This is proposition 13.2 in (Muscat 2024): given a unital Banach algebra, B(X), multiplication is a differentiable map.
The book proof is a “barebones” version, which I seek to unpack here.
1. What must be proved
The multiplication map takes two arguments, and it should be Fréchet Differentiable at the point (T, S). The definition of Fréchet Differentiable is given as Definition 12.1 in (Muscat 2024), using a single variable: for to be differentiable at a point requires there to exist a continuous and linear map such that:
where as .
As multiplication takes two arguments, the definition should be rewritten accordingly.
The map is Fréchet Differentiable at the point (T, S) if there exists a continuous, linear map such that:
where as .
Note that we have in place of . To understand this, consider the role of the term in the equation. The differentiable ( or , respectively) is the best approximation to compute the movement from a given point by the given change. The form is the more general way to write this: a map specific to the given point computes the approximation given the amount of change. In one dimension, the movement can be approximated by taking a small step (the given change) in the direction of a line (the gradient).
Norms are worth a comment:
- T, S, H and K, including their sums and products, are all members of B(X), so norms are in B(X).
- However, the arguments of the multiplication map treated as a pair uses the product norm, so .
The theme of the proof is to expand the multiplication of by and relate its parts to those in equation ():
So, comparing with () above:
- corresponds to
- corresponds to
- corresponds to
- corresponds to
The definition of Fréchet Differentiable then tells us to check that is continuous and linear, which is shown in part 2 below, and that shrinks quick enough, which is shown in part 3 below. Having proved these, we can conclude the multiplication operator is a differentiable map.
2. Map is continuous and linear
Above, we derived .
This is linear because:
it preserves addition:
it preserves scalar multiplication:
This is continuous because it is bounded:
Which means we have a bound independent of for (note use of the product norm).
Therefore, is bounded and so continuous.
3. Constraint on
The constraint on is that as . By substituting in , as derived above, and :
as .
Note line () uses the inequality (for positive and ) , and line () is based on product norm: .