Conditions for a Complex Banach Algebra with Unit to be Commutative

# Conditions for a Complex Banach Algebra with Unit to be Commutative

Lemma 1: Let $X$ be a Banach algebra over $\mathbb{C}$ with unit and let $M$ be a unit linked Banach left $X$-module. Let $Y$ be a Banach space over $\mathbb{C}$. Let $h : X \times M \to Y$ be a continuous bilinear map. Then the following are equivalent:a) $h(x, m) = h(1, xm)$ for all $x \in X$ and for all $m \in M$.b) There exists a $k > 0$ such that $\| h(x, m) \| \leq \| xm \|$ for all $x \in X$ and for all $m \in M$. |

*If $x \in X$ then we define $\exp (x) := \sum_{n=0}^{\infty} \frac{x^n}{n!}$, and if $z \in \mathbb{C}$ and $x \in X$ then we define $\exp (zx) := \sum_{n=0}^{\infty} \frac{z^nx^n}{n!}$. Note that:*

\begin{align} \quad \sum_{n=0}^{\infty} \left \| \frac{z^nx^n}{n!} \right \| = \sum_{n=0}^{\infty} \frac{[|z| \| x \|]^n}{n!} < \infty \end{align}

*Therefore the series $\sum_{n=0}^{\infty} \frac{z^nx^n}{n!}$ is absolutely convergent. Since $X$ is a Banach space, absolute convergence implies convergence, and so the series $\sum_{n=0}^{\infty} \frac{z^nx^n}{n!}$ converges in $X$, and we define that point as $\exp (zx)$.*

**Proof:**$\Rightarrow$ Since the bilinear map $h$ is continuous there exists a $k > 0$ such that $\| h(x, m) \| \leq k \| x \| \| m \|$ for all $x \in X$, $m \in M$. Since $h(x, m) = h(1, xm)$ for all $x \in X$, $m \in M$ we have that then for all $x \in X$, $m \in M$:

\begin{align} \quad \| h(x, m) \| = \| h(1, xm) \| \leq k \| 1 \| \| xm \| = k \| xm \| \end{align}

- $\Leftarrow$ Suppose there exists a $k > 0$ such that $\| h(x, m) \| \leq k \| xm \|$ for all $x \in X$, $m \in M$. Let $x \in X$ and $m \in M$. Let $f \in Y^*$. Define $F$ for all $z \in \mathbb{C}$ by:

\begin{align} \quad F(z) = f(h(\exp(-zx), exp(zx)m)) \end{align}

- It can be shown (with some complex analysis) that $F$ is an entire function. For every $z \in \mathbb{C}$ we have that

\begin{align} \quad |F(z)| &= |f(h(\exp(-zx), exp(zx)m))| \\ & \leq \| f \| \| h(\exp(-zx), exp(zx)m) \| \\ & \leq \| f \| k \| \exp(-zx) \exp(zx)m \| \\ & \leq \| f \| k \| \exp (0) m \| \\ & \leq k \| f \| \| m \| \end{align}

- So $F$ is bounded. By Liouville's theorem, $F$ must be constant on $\mathbb{C}$. If $F(z) = \sum_{n=0}^{\infty} a_nz^n$ is the power series for $F(z)$ then since $F$ is constant on $\mathbb{C}$ we have that $a_n = 0$ for all $n \geq 1$. In particular, the term $a_1 = F'(0) = 0$ and:

\begin{align} \quad F'(x) &= (f \circ h)(\exp(zx), [\exp(-zx)m]') + (f \circ h)([\exp(zx)]', \exp(-zx)m) \\ &= (f \circ h)(\exp(zx), -x \exp(-zx)m) + (f \circ h)(x \exp(zx), \exp(-zx)m) \\ \end{align}

- Plugging in $x = 0$ gives us that:

\begin{align} \quad 0 &= F'(1) \\ &= (f \circ h)(1, -xm) + (f \circ h)(x, m) \\ &= -f(h(1, xm)) + f(h(x, m)) \\ \end{align}

- Thus $f(h(x, m)) = f(h(1, xm))$. From one of the corollaries on the Corollaries to the Hahn-Banach Theorem page we have that since $X$ is a Banach algebra and $f(h(1, xm)) = f(h(x,m))$ for all $f \in X^*$ that then $h(1, xm) = h(x, m)$. But this holds for all $x \in X$ and all $m \in M$. $\blacksquare$

Theorem 2: Let $X$ be a Banach algebra over $\mathbb{C}$ with unit. If there exists a $k > 0$ such that $\| xy \| \leq k \| yx \|$ for all $x, y \in X$ then $X$ is commutative. |

**Proof:**Let $M := X$ and let $Y := X$. Since $X$ is a Banach algebra over $\mathbb{C}$ with unit we certainly have that $M$ is a unit linked Banach left $X$-module and that $Y$ is a Banach space over $\mathbb{C}$. Let $h : X \times X \to X$ be defined for all $x, y \in X$ by $h(x, y) = yx$. Then $h$ is a continuous as we proved on the Continuity of Addition, Scalar Multiplication, and Multiplication on Normed Algebras page, and $h$ is a bilinear map. Furthermore, $\| h(x, y) \| = \| yx \| \leq k \| xy \|$ for all $x, y \in X$.

- So by lemma 1 we have that $h(x, y) = h(1, xy)$ for all $x, y \in X$, i.e., $yx = xy$ for all $x, y \in X$. So $X$ is commutative. $\blacksquare$

Theorem 3: Let $X$ be a Banach algebra over $\mathbb{C}$ with unit. If there exists a $k > 0$ such that $\| x \| \leq k r(x)$ for all $x \in X$ then $X$ is commutative. |

**Proof:**Suppose that there exists a $k > 0$ such that $\| x \| \leq k r(x)$ for all $x \in X$. If $x, y \in X$ then $xy \in X$ and so by the theorem on the r(xy) = r(yx) When X is a Banach Algebra Over C page we have that

\begin{align} \quad \| xy \| \leq k r(xy) = k r(yx) = k \inf \{ \| (yx)^n \|^{1/n} : n \in \mathbb{N} \} \leq k \| yx \| \end{align}

- Since the above inequality holds for all $x, y \in X$, by Theorem 2 we have that $X$ is commutative. $\blacksquare$

Theorem 4: Let $X$ be a Banach algebra over $\mathbb{C}$ with unit. If there exists a $k > 0$ such that $\| x \|^2 \leq k \| x^2 \|$ for all $x \in X$ then $X$ is commutative. |

**Proof:**Let $x \in X$. Let $P(n)$ be the statement that the following inequality is true:

\begin{align} \quad \| x \| \leq k^{1 - 2^{-n}} \| x^{2^n} \|^{2^{-n}} \end{align}

- Then $P(1)$ says that:

\begin{align} \quad \| x \| & \leq k^{1 - 2^{-1}} \| x^2 \|^{2^{-1}} \\ \quad \| x \| & \leq k^{1/2} \| x^2 \|^{1/2} \end{align}

- We are given that $\| x \|^2 \leq k \| x^2 \|$, so square-rooting both sides of this inequality gives us that $\| x \| \leq k^{1/2} \| x^2 \|^{1/2}$, so $P(1)$ is true.

- It can be shown by induction that $P(n)$ is true for all $n \in \mathbb{N}$. So for every $x \in X$ and every $n \in \mathbb{N}$ we have that $P(n)$ is true. Therefore:

\begin{align} \quad \| x \| \leq \inf \left \{ k^{1 - 2^{-n}} \| x^{2n} \|^{2^{-n}} : n \in \mathbb{N} \right \} = \inf \{ k^{1 - 2^{-n}} : n \in \mathbb{N} \} \inf \{ \| x^{2n} \|^{2^{-n}} \} \leq \inf \{ k^{1 - 2^{-n}} : n \in \mathbb{N} \} r(x) \end{align}

- So let $K = \inf \{ k^{1 - 2^{-n}} : n \in \mathbb{N} \}$. Note $K$ always exists since $\{ k^{1 - 2^{-n}} : n \in \mathbb{N} \}$ is bounded below. Thus $\| x \| \leq K r(x)$ for all $x \in X$, so by theorem 3, $X$ is commutative. $\blacksquare$