Let $A$ be a unital commutative Banach algebra and let $\hat{a}\in C(\Phi_A)$ be the Gelfand transform of an element $a\in A$. The algebra $A$ has norm-controlled inverses, whenever there exists a function $h\colon [0,\infty)^2\to [0,\infty)$ so that $$\|a^{-1}\| \leqslant h(\|a\|, \|\widehat{a^{-1}}\|_\infty)$$ for any invertible element $a\in A$. Many naturally arising cBas have this property, however, $A=\ell_1(\mathbb Z)$ (with convolution) is a notable example of a Banach algebra that fails to have norm-controlled inverses (see Nikolski's paper). Hence my question:
I wonder if there is any relation between norm-controlled inverse and having uniformly open multiplication. It is known that $\ell_1(\mathbb Z)$ does not have uniformly open convolution, however, ${\rm BV}[0,1]$ does have norm-controlled inverses, yet the (pointwise) product is not uniformly open (see Kowalczyk–Turowska).
Suppose that $A$ is a unital semi-simple commutative Banach algebra that has uniformly open multiplication. Does it have norm-controlled inverses?
Some examples of (complex) Banach algebras that have uniformly open multiplication:
$C(X)$ for a compact, zero-dimensional Hausdorff space (Komisarski in the real case but the proof works also for the complex scalars) and $C[0,1]$ (Behrends, unpublished in the uniformly open case). I have a proof that extends this to all $C(X)$ for $X$ metric and having covering dimension 1. When $\dim X \geqslant 2$, $C(X)$ fails to have open multiplication.
$C^1[0,1]$ does have norm-controlled inverses, and it seems I can prove it has uniformly open multiplication too.
References:
A. Komisarski, A connection between multiplication in $C(X)$ and the dimension of $X$, Fund. Math. 189 (2) (2006), 149–154.
S. Kowalczyk, M. Turowska, Multiplication in the space of functions of bounded variation. J. Math. Anal. Appl., 472 (2019), 696–704.
- N. Nikolski, In search of the invisible spectrum. Ann. Inst. Fourier (Grenoble), 49 (1999), 1925–1998.
