Let V Be the Subspace of the Vector Space of All Real valued Continuous Functions That Has Basis

Trivial Subspace

These are the trivial subspace {[0,0,0]} = {0}, subspaces like Example 2 that can be geometrically represented as a line (thus "resembling" ℝ), subspaces like Example 1 that can be represented as a plane (thus "resembling" ℝ2), and the subspace ℝ3 itself.

From: Elementary Linear Algebra (Fourth Edition) , 2010

Finite Dimensional Vector Spaces

Stephen Andrilli , David Hecker , in Elementary Linear Algebra (Fourth Edition), 2010

Highlights

■

A subset of a vector space is a subspace if it is a vector space itself under the same operations.

■

The subset {0 } is a trivial subspace of any vector space.

■

Any subspace of a vector space $\mathcal{V}$ other than $\mathcal{V}$ itself is considered a proper subspace.

■

Familiar proper nontrivial subspaces of ${ℝ}^3$ are any line through the origin, any plane through the origin.

■

Familiar proper subspaces of the real-valued functions on $ℝ$ are ${\mathcal{P}}_n$ , $\mathcal{P}$ , all differentiable real-valued functions on $ℝ$ , all continuous real-valued functions on $ℝ$ .

■

Familiar proper subspaces of ${ℳ}_{nn}$ are ${\mathcal{U}}_n,{ℒ}_n,{\mathcal{D}}_n$ , the symmetric n × n matrices, the skew-symmetric n × n matrices.

■

A nonempty subset of a vector space is a subspace if it is closed under vector addition and scalar multiplication.

■

If a subset of a vector space does not contain the zero vector, it cannot be a subspace.

■

If a set of vectors is in a subspace, then any (finite) linear combination of those vectors is also in the subspace.

■

If λ is an eigenvalue for an n × n matrix A, then E _λ (eigenspace for λ) is a subspace of ${ℝ}^n$ .

■

The intersection of subspaces is a subspace.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780123747518000202

Finite Dimensional Vector Spaces

Stephen Andrilli , David Hecker , in Elementary Linear Algebra (Fifth Edition), 2016

Highlights

▪

A subset $\mathcal{W}$ of a vector space $\mathcal{V}$ is a subspace of $\mathcal{V}$ if $\mathcal{W}$ is a vector space itself under the same operations.

▪

Any subspace of a vector space $\mathcal{V}$ other than $\mathcal{V}$ itself is considered a proper subspace of $\mathcal{V}$ .

▪

The subset {0 } is a trivial subspace of any vector space.

▪

Familiar proper nontrivial subspaces of ${\mathbb{R}}^3$ are: any line through the origin, any plane through the origin.

▪

Familiar proper subspaces of the real-valued functions on $\mathbb{R}$ are: ${\mathcal{P}}_n$ , $\mathcal{P}$ , all differentiable real-valued functions on $\mathbb{R}$ , all continuous real-valued functions on $\mathbb{R}$ .

▪

Familiar proper subspaces of ${\mathcal{M}}_{nn}$ are: ${\mathcal{U}}_n$ , ${\mathcal{L}}_n$ , ${\mathcal{D}}_n$ , the symmetric n × n matrices, the skew-symmetric n × n matrices.

▪

A nonempty subset $\mathcal{W}$ of a vector space $\mathcal{V}$ is a subspace of $\mathcal{V}$ if $\mathcal{W}$ is closed under addition and scalar multiplication.

▪

If a subset S of a vector space $\mathcal{V}$ does not contain the zero vector 0, then S cannot be a subspace of $\mathcal{V}$ .

▪

If T is any set of vectors in a subspace $\mathcal{W}$ , then any linear combination of the vectors in T is also in $\mathcal{W}$ .

▪

If λ is an eigenvalue for an n × n matrix A, then E _λ (the eigenspace for λ) is a subspace of ${\mathbb{R}}^n$ .

▪

If ${\mathcal{W}}_1$ and ${\mathcal{W}}_2$ are subspaces of a vector space $\mathcal{V}$ , then ${\mathcal{W}}_1\cap {\mathcal{W}}_2,$ the intersection of these subspaces, is also a subspace of $\mathcal{V}$ .

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780128008539000049

Orthogonality

Stephen Andrilli , David Hecker , in Elementary Linear Algebra (Fifth Edition), 2016

Orthogonal Projection Onto a Subspace

Next, we present the Projection Theorem, a generalization of Theorem 1.11. Recall from Theorem 1.11 that every nonzero vector in ${\mathbb{R}}^n$ can be decomposed into the sum of two vectors, one parallel to a given vector a and another orthogonal to a.

Theorem 6.15 (Projection Theorem)

Let $\mathcal{W}$ be a subspace of ${\mathbb{R}}^n$ . Then every vector $\mathbf{v}\in {\mathbb{R}}^n$ can be expressed in a unique way as w ₁ +w ₂, where ${\mathbf{w}}_1\in \mathcal{W}$ and ${\mathbf{w}}_2\in {\mathcal{W}}^{\perp }.$

Proof

Let $\mathcal{W}$ be a subspace of ${\mathbb{R}}^n$ , and let $\mathbf{v}\in {\mathbb{R}}^n$ . We first show that v can be expressed as w ₁ +w ₂, where ${\mathbf{w}}_1\in \mathcal{W}$ , ${\mathbf{w}}_2\in {\mathcal{W}}^{\perp }.$ Then we will show that there is a unique pair w ₁,w ₂ for each v.

Let {u ₁,…,u _k } be an orthonormal basis for $\mathcal{W}$ . Expand {u ₁,…,u _k } to an orthonormal basis {u ₁,…,u _k ,u _k+1,…,u _n } for ${\mathbb{R}}^n$ . Then by Theorem 6.3, v = (v ⋅u ₁)u ₁ + ⋯ + (v ⋅u _n )u _n . Let w ₁ = (v ⋅u ₁)u ₁ + ⋯ + (v ⋅u _k )u _k and w ₂ = (v ⋅u _k+1)u _k+1 + ⋯ + (v ⋅u _n )u _n . Clearly, v =w ₁ +w ₂. Also, Theorem 6.12 implies that ${\mathbf{w}}_1\in \mathcal{W}$ and w ₂ is in ${\mathcal{W}}^{\perp }$ .

Finally, we want to show uniqueness of decomposition. Suppose that v =w ₁ +w ₂ and $\mathbf{v}={\mathbf{w}}_1^{\prime }+{\mathbf{w}}_2^{\prime },$ where ${\mathbf{w}}_1,{\mathbf{w}}_1^{\prime }\in \mathcal{W}$ and w ₂, ${\mathbf{w}}_2^{\prime }\in {\mathcal{W}}^{\perp }$ . We want to show that ${\mathbf{w}}_1={\mathbf{w}}_1^{\prime }$ and ${\mathbf{w}}_2={\mathbf{w}}_2^{\prime }.$ Now, ${\mathbf{w}}_1-{\mathbf{w}}_1^{\prime }={\mathbf{w}}_2^{\prime }-{\mathbf{w}}_2$ (why?). Also, ${\mathbf{w}}_1-{\mathbf{w}}_1^{\prime }\in \mathcal{W}$ , but ${\mathbf{w}}_2^{\prime }-{\mathbf{w}}_2\in {\mathcal{W}}^{\perp }.$ Thus, ${\mathbf{w}}_1-{\mathbf{w}}_1^{\prime }={\mathbf{w}}_2^{\prime }-{\mathbf{w}}_2\in \mathcal{W}\cap {\mathcal{W}}^{\perp }.$ By Theorem 6.11, ${\mathbf{w}}_1-{\mathbf{w}}_1^{\prime }={\mathbf{w}}_2^{\prime }-{\mathbf{w}}_2=0.$ Hence, ${\mathbf{w}}_1={\mathbf{w}}_1^{\prime }$ and ${\mathbf{w}}_2={\mathbf{w}}_2^{\prime }.$

We give a special name to the vector w ₁ in the proof of Theorem 6.15.

Definition

Let $\mathcal{W}$ be a subspace of ${\mathbb{R}}^n$ with orthonormal basis {u ₁,…,u _k }, and let $\mathbf{v}\in {\mathbb{R}}^n.$ Then the orthogonal projection of v onto $\mathcal{W}$ is the vector

${\mathbf{proj}}_{\mathcal{W}}\mathbf{v}=(\mathbf{v}\cdot {\mathbf{u}}_1){\mathbf{u}}_1+\cdots +(\mathbf{v}\cdot {\mathbf{u}}_k){\mathbf{u}}_k.$

If $\mathcal{W}$ is the trivial subspace of ${\mathbb{R}}^n$ , then ${\mathbf{proj}}_{\mathcal{W}}\mathbf{v}=0$ .

Notice that the choice of orthonormal basis for $\mathcal{W}$ in this definition does not matter. This is because if v is any vector in ${\mathbb{R}}^n$ , Theorem 6.15 asserts there is a unique expression w ₁ +w ₂ for v with ${\mathbf{w}}_1\in \mathcal{W}$ , ${\mathbf{w}}_2\in {\mathcal{W}}^{\perp },$ and we see from the proof of the theorem that ${\mathbf{w}}_1={\mathbf{proj}}_{\mathcal{W}}\mathbf{v}$ . Hence, if {z ₁,…,z _k } is any other orthonormal basis for $\mathcal{W}$ , then ${\mathbf{proj}}_{\mathcal{W}}\mathbf{v}$ is equal to (v⋅z ₁)z ₁ + ⋯ + (v⋅z _k )z _k as well. This fact is illustrated in the next example.

Example 6

Consider the orthonormal subset

$B=\left\{{\mathbf{u}}_1,{\mathbf{u}}_2\right\}=\left\{\left[\frac{8}{9},-\frac{1}{9},-\frac{4}{9}\right],\left[\frac{4}{9},\frac{4}{9},\frac{7}{9}\right]\right\}$

of ${\mathbb{R}}^3$ , and let $\mathcal{W}=\text{span}(B)$ . Notice that B is an orthonormal basis for $\mathcal{W}$ .

Also consider the orthogonal set S = {[4,1,1], [4,−5,−11]}. Now since

$\begin{array}{ll}\hfill [4,1,1]& =3\left[\frac{8}{9},-\frac{1}{9},-\frac{4}{9}\right]+3\left[\frac{4}{9},\frac{4}{9},\frac{7}{9}\right]\hfill \\ \hfill \text{and}[4,-5,-11]& =9\left[\frac{8}{9},-\frac{1}{9},-\frac{4}{9}\right]-9\left[\frac{4}{9},\frac{4}{9},\frac{7}{9}\right],\hfill \end{array}$

S is an orthogonal subset of $\mathcal{W}$ . Since $|S|=dim(\mathcal{W})$ , S is also an orthogonal basis for $\mathcal{W}$ . Hence, after normalizing the vectors in S, we obtain the following second orthonormal basis for $\mathcal{W}$ :

$C=\left\{{\mathbf{z}}_1,{\mathbf{z}}_2\right\}=\left\{\left[\frac{4}{3\sqrt{2}},\frac{1}{3\sqrt{2}},\frac{1}{3\sqrt{2}}\right],\left[\frac{4}{9\sqrt{2}},-\frac{5}{9\sqrt{2}},-\frac{11}{9\sqrt{2}}\right]\right\}.$

Let v = [1,2,3]. We will verify that the same vector for ${\mathbf{proj}}_{\mathcal{W}}\mathbf{v}$ is obtained whether B = {u ₁,u ₂} or C = {z ₁,z ₂}is used as the orthonormal basis for $\mathcal{W}$ . Now, using B yields

$\left(\mathbf{v}\cdot {\mathbf{u}}_1\right){\mathbf{u}}_1+\left(\mathbf{v}\cdot {\mathbf{u}}_2\right){\mathbf{u}}_2=-\frac{2}{3}\left[\frac{8}{9},-\frac{1}{9},-\frac{4}{9}\right]+\frac{11}{3}\left[\frac{4}{9},\frac{4}{9},\frac{7}{9}\right]=\left[\frac{28}{27},\frac{46}{27},\frac{85}{27}\right].$

Similarly, using C gives

$\begin{array}{ll}\hfill \left(\mathbf{v}\cdot {\mathbf{z}}_1\right){\mathbf{z}}_1+\left(\mathbf{v}\cdot {\mathbf{z}}_2\right){\mathbf{z}}_2& =\frac{3}{\sqrt{2}}\left[\frac{4}{3\sqrt{2}},\frac{1}{3\sqrt{2}},\frac{1}{3\sqrt{2}}\right]+\left(-\frac{13}{3\sqrt{2}}\right)\left[\frac{4}{9\sqrt{2}},-\frac{5}{9\sqrt{2}},-\frac{11}{9\sqrt{2}}\right]\hfill \\ \hfill & =\left[\frac{28}{27},\frac{46}{27},\frac{85}{27}\right].\hfill \end{array}$

Hence, with either orthonormal basis we obtain ${\mathbf{proj}}_{\mathcal{W}}\mathbf{v}=\left[\frac{28}{27},\frac{46}{27},\frac{85}{27}\right].$

The proof of Theorem 6.15 illustrates the following:

Corollary 6.16

If $\mathcal{W}$ is a subspace of ${\mathbb{R}}^n$ and $\mathbf{v}\in {\mathbb{R}}^n,$ then there are unique vectors w ₁ and w ₂, with ${\mathbf{w}}_1\in \mathcal{W}$ and ${\mathbf{w}}_2\in {\mathcal{W}}^{\perp }$ such that v =w ₁ +w ₂. Moreover, ${\mathbf{w}}_1={\mathbf{proj}}_{\mathcal{W}}\mathbf{v}$ and ${\mathbf{w}}_2=\mathbf{v}-{\mathbf{proj}}_{\mathcal{W}}\mathbf{v}={\mathbf{proj}}_{{\mathcal{W}}^{\perp }}\mathbf{v}.$

The vector w ₁ is the generalization of the projection vector proj _a b from Section 1.2 (see Exercise 16).

Example 7

Let $\mathcal{W}$ be the subspace of ${\mathbb{R}}^3$ whose vectors (beginning at the origin) lie in the plane $\mathcal{L}$ with equation 2x + y + z = 0. Let v = [−6,10,5]. (Notice that $\mathbf{v}\notin \mathcal{W}$ .) We will find ${\mathbf{proj}}_{\mathcal{W}}\mathbf{v}$ .

First, notice that [1,0,−2] and [0,1,−1] are two linearly independent vectors in $\mathcal{W}$ . (To find the first vector, choose x = 1,y = 0, and for the other, let x = 0 and y = 1.) Using the Gram-Schmidt Process on these vectors, we obtain the orthogonal basis {[1,0,−2],[−2,5,−1]}for $\mathcal{W}$ (verify!). After normalizing, we have the orthonormal basis {u ₁,u ₂}for $\mathcal{W}$ , where

${\mathbf{u}}_1=\left[\frac{1}{\sqrt{5}},0,-\frac{2}{\sqrt{5}}\right]\text{and}{\mathbf{u}}_2=\left[-\frac{2}{\sqrt{30}},\frac{5}{\sqrt{30}},-\frac{1}{\sqrt{30}}\right].$

Now,

$\begin{array}{ll}\hfill {\mathbf{proj}}_{\mathcal{W}}\mathbf{v}& =\left(\mathbf{v}\cdot {\mathbf{u}}_1\right){\mathbf{u}}_1+\left(\mathbf{v}\cdot {\mathbf{u}}_2\right){\mathbf{u}}_2\hfill \\ \hfill & =-\frac{16}{\sqrt{5}}\left[\frac{1}{\sqrt{5}},0,-\frac{2}{\sqrt{5}}\right]+\frac{57}{\sqrt{30}}\left[-\frac{2}{\sqrt{30}},\frac{5}{\sqrt{30}},-\frac{1}{\sqrt{30}}\right]\hfill \\ \hfill & =\left[-\frac{16}{5},0,\frac{32}{5}\right]+\left[-\frac{114}{30},\frac{285}{30},-\frac{57}{30}\right]\hfill \\ \hfill & =\left[-7,\frac{19}{2},\frac{9}{2}\right].\hfill \end{array}$

Notice that this vector is in $\mathcal{W}$ . Finally, $\mathbf{v}-{\mathbf{proj}}_{\mathcal{W}}\mathbf{v}=\left[1,\frac{1}{2},\frac{1}{2}\right],$ which is indeed in ${\mathcal{W}}^{\perp }$ because it is orthogonal to both u ₁ and u ₂ (verify!). Hence, we have decomposed v = [−6,10,5] as the sum of two vectors $\left[-7,\frac{19}{2},\frac{9}{2}\right]$ and $\left[1,\frac{1}{2},\frac{1}{2}\right],$ where the first is in $\mathcal{W}$ and the second is in ${\mathcal{W}}^{\perp }$ .

We can think of the orthogonal projection vector ${\mathbf{proj}}_{\mathcal{W}}\mathbf{v}$ in Example 7 as the "shadow" that v casts on the plane $\mathcal{L}$ as light falls directly onto $\mathcal{L}$ from a light source above and parallel to $\mathcal{L}$ . This concept is illustrated in Figure 6.4.

Figure 6.4. The orthogonal projection vector $\left[-7,\frac{19}{2},\frac{9}{2}\right]$ of v = [−6,10,5] onto the plane 2x + y + z = 0, pictured as a shadow cast by v from a light source above and parallel to the plane

There are two special cases of Corollary 6.16. First, if $\mathbf{v}\in \mathcal{W}$ , then ${\mathbf{proj}}_{\mathcal{W}}\mathbf{v}$ simply equals v itself. Also, if $\mathbf{v}\in {\mathcal{W}}^{\perp }$ , then ${\mathbf{proj}}_{\mathcal{W}}\mathbf{v}$ equals 0. These results are left as Exercise 13.

The next theorem assures us that orthogonal projection onto a subspace of ${\mathbb{R}}^n$ is a linear operator on ${\mathbb{R}}^n$ . The proof is left as Exercise 19.

Theorem 6.17

Let $\mathcal{W}$ be a subspace of ${\mathbb{R}}^n.$ Then the mapping L: ${\mathbb{R}}^n\to {\mathbb{R}}^n$ given by $L(\mathbf{v})={\mathbf{proj}}_{\mathcal{W}}\mathbf{v}$ is a linear operator with $ker(L)={\mathcal{W}}^{\perp }.$

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780128008539000062

Topological Algebras with Involution

In North-Holland Mathematics Studies, 2005

4.(1) Non–Banach Arens–Michael Algebras

Let A be an algebra. A modular left ideal is a left ideal, for which there is an element u ∈ A such that x - xu ∈ M, for every x ∈ A; u is often called right modular identity of M. A modular right ideal is defined similarly. An ideal is modular if it is a modular left and right ideal.

Let φ be a character of A and $M\equiv \ker \left(\phi \right)$ , there is x ∈ A with $\phi \left(x\right)\equiv \lambda \ne 0$ . Take $u\equiv {\lambda }^{-1}x$ ; then it is easily checked that u is a (right, respectively left) modular identity for M or, which amounts to the same, is an identity for the quotient algebra A/M. Hence, M is a modular ideal. If A is unital, every (left, right, 2-sided) ideal is a modular (left, right, 2-sided) ideal.

4.21 Definition

Let A be an algebra, X a vector space and L(X) the algebra of all linear operators from A in X. A representation of A on X, is a morphism ϕ from A in L(X). A nonzero representation ϕ of A in X is called algebraically irreducible if the only (linear) subspaces of X invariant under the operators ϕ(x), x ∈ X , are the trivial subspace {0} and X itself. The (Jacobson) radical of A, denoted by J _A , is the intersection of the kernels of all algebraically irreducible representations of A. The algebra A is called semisimple respectively radical, whenever J _A = {0} respectively J _A = A.

It is proved that for a given algebra A, the radical J _A is the intersection of all maximal modular left ideals of A [72, p.124, Proposition 14(ii)]. In particular (ibid., p. 125, Proposition 16(ii)),

(4.49) $J_A=\left\{x\in A:yx\in G_A^q,\forall y\in A\right\}.$

Concerning the radical of the unitization A ₁ (see 3.(3)) of A, one has (see [191, Corollary (I.3.54)]) that

(4.50) $J_A=J_{A_1}.$

For any algebra A, J _A is an ideal and, in particular, a radical algebra. On the other hand, A/J _A is a semisimple algebra (see, for instance [191, pp. 93–95]) and the same is true for all the examples in 2.4; see, in addition, Corollary 4.23. Furthermore, an algebra with zero multiplication is a radical algebra. For other examples of radical algebras and further information about them see [102, p.155], [104, p.68, 1.5 and p. 178] and [191, p.98].

4.22 Proposition

Let A[τ] be a topological algebra with $\mathfrak{M}\left(A\right)\ne \varnothing$ . Then, the following statements hold:

(1)

$J_A\subseteq \cap \left\{\ker \left(\phi \right):\phi \in \mathfrak{M}\left(A\right)\right\}$ .

(2)

The Gel'fand map $\mathcal{G}$ of A[τ] (see (4.30)) is injective if and only if ${\displaystyle \cap \left\{\ker \left(\phi \right):\phi \in \mathfrak{M}\left(A\right)\right\}=\left\{0\right\}}$ .

(3)

When A[τ_Γ] is a commutative m–convex Q–algebra (for the last term, see Definition 6.1 in Section 6), (1) becomes equality; namely, $J_A=\left\{\ker \left(\phi \right):\phi \in \mathfrak{M}\left(A\right)\right\}$ .

Proof. The assertions follow easily from the very definitions and the discussion above. For (3) also see Section 6, Theorem 6.5 and Proposition 6.10(4) and note that each maximal modular ideal in A[τ] is realized by the kernel of a character of A[τ] [262, p.72, Corollary 7.3].

For the relationship of the radical J _A with the *–radical $R_A^{\ast }$ of an involutive topological algebra A[τ], see Corollary 22.16 and Proposition 22.21.

4.23 Corollary

Every topological algebra A[τ] with $\mathfrak{M}\left(A\right)\ne \varnothing$ and injective Gel'fand map is commutative and semisimple.

The properties of the Jacobson radical we list below (Proposition 4.24) are proved in the book of F.F. Bonsall and J. Duncan [72, Proposition 25.1] in the context of Banach algebras. As A. Mallios noticed, these very proofs are of purely algebraic form. Because of the frequent use of them, throughout this book, we present their proofs.

4.24 Proposition

For a given algebra A the following statements hold:

(1)

$J_A\subseteq \left\{x\in A:r_A\left(x\right)=0\right\}$ .

(2)

$\left\{x\in Awith{r}_A\left(yx\right)=0or{r}_A\left(xy\right)=0,\forall y\in A\right\}\subseteq J_A$ .

(3)

$J_A=\left\{x\in A:r_A\left(x\right)=0\right\}$ , for every commutative advertibly complete m–convex algebra A[τ_Γ] (for the term advertibly complete algebra, see Section 6 Definition 6.1).

Proof. (1) Let x ∈ J _A . Then, from (4.49) and (4.50) we conclude that ${\lambda }^{-1}x\in G_{A_1}^q$ , for every $0\ne \lambda \in \mathbb{C}$ , whence (also see (4.7)) $s{p}_A\left(x\right)=\left\{0\right\}$ ; therefore $r_A\left(x\right)=0$ .

(2) Since, $s{p}_A\left(xy\right)\cup \left\{0\right\}=s{p}_A\left(yx\right)\cup \left\{0\right\}$ , for any x, y ∈ A [191, Proposition (II. 1.8)], one clearly gets that $r_A\left(xy\right)=r_A\left(yx\right)$ , for all x, y ∈ A. So it suffices to consider the case x ∈ A with $r_A\left(yx\right)=0$ , for every y ∈ A. Then, $yx=G_A^q$ , for each y ∈ A, therefore x ∈ J _A from (4.49).

(3) In view of (1) it suffices to show that $J\equiv \left\{x\in A:r_A\left(x\right)=0\right\}\subseteq J_A$ . From Theorem 4.6(4), (7) and (8) one has that J is an ideal. Hence, x ∈ J implies yx ∈ J, for every y ∈ A. The assertion now follows from (2).

The next result is contained in [72, Lemma 37.2].

4.25 Corollary

Every *–subalgebra B of $ℒ\left(H\right)$ , H a Hilbert space, is semisimple.

Proof. Let T ∈ J _B . Then $T^{\ast }T\in J_B$ , therefore $r_B\left(T^{\ast }T\right)=0$ from Proposition 4.24(1). Hence, (see, e.g., [279, p.37, Theorem 2.1.1] and (4.10))

${\Vert T\Vert }^2=\Vert T^{\ast }T\Vert =r_{ℒ\left(H\right)}\left(T^{\ast }T\right)\le r_B\left(T^{\ast }T\right)=0,$

(with $\Vert \cdot \Vert$ the operator norm on $ℒ\left(H\right)$ ), which implies T = 0.

4.26 Theorem

None of the Arens–Michael algebras ${\mathcal{C}}^{\infty }\left[0,1\right],{\mathcal{C}}^{\infty }\left({\mathbb{R}}^n\right),\mathfrak{D}\left({\mathbb{R}}^n\right),S\left({\mathbb{R}}^n\right),\mathcal{O}\left(\mathbb{C}\right)and{\mathcal{C}}_c\left(\mathbb{R}\right)$ (hence also ${\mathbb{C}}^{\mathbb{N}}$ ) may be a Banach algebra under its usual topology (see Examples 2.4,(1)–(6)).

Proof. The main argument for these proofs is the Singer–Wermer–Johnson theorem [72, Theorems 18.16 and 18.21], according to which a commutative semisimple Banach algebra has only trivial derivations. We recall that a derivation on an algebra A is a linear map D : A → A with the property

$D\left(xy\right)=xD\left(y\right)+D\left(x\right)y,\forall x,y\in A.$

For the algebra ${\mathcal{C}}^{\infty }\left[0,1\right]$ , a direct proof based on the preceding argument is given in [72, Corollary 18.22]. An elegant analytic proof can be found in the book of B. Aupetit [29, p.76, Corollary 4.1.12]. In this regard, also see Remark 4.27(2).

Of course, the situation for the algebra ${\mathcal{C}}^{\infty }\left({\mathbb{R}}^n\right)$ (or any algebra ${\mathcal{C}}^{\infty }\left(X\right)$ with X a 2nd countable n–dimensional ${\mathcal{C}}^{\infty }-\text{manifold}$ [262, pp. 129–131]) is exactly the same. That is, since the spectrum of ${\mathcal{C}}^{\infty }\left({\mathbb{R}}^n\right)$ is exhausted by the point evaluations ${\delta }_x,x\in {\mathbb{R}}^n$ (see Example 4.20(2)), one has that $\cap \left\{\ker \left({\delta }_x\right):x\in {\mathbb{R}}^n\right\}=\left\{0\right\}$ , therefore from Proposition 4.22(1) ${\mathcal{C}}^{\infty }\left({\mathbb{R}}^n\right)$ is semisimple. Thus, if ${\mathcal{C}}^{\infty }\left({\mathbb{R}}^n\right)$ was a Banach algebra its only derivation would be the zero one, which is clearly not true because of the maps

$f\mapsto \frac{\partial f}{\partial x_i}:{\mathcal{C}}^{\infty }\left({\mathbb{R}}^n\right)\to {\mathcal{C}}^{\infty }\left({\mathbb{R}}^n\right),i=1,\dots ,n.$

The same proof applies for the algebras $\mathfrak{D}\left({\mathbb{R}}^n\right),S\left({\mathbb{R}}^n\right)$ . In both cases semi–simplicity results from the fact that the corresponding spectrum contains all point evaluations. On the other hand, (see (2.25)) $\mathrm{supp}\left(\frac{\partial f}{\partial x_i}\right)\subseteq \mathrm{supp}\left(f\right)$ , for all $f\in \mathfrak{D}\left({\mathbb{R}}^n\right)$ , so that $\frac{\partial f}{\partial x_i}\in \mathfrak{D}\left({\mathbb{R}}^n\right)$ , for every $f\in \mathfrak{D}\left({\mathbb{R}}^n\right),i=1,\dots ,n$ . The same assertion is valid for $S\left({\mathbb{R}}^n\right)$ as follows from (2.32).

Concerning the algebra $\mathcal{O}\left(\mathbb{C}\right)$ one gets semisimplicity as in the case of ${\mathcal{C}}^{\infty }\left({\mathbb{R}}^n\right)$ , because of (4.48), while if $f\in \mathcal{O}\left(\mathbb{C}\right)$ and f′ denotes the first derivative of f, then $f^{\prime }\in \mathcal{O}\left(\mathbb{C}\right)$ [331, p.224, Corollary]. Hence, the correspondence $f\mapsto f^{\prime }:\mathcal{O}\left(\mathbb{C}\right)\to \mathcal{O}\left(\mathbb{C}\right)$ is a nonzero derivation.

The result for the algebra ${\mathcal{C}}_c\left(\mathbb{R}\right)$ is taken in a much simpler way, because if ${\mathcal{C}}_c\left(\mathbb{R}\right)$ was a Banach algebra, since it is also commutative and unital, its spectrum would be a compact space [72, 191, 327] and this contradicts (4.46).

4.27 Remarks

(1) Another example of an Arens–Michael algebra that cannot be topologized as a Banach algebra, is the cartesian product $B={\displaystyle \prod A_{\lambda }}$ of an infinite family ${\left(A_{\lambda }\right)}_{\lambda \in \Lambda }$ of Banach algebras, under the product topology (see Example 7.6(2)). This follows from a more general result according to which the cartesian product of infinitely many normed spaces, cannot be a normed space under the product topology [235, p.150, (7)].

(2) In 1998, G. Mocanu gave a version of a result of B. Aupetit [28, Theorem 1] related to the uniqueness of the complete norm in semisimple Banach algebras (see beginning of Section 2), in the context of commutative m ^*–convex Q–algebras (see [274, Theorem 4]). As a direct corollary he takes the uniqueness of the topology in commutative semisimple Fréchet Q–algebras, a known result due to R.L. Carpender [90, Theorem 5] since 1971, shown without the property Q. Based on the latter G. Mocanu proves that ${\mathcal{C}}^{\infty }\left[0,1\right]$ cannot be a Banach algebra under its usual topology [274, Corollary 10].

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/S0304020805800246

pringleopre1960.blogspot.com

Source: https://www.sciencedirect.com/topics/mathematics/trivial-subspace

Share this post