# Functional Analysis/C*-algebras

| ←Chapter 4: Geometry of Banach spaces | Functional Analysis Chapter 5: C^*-algebras |
A Banach space ${\mathcal {A}}$ over $\mathbf {C}$ is called a Banach algebra if it is an algebra and satisfies
    $\|xy\|\leq \|x\|\|y\|$.
We shall assume that every Banach algebra has the unit $1$ unless stated otherwise.
Since $\|x_{n}y_{n}-xy\|\leq \|x_{n}\|\|y_{n}-y\|+\|x_{n}-x\|\|y\|\to 0$ as $x_{n},y_{n}\to 0$, the map
    $(x,y)\mapsto xy:{\mathcal {A}}\times {\mathcal {A}}\to {\mathcal {A}}$
is continuous.
For $x\in {\mathcal {A}}$, let $\sigma (x)$ be the set of all complex numbers $\lambda$ such that $x-\lambda 1$ is not invertible.
5 Theorem For every $x\in {\mathcal {A}}$, $\sigma (x)$ is nonempty and closed and
    $\sigma (x)\subset \{s\in \mathbf {C} ||s|\leq \|x\|\}$.
Moreover,
    $r(x){\overset {\mathrm {def} }{=}}\sup\{|z||z\in \sigma (x)\}=\lim _{n\to \infty }\|x^{n}\|^{1/n}$
($r(x)$ is called the spectral radius of $x$) Proof: Let $G\subset {\mathcal {A}}$ be the group of units. Define $f:\mathbf {C} \to A$ by $f(\lambda )=\lambda 1-x$. (Throughout the proof $x$ is fixed.) If $\lambda \in \mathbf {C} \backslash \sigma (x)$, 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, $\mathbf {C} \backslash \sigma (x)$ is open and so $\sigma (x)$ is closed. Suppose that $|s|>\|x\|$ for $s\in \mathbf {C}$. By the geometric series (which is valid by Theorem 2.something), we have:
    $\left(1-{x \over s}\right)^{-1}=\sum _{n=0}^{\infty }\left({x \over s}\right)^{n}$
Thus, $1-{x \over s}$ is invertible, which is to say, $s1-x$ is invertible. Hence, $s\not \in \sigma (x)$. This complete the proof of the first assertion and gives:
    $r(x)\leq \|x\|$
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),
    $r(x)^{n}\leq \|x^{n}\|$
Next, we claim that the sequence ${x^{n} \over s^{n+1}}$ is bounded for $|s|>r(x)$. In view of the uniform boundedness principle, it suffices to show that $g\left({x^{n} \over s^{n+1}}\right)$ is bounded for every $g\in A^{*}$. But since
    $\lim _{n\to \infty }g\left({x^{n} \over s^{n+1}}\right)=0$,
this is in fact the case. Hence, there is a constant $c$ such that $\|x^{n}\|\leq c|s|^{n+1}$ for every $n$. It follows:
    $r(x)\leq \lim _{n\to \infty }\|x^{n}\|^{1/n}\leq \lim _{n\to \infty }c^{1/n}|s|=|s|$.
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 $g\in A^{*}$, the map
    $s\mapsto g((x-s)^{-1})$
is analytic in $\mathbf {C}$. Since $\lim _{s\to \infty }g((x-s)^{-1})=0$, by Liouville's theorem, we must have: $g((x-s)^{-1})=0$. Hence, $(x-s)^{-1}=0$ for every $s\in \mathbf {C}$, a contradiction. $\square$
5 Corollary (Gelfand-Mazur theorem) If every nonzero element of ${\mathcal {A}}$ is invertible, then ${\mathcal {A}}$ is isomorphic to $\mathbf {C}$. Proof: Let $x\in {\mathcal {A}}$ be a nonzero element. Since $\sigma (x)$ is non-empty, we can then find $\lambda \in \mathbf {C}$ such that $\lambda 1-x$ is not invertible. But, by hypothesis, $\lambda 1-x$ is invertible, unless $\lambda 1=x$.$\square$
Let ${\mathfrak {m}}$ be a maximal ideal of a Banach algebra. (Such ${\mathfrak {m}}$ exists by the usual argument involving Zorn's Lemma in abstract algebra). Since the complement of ${\mathfrak {m}}$ consists of invertible elements, ${\mathfrak {m}}$ is closed. In particular, ${\mathcal {A}}/{\mathfrak {m}}$ is a Banach algebra with the usual quotient norm. By the above corollary, we thus have the isomorphism:
    ${\mathcal {A}}/{\mathfrak {m}}\simeq \mathbf {C}$
Much more is true, actually. Let $\Delta ({\mathcal {A}})$ be the set of all nonzero homeomorphism $\omega :{\mathcal {A}}\to \mathbf {C}$. (The members of $\Delta ({\mathcal {A}})$ are called characters.)
5 Theorem $\Delta ({\mathcal {A}})$ is bijective to the set of all maximal ideals of ${\mathcal {A}}$.
5 Lemma Let $x\in {\mathcal {A}}$. Then $x$ is invertible if and only if $\omega (x)\neq 0$ for every $\omega \in \Delta ({\mathcal {A}})$
5 Theorem $\omega (x)=\omega ({\hat {x}})$
An involution is an anti-linear map $x\to x^{*}:A\to A$ such that $x^{**}=x$. 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
    $\|xx^{*}\|=\|x\|^{2}$ (C*-identity)
From the C*-identity follows
    $\|x^{*}\|=\|x\|$,
for $\|x\|^{2}\leq \|x^{*}\|\|x\|$ and the same for $x^{*}$ in place of $x$. In particular, $\|1\|=1$ (if $1$ exists). Furthermore, the $C^{*}$-identity is equivalent to the condition: $\|x\|^{2}\leq \|x^{*}x\|$, for this and
    $\|x^{*}x\|\leq \|x^{*}\|\|x\|$ implies $\|x\|=\|x^{*}\|$ and so $\|x\|^{2}\leq \|x^{*}x\|\leq \|x\|^{2}$.
For each $x\in {\mathcal {A}}$, let $C^{*}(x)$ be the linear span of $\{1,y_{1}y_{2}...y_{n}|y_{j}\in \{x,x^{*}\}\}$. In other words, $C^{*}(x)$ is the smallest C*-algebra that contains $x$. The crucial fact is that $C^{*}(x)$ is commutative. Moreover,
Theorem Let $x\in {\mathcal {A}}$ be normal. Then $\sigma _{A}(x)=\sigma _{C^{*}(x)}(x)$
A state on $C^{*}$-algebra ${\mathcal {A}}$ is a positive linear functional f such that $\|f\|=1$ (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 ${\mathcal {A}}$, $S$ is weak-* compact.
5 Theorem Every C^*-algebra ${\mathcal {A}}$ is *-isomorphic to $C_{0}(X)$ where $X$ is the spectrum of ${\mathcal {A}}$.
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 ${\mathcal {H}}$. Then $TT^{*}=T^{*}T$ if and only if $\|Tx\|=\|T^{*}x\|$ for all $x\in {\mathcal {H}}$.
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 $N$ 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 ${\bar {\alpha }}I$, we have:
    $0=\|(N-\alpha I)x\|=\|(N-\alpha I)^{*}x\|=\|N^{*}x-{\bar {\alpha }}x\|$.
That is, $N^{*}x={\bar {\alpha }}x$, and we thus have:
    ${\bar {\alpha }}\langle x,y\rangle =\langle N^{*}x,y\rangle =\langle x,Ny\rangle ={\bar {\beta }}\langle x,y\rangle$
If $\langle x,y\rangle$ is nonzero, we must have $\alpha =\beta$. $\square$
5 Exercise Let ${\mathcal {H}}$ be a Hilbert space with orthogonal basis $e_{1},e_{2},...$, and $x_{n}$ be a sequence with $\|x_{n}\|\leq K$. Prove that there is a subsequence of $x_{n}$ that converges weakly to some $x$ and that $\|x\|\leq K$. (Hint: Since $\langle x_{n},e_{k}\rangle$ is bounded, by Cantor's diagonal argument, we can find a sequence $x_{n_{k}}$ such that $\langle x_{n_{k}},e_{k}\rangle$ 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