36

Ozawa, Schechtman, and I finally wrote up what we know on this question. The estimate is that for every $\epsilon > 0$ there is a constant $C_\epsilon$ so that for every $n$, $\lambda(n)\le C_\epsilon n^{\epsilon}$. The paper can be downloaded from the arXiv.

18

A direct and precise way to arrive at your answer is to appeal to the theory of Fourier transforms of distributions. If I define the Fourier transform as
$$F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{ikx}\,dx,$$
then the equation $f(x+1)+f(x)=g(x)$ becomes upon Fourier transformation
$$e^{-ik}F(k)+F(k)=G(K)\Rightarrow F(k)=\frac{G(k)}{1+e^{-ik}}.$$...

answered Jun 24 '19 at 14:57

Carlo Beenakker

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17

Maybe a quicker way to see this is, if $TP_m = P_mT$ for all $m \neq n$ then $TP = PT$ where $P =\sum_{m\neq n} P_m = I - P_n$. Since $T$ commutes with $I$, it must therefore commute with $P_n$.

15

Here is perhaps the simplest example. Let $A$ be the C*-algebra of all sequences of $2 \times 2$ matrices converging to a scalar multiple of diag(1,0). Let $p$ be the constant sequence diag(1,0), and $q$ a sequence of rank 1 projections converging to diag(1,0) but never exactly equal. Then $p$ and $q$ have no upper bound at all. This example can be ...

15

Okay, I'll take this one. First let me say that the English term is "completely bounded" (or "complete isometry", etc.).
About the term "quantum". The general principle is that analyzing some aspect of a physical system typically involves very different kinds of mathematical structures, depending on whether the system is classical or quantum. Thus if you ...

14

In conversations and discussions, David Mumford mentioned several times that Gelfand's result about commutative Banach algebras had great influence for the development of schemes. He also mentioned that local coordinate systems on manifolds had influence.

13

You might take a look at Section 4 ("Toward the concept of spectrum") in the article "A mad day's work: from Grothendieck to Connes and Kontsevich: The evolution of concepts of space and symmetry" by Pierre Cartier:
http://www.ams.org/journals/bull/2001-38-04/S0273-0979-01-00913-2/
He mentions that Gelfand's work provides the motivation for the term "...

13

The shift in $\ell^2$ cannot be a product of normal operators, since its Fredholm index is nonzero, and a normal operator cannot have nonzero Fredholm index.

13

Unitization and metric completion are both left adjoint functors, as are may other "-tion" operations in mathematics, such as localization or abelianization. Specifically, there is a forgetful functor from unital algebras to nonunital algebras (including norms is not particularly important here), and unitization is its left adjoint. Conceptually this means ...

answered Nov 17 '16 at 23:54

Qiaochu Yuan

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12

Theorem Let $A$ be a unital ring and $I_1,\dots,I_n \subset A$ be 2-sided commutative ideals such that $A=I_1+\dots + I_n$. Then, $A$ is commutative.
Proof: If $A=I_1+\dots+I_n$, then $1 = x_1+\dots+x_n$ for $x_i \in I_i$. But then,
$$1 = (x_1+\dots+x_n)^{n+1} \in I_1^2 +\dots+ I_n^2$$ and we conclude that $A=I_1^2 + \dots + I_n^2$.
Now, if $I \subset A$ ...

11

The answer is yes.
Recall that if $S$ is a Suslin tree, then PFA($S$) denotes the forcing axiom for the class of all proper posets which preserve $S$. PFA($S$) is consistent with ZFC relative to a supercompact cardinal, like PFA, and shares many of the consequences of PFA, but a model of PFA($S$) naturally does not satisfy MA (since $S$ is Suslin).
Claim: ...

11

I think it's true by Dixon's theorem (JLMS 1974) which says (on the third page) that any Banach algebra consisting only of algebraic elements is nilpotent-by-finite. Thanks to this, we may assume $A$ is nilpotent. We moreover assume $A$ is infinite-dimensional and will construct an infinite-dimensional commutative subalgebra. Take the largest $n$ such that ...

11

The answer is negative. Consider the Hardy space $H^{\infty}(\mathbb{T})$ consisting of holomorphic functions admitting bounded extension to the unit circle and view it is a subalgebra of $B(L^{2}(\mathbb{T})$. Let us compute the commutant. Denote the operator of multiplication by $z$ by $M_z$. If $T$ belongs to the commutant of $H^{\infty}(\mathbb{T})$, it ...

11

Here is another approach, which like yours is not entirely rigorous, but I think gives a bit more insight into the situation.
If you are willing to assume that $f$ is not just smooth but also analytic, then by writing the $f(x+1)$ term as a Taylor expansion you can write the equation in the form
$$
(e^D+I) f = g
$$
where $D$ is the differentiation operator. ...

10

You can find examples of AF algebras without the lattice property in Section 2 of AF Algebras with a Lattice of Projections by Aldo J. Lazar here.

10

No. Some algebras satisfy the identity $xy=-yx$ but are not commutative. Such an algebra clearly satisfies your condition. Obviously, it never has an approximate identity.
To construct one, let $V,W$ be two Banach spaces and let $f: V \times V \to W$ be a continuous symplectic bilinear form. Then define a multiplication on $V + W$ where anything in $W$ ...

10

Always, according to Naĭmark, Normed Algebras, VII p. 380.

10

I guess that you mean that $B(H)$ has no character (=continuous unital algebra homomorphism into $\mathbf{C}$) if $H$ has dimension $\neq 1$ (idem for $M_n(\mathbf{C})$ for $n\neq 1$), and thus that your question assumes $\dim(X)\ge 2$ (and hence $=\infty$).
Argyros and Haydon (Acta Math, 2011: arXiv, Project Euclid unrestricted access) constructed a Banach ...

10

About the only positive result that comes to mind is the fact that homomorphisms cannot increase the spectral radius, so that if the range space is a uniform algebra then homomorphisms are necessarily contractive.
In my view and in my experience, at the level of generality considered by this question, the assertion/hope that "a positive answer to the above ...

9

Since the question was asked wrt C$*$-algebras, I guess there is room for a general remark. Suppose that $xy = 1$ where $x$ is a product of normal elements, say
$v_1v_2\cdots v_n$. Then $v_1$ is normal and has a right inverse;
therefore it is (two-sided) invertible. Thus $v_2\dots v_n y $ is
invertible, so $v_2$ has a right inverse, and thus is invertible. ...

9

I don't remember where I read this, but Gert Pedersen once said something to the effect that "When I was young, the first thing we did with any C*-algebra was to adjoin a unit, but nowadays the first thing we do is remove the unit by tensoring with the compacts." The point is that $\mathcal{A}\otimes \mathcal{K}$ is the "stabilization" of $\mathcal{A}$: it ...

9

The answer is no, in general. Here is a counterexample:
Let $A$ be the algebra of bounded linear operators on $\ell^2(\mathbb{N})$, and let $a \in A$ be the left shift on $\ell^2(\mathbb{N})$. Then the spectrum of $a$ is the closed unit disk $\overline{D}$, and the point spectrum of the operator $a$ is the open unit disk $D$.
Now we note that the notion "...

9

Yes to both.$\newcommand{\Cst}{{\rm C}^*}$
The standard example for the first is: take a discrete group $G$ and let $A$ be its full $\Cst$-algebra, $B$ its reduced $\Cst$-algebra. There is a canonical homomorphism $q:A\to B$ which is injective when restricted to $\ell^1(G)$; but $q$ is injective if and only if $G$ is amenable. So any non-amenable discrete ...

8

Posting this purely so that Ozawa's interesting question does not stay marked as unanswered, and hence lead to unnecessary effort on the part of someone reading. (Note: could someone reading this please flag the answer for moderator attention, to make it CW.)
It turns out that every uniformly bounded representation of a discrete, countable amenable group ...

8

I think Hahn-Banach can be eliminated from the usual proof, but
being a non-expert in set theory, I cannot guarantee that the proof
is completely independent of the axiom of choice.
Here is a sketch of a basic calculus proof. A function $U\to B$
from a region $U\subset C$ to a Banach space $B$ is called analytic if
every point has a neighborhood where it ...

answered Dec 19 '12 at 3:26

Alexandre Eremenko

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8

I do not believe that the Hahn-Banach theorem is necessary. At some point I had planned on writing up a blog post verifying this but I lost the motivation...
The idea is that you can prove Liouville's theorem in the Banach space setting directly without using Hahn-Banach to reduce to the case of $\mathbb{C}$ (I asked whether this was possible in this math....

answered Dec 19 '12 at 3:15

Qiaochu Yuan

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8

Here a simple example :
Let X be the cartesian product of $L^{\infty}$ and $L^{1}$ on the
interval $[0,1]$, let $P_{t}$ the canonical projection on the subspace
of functions with support $[0,t]$ and choose $Q(f_{1},f_{2}) =
(0,f_{1})$ .

8

If $G$ is a compact group, and we choose the normalized Haar measure $\mu$ on $G$, then $L^2(G)$ is a Banach algebra with convolution. Indeed, if $f$ and $h$ are in $L^2(G)$, then we use Hölder's inequality at the second step, together with the fact that the constant function 1 has $L^2$-norm equal to 1, to compute
\begin{align*} |f\ast h(x)|\leq \int_G|f(y)...

8

(What follows is largely the result of digging around online, based on knowing a few more magic words than the OP.)
Answer to the first question (I think).
Let $V$ be a separable Banach space. The standard proof that $V$ is isometrically isomorphic to a linear quotient of $\ell_1$ works by choosing a countable set $X$ that is a dense subset of the unit ...

8

The answer is no. Indeed, without loss of generality (wlog), $n=0$, so that $TP_m=P_mT$ for all $m=1,2,\dots$ (assuming here that $\mathbb{N}_0=\{0,1,\dots\}$). Since $P_0=I-\sum_{m\ge1}P_m$ and $T I=IT$, it follows that $TP_0=P_0T$. (Here, of course, $I$ is the identity operator and the operator $S:=\sum_{m\ge1}P_m$ is defined by the condition that $Sx:=\...

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