K_\mathcal{I}$, so that $\mathcal{I}\subseteq\mathcal{I}_{K_\mathcal{I}}.$ Let $f$ be an element of $\mathcal{I}_{K_\mathcal{I}}$ . Because $\mathcal{I}$ is closed, to show that $f$ is in $\mathcal{I}$ it will suffice to find for each $\epsilon>0$ a $g\in\mathcal{I}$ with $\|f-g\|_\infty<3\epsilon$. Define $U_0=f^{-1}(-\epsilon,\epsilon)$, so $U_0$ is an open set containing $K_\mathcal{I}$. For each $y\in X\setminus U_0$, because $y\notin K_\mathcal{I}$ there is an $f_y\in \mathcal{I}$ such that $f_y(y)\neq0$. Define $$g_y=\frac{f(y)}{f_y(y)}f_y$$ and $U_y=\{x\in X:|g_y(x)-f(x)|<\epsilon\}$ . Then $U_y$ is an open set containing $y$. The closed set $X\setminus U_0$ is compact, so there are finitely many points $y_1,\dots,y_n\in X\setminus U_0$ such that $U_{y_1},\ldots,U_{y_n}$ cover $X\setminus U_0$. Relabel: $U_k = U_{y_k}$ and $g_k=g_{y_k}$. Let $\varphi_0,\varphi_1,\ldots,\varphi_n$ be a partition of unity subordinate to the open cover $U_0,U_1,\ldots,U_n$. Finally, define $g=\varphi_1 g_1+\cdots+\varphi_n g_n$. That should do it.

In particular, a closed ideal is maximal if and only if the corresponding closed set is minimal, and because points are closed this means that maximal ideals correspond to points. (Maximal ideals are actually always closed in a Banach algebra, real or complex.)

Answer by Jonas Meyer for Surjective *-homs between multiplier algebras

2010-07-08T20:53:17Z

This is true if $A$ is $\sigma$-unital, and is sometimes called the "noncommutative Tietze extension theorem". A good reference is Proposition 6.8 in Lance's Hilbert C*-modules. Proposition 3.12.10 in Pedersen's C*-algebras and their automorphism groups covers the separable case, which was first proved by Akemann, Pedersen, and Tomiyama in a 1973 paper called "Multipliers of C*-algebras".

Pedersen points out in Section 3.12.11 that you can get counterexamples in the commutative case by considering non-normal locally compact Hausdorff spaces, so that Tietze's extension theorem doesn't apply.

Akemann, Pedersen, and Tomiyama are more explicit:

In fact let $X$ be a locally compact Hausdorff space which is not normal, and consider two disjoint closed sets $Y_1$ and $Y_2$ such that the function $b$ which is one on $Y_1$ and zero on $Y_2$ has no continuous extension to $X$. The restriction map of $C_0(X)$ to $C_0(Y_2\cup Y_2)$ is surjective, and $b\in M(C_0(Y_1\cup Y_2))$, but $b$ is not the image of a multiplier of $C_0(X)$.

Answer by Jonas Meyer for Can an entire non-constant function be bounded on only a finite set of lines through the origin?

2010-06-27T22:01:22Z

Newman gave an example in 1976 of a non-constant entire function bounded on each line through the origin in "An entire function bounded in every direction".

I like the second sentence of the article:

This is exactly what is needed to confuse students who have just struggled to comprehend the meaning of Liouville's theorem.

Armitage gave examples in 2007 of non-constant entire functions that go to zero in every direction in "Entire functions that tend to zero on every line". For this I have only seen the MR review. (If you don't have MathSciNet access, the link should still give you the publication information to find the article.)

Update: I just decided to take a look at the Armitage paper, and the introduction was enlightening:

Although every bounded entire (holomorphic) function on $\mathbb{C}$ is constant (Liouville’s theorem), it has been known for more than a hundred years that there exist nonconstant entire functions $f$ such that $f(z) → 0$ as $z →∞$ along every line through 0 (see, for example, Lindelöf’s book [10, pp. 119–122] of 1905). And it has been known for more than eighty years that such functions can tend to 0 along any line whatsoever (see Mittag-Leffler [11], Grandjot [8], and Bohr [4]). Further references to related work are given in Burckel’s review [5] of Newman’s note [12]. Entire functions with radial decay are used by Beardon and Minda [3] and Ullrich [14] in studies of pointwise convergent sequences of entire functions.

Armitage goes on to mention that Mittag-Leffler and Grandjot also gave explicit constructions, but states, "The examples given in what follows may nevertheless be of some interest because of their comparative simplicity." The examples are $$F(z)=\exp\left(-\int_0^\infty t^{-t}\cosh(tz^2)dt\right) - \exp\left(-\int_0^\infty t^{-t}\cosh(2tz^2)dt\right)$$ and $$G(z)=\int_0^\infty e^{i\pi t}t^{-t}\cosh(t\sqrt{z})dt\int_0^\infty e^{i\pi t}t^{-t}\cos(t\sqrt{z})dt .$$

Answer by Jonas Meyer for Is there an index for solutions to American Mathematical Monthly problems?

2010-06-06T21:46:22Z

This is somewhat redundant with Igor Pak's P.S., but I found an index for the 1918-1950 range in The Otto Dunkel memorial problem book, starting on page 80 of this pdf.

I found more: Stanley Rabinowitz and others had the same idea a couple of decades ago, and they started compiling indices, not just for the Monthly. Here are Google Books and official links for the two volumes I found:

1975-1979 (published 1999) Website

1980-1984 (published 1992) Website, Errata

Googling "Stanley Rabinowitz 1985-1989" led me to this:

The labor involved in classifying and indexing the 5-year indexes has proved to be far greater than anticipated. We hope to have the 1975-1979 Index out later this year and the 1985-1989 Index is well under way. [Stanley Rabinowitz, June 1996]

I don't know if the 1985-1989 index was ever finished, but presumably there's at least a partial draft out there somewhere. The website http://www.mathpropress.com/ of Rabinowitz's company includes contact information. They say they're looking for authors to publish, among other things, "indexes to mathematical problems or results", but the page apparently hasn't been updated since 2006.

Answer by Jonas Meyer for Simple inequality in C*-algebras

2010-05-17T04:13:12Z

Theorem 1.5 of this 1987 paper by J. Phillips says that if $f:[0,\infty)\to [0,\infty)$ is a continuous operator monotone function and $a$ and $b$ are positive operators on a Hilbert space, then $\|f(a)-f(b)\|\leq f(\|a-b\|)-f(0)$. I think that the proof is nice. Corollary 1.6 says that $\|a^{1/n}-b^{1/n}\|\leq\|a-b\|^{1/n}$, $n\geq1.$ Of course your inequality follows from taking $a=x^2$, $b=y^2$, and $n=2$.

Apparently Kittaneh and Kosaki have a similar approach in "Inequalities for the Schatten p-norm. V." Publ. Res. Inst. Math. Sci. 23 (1987), no. 2, 433--443 (MR link). I haven't read any of this article.

Perhaps I should add the following for a more general audience. A continuous function $f:[0,\infty)\to [0,\infty)$ is operator monotone if whenever $x$ and $y$ are positive operators such that $y-x$ is positive, it follows that $f(y)-f(x)$ is positive. The functions $t\mapsto t^\alpha$ are operator monotone for $0<\alpha\leq1$ (but not for $\alpha>1$).

Must we close weakly to apply the spectral theorem?

2010-05-12T21:41:32Z

Let $H$ be an infinite dimensional separable complex Hilbert space. All C*-subalgebras of $B(H)$ are assumed to be non-degenerate.

The spectral projections of a self-adjoint element $T$ of $B(H)$ lie in the weakly closed algebra generated by $T$. In the early 1970s Pedersen proved that if a C*-subalgebra $A$ of $B(H)$ contains all of the spectral projections of each of its self-adjoint elements, then $A$ is weakly closed.¹ In his words, this "characterizes von Neumann algebras (on separable Hilbert spaces) as the only C*-algebras in which the spectral theorem can be used in its full force."²

On the other hand, suppose we start with an arbitrary C*-subalgebra $A$ of $B(H)$. Must we go all the way to the weak closure to use the spectral theorem for self-adjoint elements of $A$? If not, does iterating the process of taking the smallest C*-algebra in which we can apply the spectral theorem lead to the weak closure in a finite number of steps? Pedersen asked this question over 30 years ago as a way to end a chapter on concrete C*-algebras,³ and I don't know if it has been answered. If the answer isn't known, I'd accept an answer giving a more recent reference that discusses this. For the precise question, I'll just quote Pedersen:

For any C*-subalgebra $A$ of $B(H)$ define $a(A)$ as the smallest C*-subalgebra of $B(H)$ containing all spectral projections of each self-adjoint element in $A$. It is easy to verify that $A\subset a(A)\subset A''$. If $H$ is separable is then $A''=a(A)$? This failing, is $A''=a(a...a(A)...)$ (finitely many steps)? Note that by 2.8.8 a transfinite (but countable) application of the operation $a$ will produce $A''$.

¹ G. Pedersen, C*-algebras and their automorphism groups, Corollary 2.8.8, p. 38.

² Ibid., p. 39.

³ Ibid.

Answer by Jonas Meyer for What out-of-print books would you like to see re-printed?

2010-03-15T19:33:13Z

Dixmier, C $^*$ -algebras or Les C $^*$ -algèbres et leurs représentations
Dixmier, von Neumann algebras or Les algèbres d'opérateurs dans l'espace hilbertien: algèbres de von Neumann
Pedersen, C $^*$ -algebras and their automorphism groups

What is the "Krein-Milman theorem for cones"?

2010-05-06T10:41:01Z

Update: The question is completely answered. I had overlooked a reduction to the self-adjoint case, and the latter can be proved using a Hahn-Banach separation theorem. Thanks to Matthew Daws for first pointing this out to me. Thanks also to Willie Wong for pointing out that I should have asked the author; I later did so, and he confirmed that a Hahn-Banach separation theorem was intended (and generously provided another argument). I want to point out, because this post may highlight a minor error in the book, that it is a great book.

When I hear "Krein-Milman", I think of the result at the link, a standard result in functional analysis. But recently I came across an invocation of "the Krein-Milman theorem (for cones)" which looks more like a Hahn-Banach type result, and I am having trouble tracking down what precisely the author is referring to (and why the result is true).

To make my question precise, I'll have to give the basic setup. Let $V$ be a complex vector space with a conjugate linear involution $*$ , a norm that makes $*$ an isometry, and a norm-closed cone $C\subset\{v\in V:v=v^*\}$ , meaning $C$ is closed under addition and nonnegative real scalar multiplication, and satisfies $C\cap(-C)=\{0\}$ . (Such a $V$ is called an ordered $*$ -vector space.)

To paraphrase part of Paulsen's exposition of the Choi-Effros abstract characterization of operator systems¹:

If $x$ is in $V\setminus C$, then by the Krein-Milman theorem (for cones) there exists a linear functional $s:V\to\mathbb{C}$ with $s(C)\subseteq[0,\infty)$ and $s(x)<0$.

Because $C$ is a closed convex subset of a normed (and thus locally convex) space, my first inclination was to adapt some version of Hahn-Banach. However, I can't see how to do this while keeping complex linearity and the conclusion of the above claim. (There would be no problem in the real case, e.g. by Theorem 3.4 of Rudin's Functional Analysis, 2nd edition, page 59.) I've looked across the internet and my library's bookshelves to try to find what is meant, with no luck so far.

Question 1: Does anyone have a reference for (or statement and explanation of) the "Krein-Milman theorem for cones"?

Question 2: Am I missing a straightforward argument proving the above claim, for example by an application of Hahn-Banach?

I believe I have faithfully presented enough to cover everything relevant to my question, but the spaces in Paulsen's proof actually have a lot more structure, which you can find at the link above if you think it will help. Hopefully I didn't lose something essential in an effort to not get bogged down--this is clearly a risk because I don't know what theorem is being cited.

¹ An operator system is a self-adjoint unital subspace of the algebra of bounded operators on a Hilbert space (or an abstract space that is completely order isomorphic to one of these).

Answer by Jonas Meyer for Are the norms of graphs dense in any interval?

2010-05-09T06:06:37Z

I found a reference that seems to answer your question:

Shearer, James B. On the distribution of the maximum eigenvalue of graphs, 1989. The theorem in this paper is that the set of largest eigenvalues of adjacency matrices of graphs is dense in the interval $\left[\sqrt{2+\sqrt{5}},\infty\right)$. Here's an online version.

Here's a related paper:

Hoffman, Alan J. On limit points of spectral radii of non-negative symmetric integral matrices, 1972. In this paper limit points less than $\sqrt{2+\sqrt{5}}$ are described. In particular, they form an increasing sequence starting at 2 and converging to $\sqrt{2+\sqrt{5}}$. Here's an online version. The author also posed the problem that led to Shearer's paper.

Comment by Jonas Meyer

2013-02-05T04:38:03Z

A Google search for the exact quote "every proper subgroup is finite" helps.

Comment by Jonas Meyer

2013-01-04T06:39:25Z

This had been posted on math.stackexchange.com 6 days prior: math.stackexchange.com/questions/141499/…

Comment by Jonas Meyer

2012-12-05T14:50:15Z

In a late response to John Bentin's comment, if $X$ were replaced by the space of bounded real functions, then the norms would only be seminorms, and would not satisfy $\|x\|=0\implies x=0$. Limits would not be unique.

Comment by Jonas Meyer

2012-10-16T22:06:29Z

Does Murphy show in a previous chapter (perhaps an exercise) that in a unital C*-algebra, every element is a linear combination of 4 unitary elements? (If $x=x^*$ and $\|x\|\leq 1$, then $x\pm i\sqrt{1-x^2}$ are unitary and $x$ is their average$.) $A'$ is a unital C*-algebra.

Comment by Jonas Meyer

2012-02-17T05:26:20Z

A badly scanned copy: docs.google.com/…

Comment by Jonas Meyer

2012-02-13T20:21:08Z

Both Dixmier books recently appear to have been made available in print, with cheap prices on Amazon. I'm a little confused and wary, though, because I see no evidence of this printing on the publisher's website.

Comment by Jonas Meyer

2011-08-28T02:20:28Z

The question was already asked on math.stackexchange.com a couple of weeks ago: math.stackexchange.com/questions/57532/…

Comment by Jonas Meyer

2011-08-24T22:17:44Z

Sorry, this is the wrong website for this type of question. The FAQ describes what this website is for, and the section mathoverflow.net/faq#homework gives some links to other sites where this question might be a better fit.

Comment by Jonas Meyer

2011-08-19T03:18:46Z

$(A_k)$ is a dense sequence in the C*-algebra.

Comment by Jonas Meyer

2011-08-14T05:29:31Z

(2-by-2 matrices seems simpler than $B(\ell^2)$.) You want no additional structure beyond being commutative $\star$-subrings? That is, you don't require them to be algebras (closed under scalar multiplication), and you don't require them to be closed in any topology?

Comment by Jonas Meyer

2011-08-13T09:19:11Z

Sorry, but I think this question is off-topic for this site. The FAQ describes what this website is for, and the section mathoverflow.net/faq#homework gives some links to sites where this question would be a better fit. (No, not all smooth functions are real analytic.)

Comment by Jonas Meyer

2011-08-13T08:52:44Z

André: But it does seem natural to consider whether such everywhere-defined unbounded homomorphisms exist on a given Banach algebra.

Comment by Jonas Meyer

2011-07-30T08:35:51Z

There is now a math.SE question asking for more details on this one: math.stackexchange.com/questions/54578/…

Comment by Jonas Meyer

2011-07-20T19:53:42Z

This does work for arbitrary C* algebras.

Comment by Jonas Meyer

2011-05-10T18:13:20Z

José Figueroa-O'Farrill tells a very similar story in a comment at mathoverflow.net/questions/1083/…