Functional Analysis/C*-algebras

←Chapter 4: Geometry of Banach spaces

Functional Analysis Chapter 5: C^*-algebras

. Beware this chapter is less than a draft.

A Banach space $𝒜$ over $𝐂$ is called a Banach algebra if it is an algebra and satisfies

$\Vert xy\Vert \le \Vert x\Vert \Vert y\Vert$.

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

Since $\Vert x_n{y}_n-xy\Vert \le \Vert x_n\Vert \Vert y_n-y\Vert +\Vert x_n-x\Vert \Vert y\Vert \to 0$ as $x_n,y_n\to 0$, the map

$(x,y)\mapsto xy:𝒜\times 𝒜\to 𝒜$

is continuous.

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

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

$\sigma (x)\subset \{s\in 𝐂||s|\le \Vert x\Vert \}$.

Moreover,

$r(x)\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}sup\{|z||z\in \sigma (x)\}=\underset{n\to \mathrm{\infty }}{lim}\Vert x^n{\Vert }^{1/n}$

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

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

Thus, $1-\frac{x}{s}$ is invertible, which is to say, $s1-x$ is invertible. Hence, $s\in ̸\sigma (x)$. This complete the proof of the first assertion and gives:

$r(x)\le \Vert x\Vert$

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

$r(x{)}^n\le \Vert x^n\Vert$

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

$\underset{n\to \mathrm{\infty }}{lim}g\left(\frac{x^n}{s^{n+1}}\right)=0$,

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

$r(x)\le \underset{n\to \mathrm{\infty }}{lim}\Vert x^n{\Vert }^{1/n}\le \underset{n\to \mathrm{\infty }}{lim}{c}^{1/n}|s|=|s|$.

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

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

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

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

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

$𝒜/𝔪\simeq 𝐂$

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

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

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

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

An involution is an anti-linear map $x\to x^{\ast }:A\to A$ such that $x^{\ast \ast }=x$. 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

$\Vert x{x}^{\ast }\Vert =\Vert x{\Vert }^2$ (C*-identity)

From the C*-identity follows

$\Vert x^{\ast }\Vert =\Vert x\Vert$,

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

$\Vert x^{\ast }x\Vert \le \Vert x^{\ast }\Vert \Vert x\Vert$ implies $\Vert x\Vert =\Vert x^{\ast }\Vert$ and so $\Vert x{\Vert }^2\le \Vert x^{\ast }x\Vert \le \Vert x{\Vert }^2$.

For each $x\in 𝒜$, let $C^{\ast }(x)$ be the linear span of $\{1,y_1{y}_2\mathrm{...}{y}_n|y_j\in \{x,x^{\ast }\}\}$. In other words, $C^{\ast }(x)$ is the smallest C*-algebra that contains $x$. The crucial fact is that $C^{\ast }(x)$ is commutative. Moreover,

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

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

5 Theorem Every C^*-algebra $𝒜$ is *-isomorphic to $C_0(X)$ where $X$ is the spectrum of $𝒜$.

5 Theorem If $C_0(X)$ is isomorphic to $C_0(Y)$, then it follows that $X$ and $Y$ are homeomorphic.

3 Lemma Let $T$ be a continuous linear operator on a Hilbert space $\mathscr{H}$. Then $T{T}^{\ast }=T^{\ast }T$ if and only if $\Vert Tx\Vert =\Vert T^{\ast }x\Vert$ for all $x\in \mathscr{H}$.

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

$0=\Vert (N-\alpha I)x\Vert =\Vert (N-\alpha I{)}^{\ast }x\Vert =\Vert N^{\ast }x-\overline{\alpha }x\Vert$.

That is, $N^{\ast }x=\overline{\alpha }x$, and we thus have:

$\overline{\alpha }⟨x,y⟩=⟨N^{\ast }x,y⟩=⟨x,Ny⟩=\overline{\beta }⟨x,y⟩$

If $⟨x,y⟩$ is nonzero, we must have $\alpha =\beta$. $◻$

5 Exercise Let $\mathscr{H}$ be a Hilbert space with orthogonal basis $e_1,e_2,\mathrm{...}$, and $x_n$ be a sequence with $\Vert x_n\Vert \le K$. Prove that there is a subsequence of $x_n$ that converges weakly to some $x$ and that $\Vert x\Vert \le K$. (Hint: Since $⟨x_n,e_k⟩$ is bounded, by Cantor's diagonal argument, we can find a sequence $x_{n_k}$ such that $⟨x_{n_k},e_k⟩$ is convergent for every $k$.)

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)

References

• Lecture Notes on C ∗-Algebras and K-Theory
• Part III Lecture Notes on Operators on Hilbert Space - Michaelmas 1991, Cambridge
• A Good Spectral Theorem