A Banach space
over
is called a Banach algebra if it is an algebra and satisfies
.
We shall assume that every Banach algebra has the unit
unless stated otherwise.
Since
as
, the map
![{\displaystyle (x,y)\mapsto xy:{\mathcal {A}}\times {\mathcal {A}}\to {\mathcal {A}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ec863e5f9863ebace7ef7b524a57c05921f22e0)
is continuous.
For
, let
be the set of all complex numbers
such that
is not invertible.
5 Theorem For every
,
is nonempty and closed and
.
Moreover,
![{\displaystyle r(x){\overset {\mathrm {def} }{=}}\sup\{|z||z\in \sigma (x)\}=\lim _{n\to \infty }\|x^{n}\|^{1/n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b12f1d9e67b1211b92738c43c639958bb07ca4a1)
(
is called the spectral radius of
)
Proof: Let
be the group of units. Define
by
. (Throughout the proof
is fixed.) If
, then, by definition,
or
. Similarly, we have:
. Thus,
. Since
is clearly continuous,
is open and so
is closed. Suppose that
for
. 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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/519077eb551de5ee25b6e2e11e5afbe910e561e9)
Thus,
is invertible, which is to say,
is invertible. Hence,
. This complete the proof of the first assertion and gives:
![{\displaystyle r(x)\leq \|x\|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/389f41b1008671219dbb1fbfe435fd285956a712)
Since
is compact, there is a
such that
. Since
(use induction to see this),
![{\displaystyle r(x)^{n}\leq \|x^{n}\|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a926c64378d881c146aed16bee986a138fcf6fa3)
Next, we claim that the sequence
is bounded for
. In view of the uniform boundedness principle, it suffices to show that
is bounded for every
. But since
,
this is in fact the case. Hence, there is a constant
such that
for every
. It follows:
.
Taking inf over
completes the proof of the spectral radius formula. Finally, suppose, on the contrary, that
is empty. Then for every
, the map
![{\displaystyle s\mapsto g((x-s)^{-1})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/443ad4e2bdc454af92d342406fd1ae1d15785b54)
is analytic in
. Since
, by Liouville's theorem, we must have:
. Hence,
for every
, a contradiction.
5 Corollary (Gelfand-Mazur theorem) If every nonzero element of
is invertible, then
is isomorphic to
.
Proof: Let
be a nonzero element. Since
is non-empty, we can then find
such that
is not invertible. But, by hypothesis,
is invertible, unless
.
Let
be a maximal ideal of a Banach algebra. (Such
exists by the usual argument involving Zorn's Lemma in abstract algebra). Since the complement of
consists of invertible elements,
is closed. In particular,
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} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a316eb9a7cf6d0cdda2825c271e8f70a3e255321)
Much more is true, actually. Let
be the set of all nonzero homeomorphism
. (The members of
are called characters.)
5 Theorem
is bijective to the set of all maximal ideals of
.
5 Lemma Let
. Then
is invertible if and only if
for every
5 Theorem
An involution is an anti-linear map
such that
. 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
(C*-identity)
From the C*-identity follows
,
for
and the same for
in place of
. In particular,
(if
exists). Furthermore, the
-identity is equivalent to the condition:
, for this and
implies
and so
.
For each
, let
be the linear span of
. In other words,
is the smallest C*-algebra that contains
. The crucial fact is that
is commutative. Moreover,
Theorem Let
be normal. Then
A state on
-algebra
is a positive linear functional f such that
(or equivalently
). Since
is convex and closed,
is weak-* closed. (This is Theorem 4.something.) Since
is contained in the unit ball of the dual of
,
is weak-* compact.
5 Theorem Every C^*-algebra
is *-isomorphic to
where
is the spectrum of
.
5 Theorem If
is isomorphic to
, then it follows that
and
are homeomorphic.
3 Lemma Let
be a continuous linear operator on a Hilbert space
. Then
if and only if
for all
.
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
be a normal operator. If
and
are distinct eigenvalues of
, then the respective eigenspaces of
and
are orthogonal to each other.
Proof: Let
be the identity operator, and
be arbitrary eigenvectors for
, respectively. Since the adjoint of
is
, we have:
.
That is,
, 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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8279bd6c904d59ee3833fa9717e4f9d8b83b68b7)
If
is nonzero, we must have
.
5 Exercise Let
be a Hilbert space with orthogonal basis
, and
be a sequence with
. Prove that there is a subsequence of
that converges weakly to some
and that
. (Hint: Since
is bounded, by Cantor's diagonal argument, we can find a sequence
such that
is convergent for every
.)
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)