# Functional Analysis/C*-algebras

 Functional Analysis Chapter 5: C^*-algebras
 (May 27, 2008). Beware this chapter is less than a draft.

A Banach space ${\displaystyle {\mathcal {A}}}$ over ${\displaystyle \mathbf {C} }$ is called a Banach algebra if it is an algebra and satisfies

${\displaystyle \|xy\|\leq \|x\|\|y\|}$.

We shall assume that every Banach algebra has the unit ${\displaystyle 1}$ unless stated otherwise.

Since ${\displaystyle \|x_{n}y_{n}-xy\|\leq \|x_{n}\|\|y_{n}-y\|+\|x_{n}-x\|\|y\|\to 0}$ as ${\displaystyle x_{n},y_{n}\to 0}$, the map

${\displaystyle (x,y)\mapsto xy:{\mathcal {A}}\times {\mathcal {A}}\to {\mathcal {A}}}$

is continuous.

For ${\displaystyle x\in {\mathcal {A}}}$, let ${\displaystyle \sigma (x)}$ be the set of all complex numbers ${\displaystyle \lambda }$ such that ${\displaystyle x-\lambda 1}$ is not invertible.

5 Theorem For every ${\displaystyle x\in {\mathcal {A}}}$, ${\displaystyle \sigma (x)}$ is nonempty and closed and

${\displaystyle \sigma (x)\subset \{s\in \mathbf {C} ||s|\leq \|x\|\}}$.

Moreover,

${\displaystyle r(x){\overset {\mathrm {def} }{=}}\sup\{|z||z\in \sigma (x)\}=\lim _{n\to \infty }\|x^{n}\|^{1/n}}$

(${\displaystyle r(x)}$ is called the spectral radius of ${\displaystyle x}$)
Proof: Let ${\displaystyle G\subset {\mathcal {A}}}$ be the group of units. Define ${\displaystyle f:\mathbf {C} \to A}$ by ${\displaystyle f(\lambda )=\lambda 1-x}$. (Throughout the proof ${\displaystyle x}$ is fixed.) If ${\displaystyle \lambda \in \mathbf {C} \backslash \sigma (x)}$, then, by definition, ${\displaystyle f(\lambda )\in G}$ or ${\displaystyle \lambda \in f^{-1}(G)}$. Similarly, we have: ${\displaystyle G\subset f(\sigma (x))}$. Thus, ${\displaystyle x\in f^{-1}(G)\subset \sigma (x)}$. Since ${\displaystyle f}$ is clearly continuous, ${\displaystyle \mathbf {C} \backslash \sigma (x)}$ is open and so ${\displaystyle \sigma (x)}$ is closed. Suppose that ${\displaystyle |s|>\|x\|}$ for ${\displaystyle s\in \mathbf {C} }$. By the geometric series (which is valid by Theorem 2.something), we have:

${\displaystyle \left(1-{x \over s}\right)^{-1}=\sum _{n=0}^{\infty }\left({x \over s}\right)^{n}}$

Thus, ${\displaystyle 1-{x \over s}}$ is invertible, which is to say, ${\displaystyle s1-x}$ is invertible. Hence, ${\displaystyle s\not \in \sigma (x)}$. This complete the proof of the first assertion and gives:

${\displaystyle r(x)\leq \|x\|}$

Since ${\displaystyle \sigma (x)}$ is compact, there is a ${\displaystyle a\in \sigma (x)}$ such that ${\displaystyle r(x)=a}$. Since ${\displaystyle a^{n}\in \sigma (x^{n})}$ (use induction to see this),

${\displaystyle r(x)^{n}\leq \|x^{n}\|}$

Next, we claim that the sequence ${\displaystyle {x^{n} \over s^{n+1}}}$ is bounded for ${\displaystyle |s|>r(x)}$. In view of the uniform boundedness principle, it suffices to show that ${\displaystyle g\left({x^{n} \over s^{n+1}}\right)}$ is bounded for every ${\displaystyle g\in A^{*}}$. But since

${\displaystyle \lim _{n\to \infty }g\left({x^{n} \over s^{n+1}}\right)=0}$,

this is in fact the case. Hence, there is a constant ${\displaystyle c}$ such that ${\displaystyle \|x^{n}\|\leq c|s|^{n+1}}$ for every ${\displaystyle n}$. It follows:

${\displaystyle r(x)\leq \lim _{n\to \infty }\|x^{n}\|^{1/n}\leq \lim _{n\to \infty }c^{1/n}|s|=|s|}$.

Taking inf over ${\displaystyle |s|>r(x)}$ completes the proof of the spectral radius formula. Finally, suppose, on the contrary, that ${\displaystyle \sigma (x)}$ is empty. Then for every ${\displaystyle g\in A^{*}}$, the map

${\displaystyle s\mapsto g((x-s)^{-1})}$

is analytic in ${\displaystyle \mathbf {C} }$. Since ${\displaystyle \lim _{s\to \infty }g((x-s)^{-1})=0}$, by Liouville's theorem, we must have: ${\displaystyle g((x-s)^{-1})=0}$. Hence, ${\displaystyle (x-s)^{-1}=0}$ for every ${\displaystyle s\in \mathbf {C} }$, a contradiction. ${\displaystyle \square }$

5 Corollary (Gelfand-Mazur theorem) If every nonzero element of ${\displaystyle {\mathcal {A}}}$ is invertible, then ${\displaystyle {\mathcal {A}}}$ is isomorphic to ${\displaystyle \mathbf {C} }$.
Proof: Let ${\displaystyle x\in {\mathcal {A}}}$ be a nonzero element. Since ${\displaystyle \sigma (x)}$ is non-empty, we can then find ${\displaystyle \lambda \in \mathbf {C} }$ such that ${\displaystyle \lambda 1-x}$ is not invertible. But, by hypothesis, ${\displaystyle \lambda 1-x}$ is invertible, unless ${\displaystyle \lambda 1=x}$.${\displaystyle \square }$

Let ${\displaystyle {\mathfrak {m}}}$ be a maximal ideal of a Banach algebra. (Such ${\displaystyle {\mathfrak {m}}}$ exists by the usual argument involving Zorn's Lemma in abstract algebra). Since the complement of ${\displaystyle {\mathfrak {m}}}$ consists of invertible elements, ${\displaystyle {\mathfrak {m}}}$ is closed. In particular, ${\displaystyle {\mathcal {A}}/{\mathfrak {m}}}$ is a Banach algebra with the usual quotient norm. By the above corollary, we thus have the isomorphism:

${\displaystyle {\mathcal {A}}/{\mathfrak {m}}\simeq \mathbf {C} }$

Much more is true, actually. Let ${\displaystyle \Delta ({\mathcal {A}})}$ be the set of all nonzero homeomorphism ${\displaystyle \omega :{\mathcal {A}}\to \mathbf {C} }$. (The members of ${\displaystyle \Delta ({\mathcal {A}})}$ are called characters.)

5 Theorem ${\displaystyle \Delta ({\mathcal {A}})}$ is bijective to the set of all maximal ideals of ${\displaystyle {\mathcal {A}}}$.

5 Lemma Let ${\displaystyle x\in {\mathcal {A}}}$. Then ${\displaystyle x}$ is invertible if and only if ${\displaystyle \omega (x)\neq 0}$ for every ${\displaystyle \omega \in \Delta ({\mathcal {A}})}$

5 Theorem ${\displaystyle \omega (x)=\omega ({\hat {x}})}$

An involution is an anti-linear map ${\displaystyle x\to x^{*}:A\to A}$ such that ${\displaystyle x^{**}=x}$. Prototypical examples are the complex conjugation of functions and the operation of taking the adjoint of a linear operator. These examples explain why we require an involution to be anti-linear.

Now, the interest of study in this chapter. A Banach algebra with an involution is called a C*-algebra if it satisfies

${\displaystyle \|xx^{*}\|=\|x\|^{2}}$ (C*-identity)

From the C*-identity follows

${\displaystyle \|x^{*}\|=\|x\|}$,

for ${\displaystyle \|x\|^{2}\leq \|x^{*}\|\|x\|}$ and the same for ${\displaystyle x^{*}}$ in place of ${\displaystyle x}$. In particular, ${\displaystyle \|1\|=1}$ (if ${\displaystyle 1}$ exists). Furthermore, the ${\displaystyle C^{*}}$-identity is equivalent to the condition: ${\displaystyle \|x\|^{2}\leq \|x^{*}x\|}$, for this and

${\displaystyle \|x^{*}x\|\leq \|x^{*}\|\|x\|}$ implies ${\displaystyle \|x\|=\|x^{*}\|}$ and so ${\displaystyle \|x\|^{2}\leq \|x^{*}x\|\leq \|x\|^{2}}$.

For each ${\displaystyle x\in {\mathcal {A}}}$, let ${\displaystyle C^{*}(x)}$ be the linear span of ${\displaystyle \{1,y_{1}y_{2}...y_{n}|y_{j}\in \{x,x^{*}\}\}}$. In other words, ${\displaystyle C^{*}(x)}$ is the smallest C*-algebra that contains ${\displaystyle x}$. The crucial fact is that ${\displaystyle C^{*}(x)}$ is commutative. Moreover,

Theorem Let ${\displaystyle x\in {\mathcal {A}}}$ be normal. Then ${\displaystyle \sigma _{A}(x)=\sigma _{C^{*}(x)}(x)}$

A state on ${\displaystyle C^{*}}$-algebra ${\displaystyle {\mathcal {A}}}$ is a positive linear functional f such that ${\displaystyle \|f\|=1}$ (or equivalently ${\displaystyle f(1)=1}$). Since ${\displaystyle S}$ is convex and closed, ${\displaystyle S}$ is weak-* closed. (This is Theorem 4.something.) Since ${\displaystyle S}$ is contained in the unit ball of the dual of ${\displaystyle {\mathcal {A}}}$, ${\displaystyle S}$ is weak-* compact.

5 Theorem Every C^*-algebra ${\displaystyle {\mathcal {A}}}$ is *-isomorphic to ${\displaystyle C_{0}(X)}$ where ${\displaystyle X}$ is the spectrum of ${\displaystyle {\mathcal {A}}}$.

5 Theorem If ${\displaystyle C_{0}(X)}$ is isomorphic to ${\displaystyle C_{0}(Y)}$, then it follows that ${\displaystyle X}$ and ${\displaystyle Y}$ are homeomorphic.

3 Lemma Let ${\displaystyle T}$ be a continuous linear operator on a Hilbert space ${\displaystyle {\mathcal {H}}}$. Then ${\displaystyle TT^{*}=T^{*}T}$ if and only if ${\displaystyle \|Tx\|=\|T^{*}x\|}$ for all ${\displaystyle x\in {\mathcal {H}}}$.

Continuous linear operators with the above equivalent conditions are said to be normal. For example, an orthogonal projection is normal. See w:normal operator for additional examples and the proof of the above lemma.

3 Lemma Let ${\displaystyle N}$ be a normal operator. If ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are distinct eigenvalues of ${\displaystyle N}$, then the respective eigenspaces of ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are orthogonal to each other.
Proof: Let ${\displaystyle I}$ be the identity operator, and ${\displaystyle x,y}$ be arbitrary eigenvectors for ${\displaystyle \alpha ,\beta }$, respectively. Since the adjoint of ${\displaystyle \alpha I}$ is ${\displaystyle {\bar {\alpha }}I}$, we have:

${\displaystyle 0=\|(N-\alpha I)x\|=\|(N-\alpha I)^{*}x\|=\|N^{*}x-{\bar {\alpha }}x\|}$.

That is, ${\displaystyle N^{*}x={\bar {\alpha }}x}$, and we thus have:

${\displaystyle {\bar {\alpha }}\langle x,y\rangle =\langle N^{*}x,y\rangle =\langle x,Ny\rangle ={\bar {\beta }}\langle x,y\rangle }$

If ${\displaystyle \langle x,y\rangle }$ is nonzero, we must have ${\displaystyle \alpha =\beta }$. ${\displaystyle \square }$

5 Exercise Let ${\displaystyle {\mathcal {H}}}$ be a Hilbert space with orthogonal basis ${\displaystyle e_{1},e_{2},...}$, and ${\displaystyle x_{n}}$ be a sequence with ${\displaystyle \|x_{n}\|\leq K}$. Prove that there is a subsequence of ${\displaystyle x_{n}}$ that converges weakly to some ${\displaystyle x}$ and that ${\displaystyle \|x\|\leq K}$. (Hint: Since ${\displaystyle \langle x_{n},e_{k}\rangle }$ is bounded, by Cantor's diagonal argument, we can find a sequence ${\displaystyle x_{n_{k}}}$ such that ${\displaystyle \langle x_{n_{k}},e_{k}\rangle }$ is convergent for every ${\displaystyle k}$.)

5 Theorem (Von Neumann double commutant theorem) M is equal to its double commutant if and only if it is closed in either weak-operator topology or strong-operator topology.
Proof: (see w:Von Neumann bicommutant theorem)