Normed and Banach Algebras

Submultiplicative Norm

A norm over a vector space $X$ is submultiplicative if $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$ for all $a,b\in X$.

Normed Algebra & Banach Algebra

Let $\mathcal{A}$ be an algebra (over $\text{C}$) equipped with a submultiplicative norm, then $\text{A}$ is called a normed algebra. If it is also Banach (i.e. complete), then it is called a Banach algebra.

e.g.

• Suppose $X$ is a Banach space, and $B(X)$ is the algebra of bounded linear operators on $X$ with operator norm $\Vert \cdot {\Vert }_{\infty }$. Then $(B(X),\Vert \cdot {\Vert }_{\infty })$ is a Banach algebra.

Generated Algebra

Ideals

Theorem

Let $A$ be a Banach algebra, $I$ be a closed ideal in $A$, then $A/I$ is a Banach algebra, with the norm defined as $\Vert a+I\Vert ={\inf }_{i\in I}\Vert a+i\Vert$ for all $a\in A$.

Spectrum

Recall that the set of invertible elements in a algebra forms a group:

An element of a ring $R$ is called a unit if it is invertible with respect to multiplication. The set of invertible elements is a group called the the group of units in $R$ and denotes $R^{\times }$.

Link to original

Here, for any unital algebra $A$, we denote the set of invertible elements in $A$ as $\mathrm{Inv}(A):=A^{\times }$.

Spectrum

Suppose $A$ is a normed algebra. Then the spectrum of $a\in A$ is the set ${\sigma }_A(a)=\{\lambda \in \text{C}:a-\lambda \cdot 1\text{ is not invertible}\}.$

e.g.

• Suppose $A:=M_{n\times n}(\text{C})$, the algebra of $n\times n$ complex matrices, then the spectrum of $A$ is the set of all eigenvalues of $A$.
• $\Omega$ is a compact Hilbert space, then the spectrum of $f\in C(\Omega )$, for which $C(\Omega )$ is the algebra of all continuous maps $\Omega \to \text{C}$, is the range of $f$.
• Consider the algebra $\text{C}[z]$. The spectrum of $z$ is the empty set.

Proposition

The spectrum is a closed subset of $\text{C}$.

Neumann Theorem

Let $A$ be a unital Banach algebra, and $a\in A$ such that $\Vert a\Vert <1$. Then $(1-a)$ is invertible and $(1-a{)}^{-1}={\sum }_{n=0}^{\infty }{a}^n.$

Theorem

Let $A$ be a unital Banach algebra, then $\text{Inv}(A)$ is open in $A$, and the mapping $\text{Inv}(A)\to A$, $a\mapsto a^{-1}$ is Fréchet differentiable.

Lemma

If $a$ is an element of a unital Banach algebra $A$, then the spectrum $\sigma (a)$ is a closed subset of $\text{C}$ and $|\lambda |\le \Vert a\Vert$ for any $\lambda \in \sigma (a)$.

Proof We first prove that $|\lambda |\le \Vert a\Vert$ by contradiction. Assume $|\lambda |>\Vert a\Vert$ for some $\lambda \in \sigma (a)$. Then, by the Neumann theorem, we have $(1-\frac{a}{\lambda })$ is invertible, hence $a-\lambda \cdot 1=(1-\frac{a}{\lambda })\cdot (-\lambda )$ is also invertible. This contradicts the definition of $\sigma (a)$. Now we show that $\sigma (a)$ is closed. Define the map $\phi :\text{C}\to A$, $\lambda \mapsto a-\lambda \cdot 1$, which is continuous. Then $\text{C}\setminus \sigma (a)={\phi }^{-1}(\text{Inv}(A))$. Since $\text{Inv}(A)$ is open in $A$, it follows that $\sigma (a)$ is closed in $\text{C}$. $\square$

Lemma

If $a$ is an element of a unital Banach algebra $A$, then the map $F:\text{C}\setminus \sigma (a)\to A$, $\lambda \mapsto (a-\lambda \cdot 1{)}^{-1}$ is differentiable.

Proof Observe that $F=(a\mapsto a^{-1})\circ (\lambda \mapsto a-\lambda \cdot 1)$, which is a composition of two differentiable maps, hence $F$ is differentiable. $\square$

Gelfand Theorem

If $a$ is an element of a unital Banach algebra $A$, then the spectrum $\sigma (a)$ is nonempty.

Proof Assume that $\sigma (a)=\varnothing$ (i.e. $a-\lambda \cdot 1$ is invertible for all $\lambda \in \text{C}$) and we shall obtain a contradiction. For all $\lambda \in \text{C}$ with $|\lambda |>2\Vert a\Vert$, we have $\Vert {\lambda }^{-1}a\Vert <\frac{1}{2}<1$, so Neumann theorem implies that $(1-{\lambda }^{-1}a)$ is invertible, and $(1-{\lambda }^{-1}a{)}^{-1}={\sum }_{n=0}^{\infty }({\lambda }^{-1}a{)}^n.$Therefore, $\Vert (1-{\lambda }^{-1}a{)}^{-1}-1\Vert =\Vert {\sum }_{n=1}^{\infty }({\lambda }^{-1}a{)}^n\Vert \le {\sum }_{n=1}^{\infty }\Vert {\lambda }^{-1}a{\Vert }^n=\frac{\Vert {\lambda }^{-1}a\Vert }{1-\Vert {\lambda }^{-1}a\Vert }\le 2\Vert {\lambda }^{-1}a\Vert <1,$where the first inequality follows from the triangle inequality. Moreover, from the above lemma, we know that the $\lambda \mapsto (a-\lambda \cdot 1{)}^{-1}$ is differentiable on $\text{C}\setminus \sigma (a)$, so it is continuous on $\text{C}\setminus \sigma (a)$.