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I am currently reading through Chapter 15 of "Complete Normed Algebras" by F.F. Bonsall and J. Duncan. I have come across the following proposition on page 76:

Proposition 5 Let $ A $ be a complex Banach algebra with unit, $ M $ a unit linked Banach left $ A $-module, $ X $ a complex Banach space, and $ h : A \times M \to X $ a continuous bilinear mapping. The following conditions are equivalent.

(i) $ h(a, m) = h(1, am) $ ($ a \in A $, $ m \in M $).

(ii) There exists $ \kappa > 0 $ with $ \| h(a, m) \| \leq \kappa \| am \| $ ( $ a \in A $, $ m \in M $).

The proof given in the book is as follows:

The direction $ (i) \Rightarrow (ii) $ is easy. The direction $ (ii) \Rightarrow (i) $ is given in the book as follows:

Proof: Let condition $ (ii) $ hold and let $ f \in X^* $. Given $ a \in A $, $ m \in M $, let $ F $ be defined on $ \mathbb{C} $ by

$$ F(z) = (f \circ h)(\exp(-za), (\exp(za))m) $$

Then $ F $ is an entire function and, for $ z \in \mathbb{C} $,

$$ |F(z)| \leq \| f \| \| h(\exp(-za), (\exp(za))m) \| \leq \kappa \| f \| \| m \| $$

By Liouville's theorem, $ F $ is constant, and so the coefficient of $ z $ in the power series expansion of $ F $ is zero, i.e.

$$ f(h(1, am)) - f(h(a, m)) = 0 $$

But $ f \in X^* $ is arbitrary. $ \blacksquare $

My questions are:

  1. What is the definition of $ \exp (za) $? I cannot find this in the text.

  2. Liouville's theorem in complex analysis states that if $ F $ is a bounded entire function then $ F $ is constant. I am not sure why the given $ F $ is an entire function though (perhaps because I am not sure what $ \exp (za) $ means).

  3. By Liouville's theorem $ F(z) $ is constant. If $ F(z) $ has power series $ \sum_{n=0}^{\infty} a_nz^n $ then $ F $ being constant implies that $ a_1, a_2, ... = 0 $, and so $ F(z) = F(0) = f(\exp(0), \exp(0)m) \overset{?} = f(1, m)$? I am not sure how the authors conclude that $ f(h(1, am)) - f(h(a, m)) = 0 $.

Any help would be greatly appreciated. Thanks!

LMW
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1 Answers1

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In any unital Banach algebra $A$, we define the exponential function just as we would in any analysis class: $$\exp(a)=\sum_{n=0}^\infty \frac{a^n}{n!}\qquad(a\in A).$$ Showing that $F$ is analytic can be a bit cumbersome, but one way to go about it is to define what it means for a function $\mathbb C\to X$ to be analytic at a point (it has a continuous derivative in a nieghborhood of the point) for any Banach space $X$, then showing the composition of a bounded linear (resp. bilinear) map with an analytic map is analytic.

Next, at the end we don't have $f(h(1, am))$ and $f(h(a, m))$ are values of $f$, rather $$f(h(1, am)) - f(h(a, m))=F'(0)=0 $$

Aweygan
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  • Thank you again for the help! I have one question though. How is $ 0 = F'(1) = f(h(1, am)) - f(h(a, m)) $ obtained? Is this derived somehow by a version of the Mean Value theorem? If so, is it obvious, because I'm having difficult seeing it. Thanks again! – LMW Jun 11 '18 at 18:07
  • Sorry, I made an error, it should be $F'(0)$, not $F'(1)$. Well the function $F$ is constant, so $0=F'(0)$, no problem there. What we have is that for $f(z)=h(k_1(z),k_2(z)))$, where $h:X_1\times X_2\to\mathbb C$ is a continuous, bilinear, and $k_i:\mathbb C\to X_i$ is analytic, then $f'(z)=h(k_1'(z),k_2(z))+h(k_1(z),k_2'(z))$ (similar to the product rule from calculus). – Aweygan Jun 11 '18 at 18:18
  • Now the derivative of $z\mapsto\exp(za)$ is $z\mapsto a\exp(za)$ (similarly for $z\mapsto \exp(za)m$), so just plug everything in. – Aweygan Jun 11 '18 at 18:20
  • Ok, I think I'm almost there. So $ F'(z) = f(h(-a\exp(-za), \exp(za)m)) + f(h(\exp(-za), a \exp(za)m) $, and plugging in $ z = 0 $ gives $ 0 = F'(0) = f(h(-a, m)) + f(h(1, am)) $ and so $ 0 = f(h(1, am)) - f(h(a,m)) $? It has been awhile since I have looked at Calculus. Sorry if this is trivial - I just want to be absolutely sure! – LMW Jun 11 '18 at 19:06
  • Yep you got it. – Aweygan Jun 11 '18 at 19:06