Let us say that a Hamel basis $H$ in an algebra $A$ is closed under multiplication, if $ab\in H$ whenever $a,b\in H$. It is an easy observation that if $A$ has such a basis then there it also has a character (a linear-multiplicative functional; see Amer. Math. Monthly 124 (2017), no. 7, 651–653.)

All characters on Banach algebras are automatically continuous. When $A=c_0$, then they are just point evaluations.

Does $c_0$ (with the pointwise product) have a Hamel basis closed under multiplication?

One could also ask:

Is there an infinite-dimensional (complex) abelian Banach algebra that has a Hamel basis closed under multiplication?

  • $\begingroup$ While this doesn't solve your questions, perhaps one could get some mileage from the fact that each $f\in C(X)$ can be written as a linear combination of four unitaries? Admittedly, I don't know if one can extract a subsemigroup of the unitary group of C(X) which forms a Hamel basis, but perhaps one can do some (transfinite) induction to build such a subsemigroup... $\endgroup$ – Yemon Choi Nov 27 '17 at 21:14
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    $\begingroup$ recreational-mathematics? Really? $\endgroup$ – Gerry Myerson Nov 27 '17 at 21:31
  • $\begingroup$ $c_0$ certainly doesn't. The general case remains unclear (to me) $\endgroup$ – fedja Nov 27 '17 at 23:24
  • $\begingroup$ Maybe it’s recreational in the sense that one is trying to re-create the characters from the basis? $\endgroup$ – Anthony Quas Nov 28 '17 at 1:48
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    $\begingroup$ @GerryMyerson In a sense it is: I know much harder problems (both solved and unsolved) that you most likely would not hesitate to put under that tag. This particular one is not too challenging, but the answer is somewhat disappointing (see below). $\endgroup$ – fedja Nov 28 '17 at 3:29

Assume that a Banach algebra $B$ has such a basis $H$. Take any $h\in H$. Consider the element $x=ah+a^2h^2+a^3h^3+\dots$ where $a>0$ is chosen small enough to make the series converge in $B$. Assume $x=\sum_{j=1}^n c_jh_j$ for some $h_j\in H$. Write $$ x=ah+a^2h^2+\dots+a^mh^m+a^mh^mx=ah+a^2h^2+\dots+a^mh^m+\sum_j c_ja^m(h^mh_j) $$ If all the basis elements $h^k\in H$, $k=1,\dots,m$ are distinct, then the elements $h^mh_j\in H$ in the last sum can cancel only $n$ of them and we get a representation for $x$ of length $\ge m-n>n$ if $m>2n$. Thus some two of the powers must coincide and, thereby, $h^k$, $k\ge 1$ span only a finite dimensional space. Since it is true for every $h\in H$, we conclude that the powers of every element of $B$ span only a finite-dimensional space. This excludes any "decent" infinite-dimensional Banach algebra ($c_0$ in particular).

However something survives. The most degenerate example is the following. Take some Banach algebra, say $B=c_0$, an idempotent element $y\in B$ (of norm $1$, say, though it doesn't matter), say $y=(1,0,0,0,\dots)$, some linear multiplicative functional $\psi$ on $B$ such that $\psi(y)=1$, which would be the evaluation of the first coordinate in that particular example, and redefine the product in $B$ to $$ ab=\psi(a)\psi(b)y $$
(this is not a very exciting product, of course, but it is a product nevertheless). Now just take any maximal linearly independent set containing $y$ in $B\setminus\operatorname{ker}\psi$ and normalize its elements $h$ by replacing them by $\frac 1{\psi(h)}h$. You have your basis.

  • $\begingroup$ The conclusion that the powers of any element span a f.-d. space works only for Abelian Banach algebras, right? $\endgroup$ – Ilya Bogdanov Nov 28 '17 at 7:24
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    $\begingroup$ Thank you, fedja. I am surprised by the simplicity of your counter-example. Very nice! $\endgroup$ – Tomek Kania Nov 28 '17 at 12:45
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    $\begingroup$ @IlyaBogdanov As written, yes, but if you apply the geometric progression trick to a sum of basis elements instead of a single one and use the size estimates in addition to the mere length count (note that for positive $a$ there cannot be any cancellation until you reach the tail sum and that one can be made to have as small sum of coefficients as you want), you will find out that there may be only finitely many different products for any finite set of basis elements even in the non-commutative case. So the conclusion still holds there. $\endgroup$ – fedja Nov 28 '17 at 13:51
  • $\begingroup$ @IlyaBogdanov If you want to play with this question a bit more, the story gets more interesting if you relax the condition that every product of basis elements should be a basis element to the condition that every product of basis elements should be a multiple of a basis element ($0$ multiples are allowed). This gives one more freedom but, probably, is still restrictive enough to exclude $c_0$ (I haven't thought of that, so no warranty here). $\endgroup$ – fedja Nov 28 '17 at 14:21

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