A New Equivalent Definition of Banach Algebras: Insights from a Categorical Perspective on Norm Submultiplicativity

A Categorical Perspective on Norm Submultiplicativity in Banach Algebras via Endomorphism EmbeddingIntroductionTraditional PerspectiveA Categorical Perspective and an Equivalent Definition of Banach AlgebraBanach Module: Some ExamplesConclusion

A Categorical Perspective on Norm Submultiplicativity in Banach Algebras via Endomorphism Embedding

Introduction

An unital Banach Algebra $\mathcal{B}$$\mathcal{B}$ is a Banach space with $\mathbb{F}$$\mathbb{F}$-algebra structure satisfying $\parallel x\cdot y\parallel \le \parallel x\parallel \parallel y\parallel$$\|x\cdot y\|\leq\|x\|\|y\|$.

In this blog, we only consider $\mathbb{F}=\mathbb{R}$$\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$$\mathbb{C}$. This restriction is natural since if $\mathrm{Char}(\mathbb{F})=0$$\mathrm{Char}(\mathbb{F})=0$ and $\mathbb{F}$$\mathbb{F}$ is complete and Archimedean, then $\mathbb{F}\cong \mathbb{R}$$\mathbb{F}\cong\mathbb{R}$ or $\mathbb{C}$$\mathbb{C}$.

Traditional Perspective

Some textbooks explain that we require $\parallel xy\parallel \le \parallel x\parallel \parallel y\parallel$$\|xy\|\leq\|x\|\|y\|$ to ensure the continuity of either:

$$\begin{array}{}\text{(1)}& \cdot :\mathcal{B}\times \mathcal{B}\to \mathcal{B}\end{array}$$

or

$$\begin{array}{}\text{(2)}& \mu :\mathcal{B}\otimes \mathcal{B}\to \mathcal{B},a\otimes b\mapsto a\cdot b\end{array}$$

making $\mathcal{B}$$\mathcal{B}$ a topological ring. However, this condition is stronger than necessary for mere continuity.

A Categorical Perspective and an Equivalent Definition of Banach Algebra

Let us present an alternative viewpoint. Recall that for any ring $R$$R$, we have a canonical embedding:

$$\begin{array}{}\text{(3)}& \iota :R\to {\mathrm{End}}_{\mathrm{Ab}}(R),a\mapsto {\iota }_a,{\iota }_a(x)=a\cdot x\end{array}$$

For a Banach algebra $\mathcal{B}$$\mathcal{B}$, this becomes:

$$\begin{array}{}\text{(4)}& \iota :\mathcal{B}\to {\mathrm{End}}_{\mathrm{Ban}}(\mathcal{B})\end{array}$$

A fundamental property of bounded operators is that for $T\in \mathrm{Ban}(X,Y)$$T\in\mathrm{Ban}(X,Y)$ and $S\in \mathrm{Ban}(Y,Z)$$S\in\mathrm{Ban}(Y,Z)$, we have:

$$\begin{array}{}\text{(5)}& \parallel S\circ T\parallel \le \parallel S\parallel \parallel T\parallel \end{array}$$

Therefore:

$$\begin{array}{}\text{(6)}& \parallel {\iota }_x\circ {\iota }_y\parallel \le \parallel {\iota }_x\parallel \parallel {\iota }_y\parallel \end{array}$$

Proposition. $\parallel {\iota }_x\parallel =\parallel x\parallel$$\|\iota_x\|=\|x\|$​, hence every Banach Algebra is a sub Banach Algebra of some ${\mathrm{End}}_{\mathrm{Ban}}(X)$$\mathrm{End_{Ban}}(X)$​.

Proof.

Lemma. $\parallel {\iota }_1\parallel =1=\parallel 1\parallel$$\|\iota_1\|=1=\|1\|$.

Proof. ${\iota }_1=\mathrm{id}$$\iota_1=\mathrm{id}$, and $\parallel \mathrm{id}\parallel =1$$\|\mathrm{id}\|=1$. For $1\in \mathbb{C}$$1\in\mathbb C$, $\parallel x\parallel =\parallel 1\cdot x\parallel =\parallel 1\parallel \parallel x\parallel ⟹\parallel 1\parallel =1◻$$\|x\|=\|1\cdot x\|=\|1\|\|x\|\implies\|1\|=1\quad\square$

$$\begin{array}{}\text{(7)}& \begin{array}{c}\parallel {\iota }_x\parallel =\underset{\parallel y\parallel =1}{sup}\parallel x\cdot y\parallel \ge \parallel x\cdot 1\parallel =\parallel x\parallel ⟹\parallel {\iota }_x\parallel \ge \parallel x\parallel \end{array}\end{array}$$

By $\parallel x\cdot y\parallel \le \parallel x\parallel \parallel y\parallel$$\|x\cdot y\|\leq\|x\|\|y\|$, we have:

$$\begin{array}{}\text{(8)}& \parallel {\iota }_x(y)\parallel =\parallel x\cdot y\parallel \le \parallel x\parallel \parallel y\parallel ⟹\parallel {\iota }_x\parallel \le \parallel x\parallel ◻\end{array}$$

Thus, the inequality $\parallel x\cdot y\parallel \le \parallel x\parallel \parallel y\parallel$$\|x\cdot y\|\leq\|x\|\|y\|$ simply states that:

$†.$$\dagger.$ $\iota$$\iota$ is norm-preserving map

$†.$$\dagger.$ $(\mathcal{B},\iota )$$(\mathcal{B},\iota)$ is a sub-Banach algebra of ${\mathrm{End}}_{\mathrm{Ban}}(\mathcal{B})$$\mathrm{End}_{\mathrm{Ban}}(\mathcal{B})$​​.

Proposition. In fact, we have $\parallel {\iota }_x\parallel =\parallel x\parallel ⟺\parallel x\cdot y\parallel \le \parallel x\parallel \parallel y\parallel$$\|\iota_x\|=\|x\|\iff \|x\cdot y\|\le\|x\|\|y\|$​.

Proof. We already see that $\parallel x\cdot y\parallel \le \parallel x\parallel \parallel y\parallel ⟹\parallel {\iota }_x\parallel =\parallel x\parallel$$\|x\cdot y\|\leq\|x\|\|y\|\implies \|\iota_x\|=\|x\|$. Let us prove the converse.

Assume that $\parallel {\iota }_x\parallel =\parallel x\parallel$$\|\iota_x\|=\|x\|$, then

$$\begin{array}{}\text{(9)}& \parallel {\iota }_x\parallel =\underset{y\in X}{sup}{\displaystyle \frac{\parallel x\cdot y\parallel }{\parallel y\parallel }}=\parallel x\parallel ⟹\parallel x\cdot y\parallel \le \parallel x\parallel \parallel y\parallel \end{array}$$

Or just consider that

$$\begin{array}{}\text{(10)}& \parallel x\cdot y\parallel =\parallel {\iota }_{x\cdot y}\parallel =\parallel {\iota }_x\circ {\iota }_y\parallel \le \parallel {\iota }_x\parallel \parallel {\iota }_y\parallel =\parallel x\parallel \parallel y\parallel ◻\end{array}$$

Hence we get an equivalent definition of Banach Algebra.

Readers could compare it with how we get the equivalent definition of Boolean Ring. Click here

Remark.

Readers may draw a parallel with the fact that every ${\mathrm{End}}_{\mathrm{Ab}}(M)$$\mathrm{End_{Ab}}(M)$ is a ring and every ring $R$$R$ can be viewed as a subring of the endomorphism ring ${\mathrm{End}}_{\mathrm{Ab}}(R)$$\mathrm{End_{Ab}}(R)$. That is, study ring is just study ${\mathrm{End}}_{\mathrm{Ab}}(M)$$\mathrm{End_{Ab}}(M)$. This observation motivates the study of $R$$R$ -modules in the context of ring theory.

Analogously, viewing every ${\mathrm{End}}_{\mathrm{Ban}}(\mathcal{B})$$\mathrm{End_{Ban}}(\mathcal{B})$ is a Banach Algebra and every Banach algebra $\mathcal{B}$$\mathcal{B}$ as a sub-Banach algebra of ${\mathrm{End}}_{\mathrm{Ban}}(\mathcal{B})$$\mathrm{End_{Ban}}(\mathcal{B})$ inspires us to investigate $\mathcal{B}$$\mathcal{B}$-modules in the context of Banach algebras.

For an $R$$R$ module, the underline space is an abelian group $M$$M$ with $f:R\to {\mathrm{End}}_{\mathrm{Ab}}(M)$$f:R\to\mathrm{End}_{\mathrm{Ab}}(M)$. For a Banach Module, we should consider a Banach Algebra homomorphism $f:\mathcal{B}\to {\mathrm{End}}_{\mathrm{Ban}}(X)$$f:\mathcal B\to\mathrm{End}_{\mathrm{Ban}}(X)$. In other word, the module action is continuous and satisfies that $\parallel b\cdot x\parallel \le \parallel b\parallel \parallel x\parallel$$\|b\cdot x\|\le\|b\|\|x\|$​.

Remark. In general, ${\mathrm{End}}_{R-\mathrm{Mod}}(M)$$\mathrm{End}_{R-\mathrm{Mod}}(M)$ is a $R$$R$-algebra since we have $\iota :R\to {\mathrm{End}}_{R-\mathrm{Mod}}(M)$$\iota:R\to \mathrm{End}_{R-\mathrm{Mod}}(M)$ and $fr=rf$$fr=rf$.

Banach Module: Some Examples

1.

Let $X$$X$ be a Banach space and consider $\mathrm{Ban}(X)$$\mathrm{Ban}(X)$. We claim that $\mathrm{Ban}(X)$$\mathrm{Ban}(X)$ is a Banach algebra.

Clearly, $\mathrm{Ban}(X)$$\mathrm{Ban}(X)$ is a Banach space with the operator norm:

$$\begin{array}{}\text{(11)}& \parallel T\parallel :=\underset{x\in X,x\ne 0}{sup}{\displaystyle \frac{\parallel Tx\parallel }{\parallel x\parallel }}\end{array}$$

This immediately gives us:

$$\begin{array}{}\text{(12)}& \parallel Tx\parallel \le \parallel T\parallel \parallel x\parallel \end{array}$$

From this, we can derive:

$$\begin{array}{}\text{(13)}& \parallel T\circ Sx\parallel =\parallel T(Sx)\parallel \le \parallel T\parallel \parallel Sx\parallel \le \parallel T\parallel \parallel S\parallel \parallel x\parallel ⟹\parallel T\circ S\parallel \le \parallel T\parallel \parallel S\parallel \end{array}$$

Thus, $\mathrm{Ban}(X)$$\mathrm{Ban}(X)$ is a Banach algebra, and it naturally makes $X$$X$ a module over this Banach algebra. The scalar multiplication:

$$\begin{array}{}\text{(14)}& \cdot :\mathrm{Ban}(X)\times X\to X,(T,x)\mapsto Tx\end{array}$$

is continuous because $\parallel Tx\parallel \le \parallel T\parallel \parallel x\parallel$$\|Tx\|\leq\|T\|\|x\|$​.

2

Let $A$$A$ be a unital Banach Algebra and $X$$X$ be a Banach space, then define a module structure on $A{\otimes }_{\mathbb{F}}X$$A\otimes_{\mathbb F} X$ by

$$\begin{array}{}\text{(15)}& a\cdot (b\otimes x)=ab\otimes x\end{array}$$

Here the norm on tensor product is defined by:

$$\begin{array}{}\text{(16)}& \parallel u{\parallel }_{\pi }=inf\{\sum _{i=1}^n\parallel x_i\parallel \parallel y_i\parallel :u=\sum _{i=1}^n{x}_i\otimes y_i,x_i\in X,y_i\in Y\}\end{array}$$

https://marco-yuze-zheng.blogspot.com/2024/10/introduction-to-tensor-5-tensor-product.html

Conclusion

This categorical perspective reveals that the submultiplicativity condition in Banach algebras is not merely a technical requirement for continuity, but rather a natural consequence of viewing the algebra through its canonical representation as bounded operators.