Conditions for a Complex Banach Algebra with Unit to be Commutative

Lemma 1: Let $\mathfrak{A}$ be a Banach algebra over $\mathbb{C}$ with unit and let $M$ be a unit linked Banach left $\mathfrak{A}$-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) $h(a, m) = h(1, am)$ for all $a \in \mathfrak{A}$ and for all $m \in M$. b) There exists a $k > 0$ such that $\| h(a, m) \| \leq \| am \|$ for all $a \in \mathfrak{A}$ and for all $m \in M$.

If $a \in \mathfrak{A}$ then we define $\exp (a) := \sum_{n=0}^{\infty} \frac{a^n}{n!}$, and if $z \in \mathbb{C}$ and $a \in \mathfrak{A}$ then we define $\exp (za) := \sum_{n=0}^{\infty} \frac{z^na^n}{n!}$. Note that:

(1)

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

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

• Proof: $\Rightarrow$ Since the bilinear map $h$ is continuous there exists a $k > 0$ such that $\| h(a, m) \| \leq k \| a \| \| m \|$ for all $a \in \mathfrak{A}$, $m \in M$. Since $h(a, m) = h(1, am)$ for all $a \in \mathfrak{A}$, $m \in M$ we have that then for all $a \in \mathfrak{A}$, $m \in M$:

(2)

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

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

(3)

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

(4)

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

(5)

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

• So:

(6)

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

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

Theorem 2: Let $\mathfrak{A}$ be a Banach algebra over $\mathbb{C}$ with unit. If there exists a $k > 0$ such that $\| ab \| \leq k \| ba \|$ for all $a, b \in \mathfrak{A}$ then $\mathfrak{A}$ is commutative.

• Proof: Let $M := \mathfrak{A}$ and let $Y := \mathfrak{A}$. Since $\mathfrak{A}$ is a Banach algebra over $\mathbb{C}$ with unit we certainly have that $M$ is a unit linked Banach left $\mathfrak{A}$-module and that $Y$ is a Banach space over $\mathbb{C}$. Let $h : \mathfrak{A} \times \mathfrak{A} \to \mathfrak{A}$ be defined for all $a, b \in \mathfrak{A}$ by $h(a, b) = ba$. 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, $\| h(a, b) \| = \| ba \| \leq k \| ab \|$ for all $a, b \in \mathfrak{A}$.

• So by Lemma 1 we have that $h(a, b) = h(1, ab)$ for all $a, b \in \mathfrak{A}$, i.e., $ba = ab$ for all $a, b \in \mathfrak{A}$. So $\mathfrak{A}$ is commutative. $\blacksquare$

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

• Proof: Suppose that there exists a $k > 0$ such that $\| a \| \leq k r(a)$ for all $a \in \mathfrak{A}$. If $a, b \in \mathfrak{A}$ then $ab \in \mathfrak{A}$ and so by the Theorem on the r(xy) = r(yx) When A is a Banach Algebra Over C page we have that

(7)

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

• Since the above inequality holds for all $a, b \in \mathfrak{A}$, by Theorem 2 we have that $\mathfrak{A}$ is commutative. $\blacksquare$

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

• Proof: Let $a \in \mathfrak{A}$. Let $P(n)$ be the statement that the following inequality is true:

(8)

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

• Then $P(1)$ says that:

(9)

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

• We are given that $\| a \|^2 \leq k \| a^2 \|$, so square-rooting both sides of this inequality gives us that $\| a \| \leq k^{1/2} \| a^2 \|^{1/2}$, 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 $a \in \mathfrak{A}$ and every $n \in \mathbb{N}$ we have that $P(n)$ is true. Therefore:

(10)

\begin{align} \quad \| a \| \leq \inf \left \{ k^{1 - 2^{-n}} \| a^{2n} \|^{2^{-n}} : n \in \mathbb{N} \right \} = \inf \{ k^{1 - 2^{-n}} : n \in \mathbb{N} \} \inf \{ \| a^{2n} \|^{2^{-n}} \} \leq \inf \{ k^{1 - 2^{-n}} : n \in \mathbb{N} \} r(a) \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 $\| a \| \leq K r(a)$ for all $a \in \mathfrak{A}$, so by Theorem 3, $\mathfrak{A}$ is commutative. $\blacksquare$