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18 votes

Rigorous justification for this formal solution to $f(x+1)+f(x)=g(x)$

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}^\...
Carlo Beenakker's user avatar
17 votes
Accepted

Operator that commutes with projections

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 $P_n$.
Nik Weaver's user avatar
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16 votes
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Quantum functional analysis

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 ...
Nik Weaver's user avatar
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15 votes
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Conceptually, what does unitization do?

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 ...
Qiaochu Yuan's user avatar
14 votes

is every element in a C* algebra a product of normal elements?

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.
Michael Renardy's user avatar
13 votes
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Criterion for a Banach algebra to be finite dimensional

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 $A$ ...
Narutaka OZAWA's user avatar
13 votes
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Examples of amenable Banach algebras which have non-amenable subalgebra

Mateusz's answer mentions lots of good mathematics but I feel obliged to point out that the fundamental example which answers your original question in the negative is $M_2({\bf C})$. (Banach algebras ...
Yemon Choi's user avatar
  • 25.5k
12 votes
Accepted

Bicommutant theorem for commutative operator algebras

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}(\...
Mateusz Wasilewski's user avatar
12 votes

When are homomorphisms between Banach algebras contractions?

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 ...
Yemon Choi's user avatar
  • 25.5k
11 votes
Accepted

About the existence of characters on $B(X)$

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 ...
YCor's user avatar
  • 61k
11 votes

Rigorous justification for this formal solution to $f(x+1)+f(x)=g(x)$

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 ...
Dan Romik's user avatar
  • 2,490
11 votes

Examples of amenable Banach algebras which have non-amenable subalgebra

There are plenty of examples, even for $C^{\ast}$-algebras. By the results of Connes and Haagerup, for $C^{\ast}$-algebras amenability is equivalent to nuclearity, so I will work with nuclearity, ...
Mateusz Wasilewski's user avatar
11 votes

Two inequalities in $C^*$ algebras

As observed, the quadratic term may be equivalently removed from the inequality due to different homogeneity; then $x^*a^*ax+a^*x^*xa\leq a^*x^*ax+x^*a^*xa$ can be rewritten $[a,x]^*[a,x]\le0$, so ...
Pietro Majer's user avatar
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11 votes
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Are nearby subalgebras of matrix algebras conjugate?

I'll describe a 1-parameter family of nonisomorphic 4-dimensional subalgebras of $M_4(K)$. Consider, for $t\in K^*$, the matrices $$X=\begin{pmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 0 &...
YCor's user avatar
  • 61k
10 votes
Accepted

is every element in a C* algebra a product of normal elements?

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 ...
David Handelman's user avatar
10 votes

About the existence of characters on $B(X)$

Examples were known before the Argyros-Haydon space mentioned in Yves Cornulier's answer. For instance, if $J$ denotes the James space, then the image of the canonical map $J\to J^{**}$ has ...
Yemon Choi's user avatar
  • 25.5k
10 votes
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$K$-theory and surjective norm-decreasing $*$-homomorphisms between $C^*$-algebras

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 ...
Yemon Choi's user avatar
  • 25.5k
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" ...
Jochen Glueck's user avatar
9 votes

Conceptually, what does unitization do?

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 ...
Nik Weaver's user avatar
  • 42.4k
9 votes
Accepted

Removing the interior of spectrums

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 ...
Jochen Glueck's user avatar
8 votes
Accepted

What algebras are quotients of $\ell_1(\mathbf{N})$?

(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 ...
Yemon Choi's user avatar
  • 25.5k
8 votes

English translation of Hilbert's work

Yes, there is a translation, but not an official one and only of the "Erste Mitteilung", not the whole book: The pdf FREDHOLM, HILBERT, SCHMIDT Three Fundamental Papers on Integral Equations, ...
Dirk's user avatar
  • 12.4k
8 votes

Operator that commutes with projections

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=...
Iosif Pinelis's user avatar
8 votes
Accepted

A generalization of unsolvable equation $ab-ba=1$ in a Banach algebra

In the algebra of real $2\times 2$ matrices, take $$ a=\left[ \begin{array}{cc} 0 & 1\\ 0 & 0\end{array} \right],\ b=\left[ \begin{array}{cc} 0 & 0\\ 1 & 0\end{array}\right],\ ...
Richard Stanley's user avatar
8 votes
Accepted

A precise definition of contractible Banach algebras

$A$ is contractible if $H^1(A,X)=0$ for all Banach $A$-bimodules $X$ (here $H^1$ denotes continuous Hochschild cohomology for Banach algebras, as defined in the works of Johnson or Helemskii). It is ...
Yemon Choi's user avatar
  • 25.5k
8 votes
Accepted

Banach algebra $A$ without an approximate identity but $A^2=A$

For finite-dimensional algebra, to have an approximate unit is the same as having a unit. The non-unital algebra $A$ of matrices $\begin{pmatrix}0 & x\\ 0 & y\end{pmatrix}$ has no unit (...
YCor's user avatar
  • 61k
8 votes

$K$-theory and surjective norm-decreasing $*$-homomorphisms between $C^*$-algebras

There are even commutative counterexamples. Let $A = C[0,2]$ and let $A_0$ be the $*$-subalgebra of all polynomials in $x$. Then let $p: C[0,2] \to C[0,1]$ be the restriction map. (My first example ...
Nik Weaver's user avatar
  • 42.4k
8 votes

Connectedness of Invertible elements in a non- commutative C*- algebra

The group $K_1(A)$ is the invertibles in $\lim M_n(A)$ modulo the path component of the identity. (Here the embedding of $M_n(A)$ into $M_{n+1}(A)$ goes by adding an extra column and row of zeros.) So ...
Nik Weaver's user avatar
  • 42.4k
8 votes

If $A\hat\otimes B$ has identity then so are $A$ and $B$

The answer is NO, if you count $0$ as a unital Banach algebra. Otherwise, it's YES. Let $e \in A \hat\otimes B$ be a unit and take $b_0\in B$ and $g\in B^*$ with $g(b_0)=1$. Let $g\cdot b_0\in B^*$ be ...
Narutaka OZAWA's user avatar
8 votes

Are nearby subalgebras of matrix algebras conjugate?

Here's another example. It's quite distinct from my previous answer since in my previous answer the homomorphisms $f_t$ are non-injective and have images that are not isomorphic (and hence non-...
YCor's user avatar
  • 61k

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