$C^{*}$-algebras isomorphically representable on $l^{p}$

TL;DR

This paper characterizes when $C^{*}$-algebras can be represented on $l^{p}$ spaces for $p eq 2$, showing such representations are tied to residual finite dimensionality and involve a specific compactness property of homomorphisms.

Contribution

It establishes a characterization of $C^{*}$-algebras representable on $l^{p}$ spaces via a compactness property and residual finite dimensionality.

Findings

Homomorphisms into $B(l^{p}(J))$ satisfy a compactness property.

A $C^{*}$-algebra is isomorphic to a subalgebra of $B(l^{p}(J))$ iff it is residually finite dimensional.

Representation on $l^{p}$ spaces is characterized by residual finite dimensionality.

Abstract

Let $p\in(1,\infty)\backslash\{2\}$. We show that every homomorphism from a $C^{*}$-algebra $\mathcal{A}$ into $B(l^{p}(J))$ satisfies a compactness property where $J$ is any set. As a consequence, we show that a $C^{*}$-algebra $\mathcal{A}$ is isomorphic to a subalgebra of $B(l^{p}(J))$, for some set $J$, if and only if $\mathcal{A}$ is residually finite dimensional.

Equations

$$\frac{1}{C}\Vert a\Vert \le \Vert \varphi (a)\Vert \le C\Vert a\Vert ,$$

$$g_n(x)=\{\begin{array}{ll}\frac{1}{\sqrt{x}},& \frac{1}{n}\le x\le 1\\ n\sqrt{n}x,& 0\le x\le \frac{1}{n}\end{array}.$$

$$x{g}_n(x{)}^2=\{\begin{array}{ll}1,& \frac{1}{n}\le x\le 1\\ n^3{x}^3,& 0\le x\le \frac{1}{n}\end{array}.$$

$$x{g}_n(x)=\{\begin{array}{ll}\sqrt{x},& \frac{1}{n}\le x\le 1\\ n\sqrt{n}{x}^2,& 0\le x\le \frac{1}{n}\end{array}$$

$$\omega (a^2)\le \omega (a{)}^{\frac{2}{3}}\omega (a^4{)}^{\frac{1}{3}}.$$

$$\omega (f(a))=\int f(x)d\mu (x),$$

$$\omega (a^2)=\int x^{2}d\mu (x)\le {\left(\int xd\mu (x)\right)}^{\frac{2}{3}}{\left(\int x^{4}d\mu (x)\right)}^{\frac{1}{3}}=\omega (a{)}^{\frac{2}{3}}\omega (a^4{)}^{\frac{1}{3}}.$$

$$\omega (a)=y_0^{\ast }(\varphi (a)x_0),$$

$$a_{\delta }=\left|\sum _{k=1}^n{\delta }_k{b}_k\right|\in \mathcal{A}.$$

$$\omega (a_{\delta }^2)\le \omega (a_{\delta }{)}^{\frac{2}{3}}\omega (a_{\delta }^4{)}^{\frac{1}{3}}.$$

$$\mathbb{E}\omega (a_{\delta }^2)\le [\mathbb{E}\omega (a_{\delta }){]}^{\frac{2}{3}}[\mathbb{E}\omega (a_{\delta }^4){]}^{\frac{1}{3}},$$

$$\mathbb{E}\omega (a_{\delta }^2)$$

$$n\gamma \le [\mathbb{E}\omega (a_{\delta }){]}^{\frac{2}{3}}[\mathbb{E}\omega (a_{\delta }^4){]}^{\frac{1}{3}}.$$

$$a_{\delta }^4={\left[{\left(\sum _{k=1}^n{\delta }_k{b}_k\right)}^{\ast }\left(\sum _{k=1}^n{\delta }_k{b}_k\right)\right]}^2=\sum _{1\le k_1,\dots ,k_4\le n}{\delta }_{k_1}{\delta }_{k_2}{\delta }_{k_3}{\delta }_{k_4}{b}_{k_1}^{\ast }{b}_{k_2}{b}_{k_3}^{\ast }{b}_{k_4}.$$

$$\mathbb{E}\omega (a_{\delta }^4)=\sum _{1\le k_1,\dots ,k_4\le n}\mathbb{E}({\delta }_{k_1}{\delta }_{k_2}{\delta }_{k_3}{\delta }_{k_4})\omega (b_{k_1}^{\ast }{b}_{k_2}{b}_{k_3}^{\ast }{b}_{k_4})\le \sum _{1\le k_1,\dots ,k_4\le n}\mathbb{E}({\delta }_{k_1}{\delta }_{k_2}{\delta }_{k_3}{\delta }_{k_4}).$$

$$(k_1=k_2\text{ and }{k}_3=k_4)\text{ or }(k_1=k_3\text{ and }{k}_2=k_4)\text{ or }(k_1=k_4\text{ and }{k}_2=k_3).$$

$$\mathbb{E}\omega (a_{\delta })\ge \frac{{\gamma }^{\frac{3}{2}}}{3^{\frac{1}{2}}}{n}^{\frac{1}{2}}.$$

$$\omega (a_{\delta })=\omega \left(\left|\sum _{k=1}^n{\delta }_k{b}_k\right|\right)\le \underset{c\in \mathcal{A},\Vert c\Vert \le 1}{\sup }\left|\omega \left(c\sum _{k=1}^n{\delta }_k{b}_k\right)\right|.$$

$$\left|\omega \left(c\sum _{k=1}^n{\delta }_k{b}_k\right)\right|$$

$$\frac{{\gamma }^{\frac{3}{2}}}{3^{\frac{1}{2}}}{n}^{\frac{1}{2}}\le \Vert y_0^{\ast }\Vert \Vert \varphi \Vert (\Vert \varphi \Vert \Vert x_0\Vert n^{\frac{1}{p}}+1).$$

$$\omega (a)=y_0^{\ast }(\varphi (a)x_0),$$

$$a\cdot b=ba.$$

$${\varphi }_1(a)y^{\ast }=y^{\ast }\circ \varphi (a),$$

$${\omega }_1(a)=\omega (a)=x_0^{\ast \ast }(\varphi (a)y_0^{\ast }),$$

$$\omega (a)=y_0^{\ast }(\varphi (a)x_0),a\in \mathcal{A},$$

$$\underset{n\to \infty }{\lim }\Vert \varphi (v_n)x_0\Vert =\underset{u\in \mathcal{U}(\mathcal{A})}{\sup }\Vert \varphi (u)x_0\Vert .$$

$$\omega (a)=y_0^{\ast }(\varphi (a)x_0)=\underset{n\to \infty }{\lim }{x}_n^{\ast }(\varphi (v_{n}a)x_0),$$

$$\omega (1)=\underset{n\to \infty }{\lim }{x}_n^{\ast }(\varphi (v_n)x_0)=\underset{n\to \infty }{\lim }\Vert \varphi (v_n)x_0\Vert =\underset{u\in \mathcal{U}(\mathcal{A})}{\sup }\Vert \varphi (u)x_0\Vert ,$$

$$|\omega (u_0)|=\underset{n\to \infty }{\lim }|x_n^{\ast }(\varphi (v_n{u}_0)x_0)|\le \underset{u\in \mathcal{U}(\mathcal{A})}{\sup }\Vert \varphi (u)x_0\Vert .$$

$$\Vert a_k\Vert \ge \frac{\Vert \varphi (a_k)x_0\Vert }{\Vert \varphi \Vert \Vert x_0\Vert }\ge \frac{ϵ}{\Vert \varphi \Vert \Vert x_0\Vert },$$

Full text

1. Introduction

For $1\leq p<\infty$ and a set $J$, let $l^{p}(J)$ be the space $\{f\colon J\to\mathbb{C}\colon\sum_{j\in J}|f(j)|^{p}<\infty\}$ with norm $\|f\|=(\sum_{j\in J}|f(j)|^{p})^{\frac{1}{p}}$. Two Banach algebras $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$ are isomorphic if there exist a bijective homomorphism $\phi\colon\mathcal{A}_{1}\to\mathcal{A}_{2}$ and $C>0$ such that

[TABLE]

for all $a\in\mathcal{A}_{1}$. The algebras $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$ are isometrically isomorphic if moreover, $\phi$ can be chosen so that $\|\phi(a)\|=\|a\|$ for all $a\in\mathcal{A}_{1}$.

Gardella and Thiel [2] showed that for $p\in[1,\infty)\backslash\{2\}$, a $C^{*}$-algebra $\mathcal{A}$ is isometrically isomorphic to a subalgebra of $B(l^{p}(J))$, for some set $J$, if and only if $\mathcal{A}$ is commutative. So it is natural to consider the question whether this result holds if we relax the condition of isometrically isomorphic to isomorphic. In this paper, we show that for $p\in(1,\infty)\backslash\{2\}$, a $C^{*}$-algebra $\mathcal{A}$ is isomorphic to a subalgebra of $B(l^{p}(J))$, for some set $J$, if and only if $\mathcal{A}$ is residually finite dimensional (Corollary 2.2). We prove this by showing that every homomorphism from a $C^{*}$-algebra $\mathcal{A}$ into $B(l^{p}(J))$ satisfies a compactness property (Theorem 2.1).

The proofs of the main results Theorem 2.1 and Corollary 2.2 in this paper are quite different from the proof of Gardella-Thiel’s result. Lamperti’s characterization [5] of isometries on $L^{p}$, for $p\neq 2$, plays a crucial role in the proof of Gardella-Thiel’s result, while uniform convexity of $l^{p}$, for $1<p<\infty$, and an argument in probability that imitates the proof of Khintchine’s inequality [6, Theorem 2.b.3], for $p=1$, are used in the proof of Theorem 2.1.

2. Main results and proofs

Throughout this paper, the scalar field is $\mathbb{C}$; for algebras $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$, a homomorphism $\phi\colon\mathcal{A}_{1}\to\mathcal{A}_{2}$ is a bounded linear map such that $\phi(a_{1}a_{2})=\phi(a_{1})\phi(a_{2})$ for all $a_{1},a_{2}\in\mathcal{A}$; for an element $a$ of a $C^{*}$-algebra, $|a|=\sqrt{a^{*}a}$; the algebra of bounded linear operators on a Banach space $\mathcal{X}$ is denoted by $B(\mathcal{X})$ and the dual of $\mathcal{X}$ is denoted by $\mathcal{X}^{*}$; for $1\leq p\leq\infty$, the $l^{p}$ direct sum of Banach spaces $\mathcal{X}_{\alpha}$, for $\alpha\in\Lambda$, is denoted by $(\oplus_{\alpha\in\Lambda}\mathcal{X}_{\alpha})_{l^{p}}$. Two Banach spaces $\mathcal{X}_{1}$ and $\mathcal{X}_{2}$ are isomorphic if there is an invertible operator $S\colon\mathcal{X}_{1}\to\mathcal{X}_{2}$. A $C^{*}$-algebra $\mathcal{A}$ is residually finite dimensional if for every $a\in\mathcal{A}$, there is a $*$-representation $\phi$ of $\mathcal{A}$ on a finite dimensional space such that $\phi(a)\neq 0$.

Theorem 2.1**.**

Let $p\in(1,\infty)\backslash\{2\}$. Let $J$ be a set. Let $\mathcal{A}$ be a $C^{*}$-algebra. Let $\phi\colon\mathcal{A}\to B(l^{p}(J))$ be a homomorphism. Then

1. (i)

the norm closure of $\{\phi(a)x\colon a\in\mathcal{A},\,\|a\|\leq 1\}$ in $l^{p}(J)$ is norm compact for every $x\in l^{p}(J)$; and 2. (ii)

$\mathcal{A}/\text{ker }\phi$* is a residually finite dimensional $C^{*}$-algebra.*

Corollary 2.2**.**

Let $p\in(1,\infty)\backslash\{2\}$. A $C^{*}$-algebra $\mathcal{A}$ is isomorphic to a subalgebra of $B(l^{p}(J))$, for some set $J$, if and only if $\mathcal{A}$ is residually finite dimensional.

Theorem 2.1 and Corollary 2.2 will be proved at the end of this section after a series of lemmas are proved. Theorem 2.1 has an easier proof when $\phi$ is contractive. Indeed, if $\phi\colon\mathcal{A}\to B(l^{p}(J))$ is a contractive homomorphism, then the range of $\phi$ is in the algebra of diagonal operators on $l^{p}(J)$ by [8, Proposition 2.12] (or by [2, Lemma 5.2] when $J$ is countable). Thus, $\{\phi(a)x\colon a\in\mathcal{A},\,\|a\|\leq 1\}$ is norm relatively compact, for every $x\in l^{p}(J)$, and $\mathcal{A}/\text{ker }\phi$ is commutative.

It is not known if Theorem 2.1 and Corollary 2.2 hold for $p=1$. However, throughout their proofs, we use, in an essential way, the assumption that $p$ is in the reflexive range. For example, in the proof of Theorem 2.1(i), we use the fact that every bounded sequence in $l^{p}(J)$ has a weakly convergent subsequence. In the proof of Corollary 2.2, we use a classical result of Pełczyński that the $l^{p}$ direct sum of finite dimensional Hilbert spaces is isomorphic to $l^{p}(J)$ for some set $J$. This result of Pełczyński holds only when $p$ is in the reflexive range.

The structure of the proof of Theorem 2.1(i) goes as follows: If the closure of $\{\phi(a)x_{0}\colon a\in\mathcal{A},\,\|a\|\leq 1\}$ is not compact for some $x_{0}\in l^{p}(J)$, then we can find a bounded sequence in $(b_{k})_{k\in\mathbb{N}}$ in $\mathcal{A}$ such that $\phi(b_{k})x_{0}\to 0$ weakly, as $k\to\infty$, and $\inf_{k\in\mathbb{N}}\|\phi(b_{k})x_{0}\|>0$. Assume that $p>2$. In Lemma 2.5, we show that $\phi(b_{k})\to 0$ weakly implies that $\omega(b_{k}^{*}b_{k})\to 0$ for all positive linear functional $\omega:\mathcal{A}\to\mathbb{C}$ of the form $\omega(a)=y_{0}^{*}(\phi(a)x_{0})$. This is proved by considering $\sum_{k=1}^{n}\delta_{k}b_{k}$ for random $\delta_{1},\ldots,\delta_{n}$ in $\{-1,1\}$ and by exploiting $p>2$. Lemma 2.9 says that when $y_{0}^{*}\in(l^{p}(J))^{*}$ is suitably chosen, $\omega(b_{k}^{*}b_{k})\to 0$ implies that $\|\phi(b_{k})x_{0}\|\to 0$, which contradicts with $\inf_{k\in\mathbb{N}}\|\phi(b_{k})x\|>0$. This is proved by using uniform convexity of $l^{p}(J)$.

Theorem 2.1(ii) follows from Theorem 2.1(i) by using a GNS type construction and a classical result about compact unitary representations of groups on Hilbert spaces.

The following two lemmas are needed for the proof of Lemma 2.5.

Lemma 2.3**.**

Let $\mathcal{A}$ be a unital $C^{*}$-algebra. Let $a\in\mathcal{A}$. Then there exists a sequence $(c_{n})_{n\in\mathbb{N}}$ in $\mathcal{A}$ such that $\|c_{n}\|\leq 1$ for all $n\in\mathbb{N}$ and $|a|=\lim_{n\to\infty}c_{n}a$.

Proof.

Without loss of generality, we may assume that $\|a\|\leq 1$. For $n\in\mathbb{N}$, define $g_{n}\in C[0,1]$ by

[TABLE]

Take $c_{n}=g_{n}(a^{*}a)a^{*}$. Then $c_{n}c_{n}^{*}=g_{n}(a^{*}a)a^{*}ag_{n}(a^{*}a)$. Note that

[TABLE]

Thus, $0\leq xg_{n}(x)^{2}\leq 1$ for all $x\in[0,1]$ and so $0\leq c_{n}c_{n}^{*}\leq 1$. Hence $\|c_{n}\|\leq 1$.

We have

[TABLE]

and so $|xg_{n}(x)-\sqrt{x}|\leq\frac{1}{\sqrt{n}}$ for all $x\in[0,1]$. Since $c_{n}a=g_{n}(a^{*}a)a^{*}a$, it follows that $\|c_{n}a-\sqrt{a^{*}a}\|\leq\frac{1}{\sqrt{n}}$. Thus, the result follows. ∎

Lemma 2.4**.**

Let $\mathcal{A}$ be a unital $C^{*}$-algebra. Let $\omega$ be a positive linear functional on $\mathcal{A}$. Let $a\in\mathcal{A}$. If $a\geq 0$ then

[TABLE]

Proof.

There exists a measure $\mu$ on $[0,\|a\|]$ such that

[TABLE]

for all $f\in C[0,\|a\|]$. So

[TABLE]

∎

Lemma 2.5**.**

Let $2<p<\infty$. Let $J$ be a set. Let $\mathcal{A}$ be a unital $C^{*}$-algebra. Let $\phi\colon\mathcal{A}\to B(l^{p}(J))$ be a unital homomorphism. Let $x_{0}\in l^{p}(J)$. Let $y_{0}^{*}$ be a bounded linear functional on $l^{p}(J)$. Define $\omega\colon\mathcal{A}\to\mathbb{C}$ by

[TABLE]

for $a\in\mathcal{A}$. Assume that $\omega$ is a positive linear functional. Let $(b_{k})_{k\in\mathbb{N}}$ be a sequence in $\mathcal{A}$ such that $\|b_{k}\|\leq 1$ for all $k\in\mathbb{N}$ and $\phi(b_{k})x_{0}\to 0$ weakly as $k\to\infty$. Then $\omega(b_{k}^{*}b_{k})\to 0$ as $k\to\infty$.

Proof.

By contradiction, suppose that $\omega(b_{k}^{*}b_{k})$ does not converge to 0. Passing to a subsequence, we have that there exists $\gamma>0$ such that $\omega(b_{k}^{*}b_{k})\geq\gamma$ for all $k\in\mathbb{N}$.

Since $\|\phi(b_{k})x_{0}\|\leq\|\phi\|\|x_{0}\|$ and $\phi(b_{k})x_{0}\to 0$ weakly, passing to a further subsequence, we may assume that there are $z_{1},z_{2},\ldots$ in $l^{p}(J)$ with disjoint supports such that $\|z_{k}\|\leq\|\phi\|\|x_{0}\|$ and $\|\phi(b_{k})x_{0}-z_{k}\|\leq\frac{1}{2^{k}}$ for all $k\in\mathbb{N}$.

Let $n\in\mathbb{N}$. For each $\delta=(\delta_{1},\ldots,\delta_{n})\in\{-1,1\}^{n}$, let

[TABLE]

By Lemma 2.4,

[TABLE]

Thus,

[TABLE]

where $\mathbb{E}$ denotes expectation over $\delta=(\delta_{1},\ldots,\delta_{n})$ uniformly distributed on $\{-1,1\}^{n}$.

Note that

[TABLE]

Therefore,

[TABLE]

We have

[TABLE]

Since $\|b_{k}\|\leq 1$, it follows that

[TABLE]

Note that $\mathbb{E}(\delta_{k_{1}}\delta_{k_{2}}\delta_{k_{3}}\delta_{k_{4}})=0$ unless the following occurs:

[TABLE]

Thus, $\mathbb{E}\omega(a_{\delta}^{4})\leq 3n^{2}$. So by (2.1), we have $n\gamma\leq 3^{\frac{1}{3}}n^{\frac{2}{3}}[\mathbb{E}\omega(a_{\delta})]^{\frac{2}{3}}$. Hence,

[TABLE]

Fix $\delta\in\{-1,1\}^{n}$. By Lemma 2.3,

[TABLE]

For $c\in\mathcal{A}$ with $\|c\|\leq 1$,

[TABLE]

where the last two inequalities follow from the fact that $z_{1},z_{2},\ldots$ have disjoint supports, $\|z_{k}\|\leq\|\phi\|\|x_{0}\|$ and $\|\phi(b_{k})x_{0}-z_{k}\|\leq\frac{1}{2^{k}}$. Thus, $\omega(a_{\delta})\leq\|y_{0}^{*}\|\|\phi\|(\|\phi\|\|x_{0}\|n^{\frac{1}{p}}+1)$ for all $\delta\in\{-1,1\}^{n}$. So by (2.2),

[TABLE]

Since $n$ can be chosen to be arbitrarily large and $p>2$, an absurdity follows. ∎

For $1<p<2$, we have the following result, where the order of $b_{k}^{*}$ and $b_{k}$ are switched, by using the dual $l^{p}$ space in Lemma 2.5.

Lemma 2.6**.**

Let $1<p<2$. Let $J$ be a set. Let $\mathcal{A}$ be a unital $C^{*}$-algebra. Let $\phi\colon\mathcal{A}\to B(l^{p}(J))$ be a unital homomorphism. Let $x_{0}\in l^{p}(J)$. Let $y_{0}^{*}$ be a bounded linear functional on $l^{p}$. Define $\omega\colon\mathcal{A}\to\mathbb{C}$ by

[TABLE]

for $a\in\mathcal{A}$. Let $(b_{k})_{k\in\mathbb{N}}$ be a sequence in $\mathcal{A}$ such that $\|b_{k}\|\leq 1$ for all $k\in\mathbb{N}$ and that the sequence $y_{0}^{*}\circ\phi(b_{k})$ of bounded linear functionals on $l^{p}(J)$ converges to [math] weakly as $k\to\infty$. Assume that $\omega$ is a positive linear functional. Then $\omega(b_{k}b_{k}^{*})\to 0$ as $k\to\infty$.

Proof.

Let $\mathcal{A}_{1}$ be the unital $C^{*}$-algebra consisting of the same elements as $\mathcal{A}$ but with reverse order multiplication

[TABLE]

Define a unital homomorphism $\phi_{1}\colon\mathcal{A}_{1}\to B((l^{p}(J))^{*})$ by

[TABLE]

for all $a\in\mathcal{A}_{1},\,y^{*}\in(l^{p}(J))^{*}$. Define $\omega_{1}\colon\mathcal{A}_{1}\to\mathbb{C}$ by

[TABLE]

for all $a\in\mathcal{A}_{1}$, where $x_{0}^{**}$ is the image of $x_{0}$ in the bidual $(l^{p})^{**}$. By Lemma 2.5, the result follows. ∎

The following two lemmas are needed for the proof of Lemma 2.9.

Lemma 2.7** ([1]).**

Let $1<p<\infty$. Let $J$ be a set. For every $\epsilon>0$, there exists $\gamma>0$ such that for all $x,y\in l^{p}(J)$ satisfying $\|x\|,\|y\|\leq 1$ and $\|x+y\|>2-\gamma$, we have $\|x-y\|<\epsilon$.

Lemma 2.8** ([9]).**

Let $\mathcal{A}$ be a unital $C^{*}$-algebra. Then the closed unital ball of $\mathcal{A}$ is the closed convex hull of the set of all unitary elements of $\mathcal{A}$.

Lemma 2.9**.**

Let $1<p<\infty$. Let $J$ be a set. Let $\mathcal{A}$ be a unital $C^{*}$-algebra. Let $\phi\colon\mathcal{A}\to B(l^{p}(J))$ be a unital homomorphism. Let $x_{0}\in l^{p}(J)$. Then there exists $y_{0}^{*}\in(l^{p}(J))^{*}$ such that $\omega\colon\mathcal{A}\to\mathbb{C}$,

[TABLE]

defines a positive linear functional and for every $\epsilon>0$, there exists $\gamma>0$ such that whenever $a\in\mathcal{A}$ satisfies $\|a\|\leq 1$ and $\omega(a^{*}a)<\gamma$, we have $\|\phi(a)x_{0}\|<\epsilon$.

Proof.

Let $\mathcal{U}(\mathcal{A})$ be the set of all unitary elements of $\mathcal{A}$. Let $(v_{n})_{n\in\mathbb{N}}$ be a sequence in $\mathcal{U}(\mathcal{A})$ such that

[TABLE]

For each $n\in\mathbb{N}$, let $x_{n}^{*}$ be a bounded linear functional on $l^{p}(J)$ such that $\|x_{n}^{*}\|=1$ and $x_{n}^{*}(\phi(v_{n})x_{0})=\|\phi(v_{n})x_{0}\|$. Then $x_{n}^{*}\circ\phi(v_{n})$ is a bounded sequence in $(l^{p}(J))^{*}$. Passing to a subsequence, we may assume that $x_{n}^{*}\circ\phi(v_{n})$ converges weakly to a bounded linear functional $y_{0}^{*}\in(l^{p}(J))^{*}$ as $n\to\infty$. Thus, $\omega\colon\mathcal{A}\to\mathbb{C}$,

[TABLE]

for $a\in\mathcal{A}$, defines a bounded linear functional on $\mathcal{A}$. Note that

[TABLE]

and for every $u_{0}\in\mathcal{U}(\mathcal{A})$,

[TABLE]

So by Lemma 2.8, we have $\|\omega\|\leq\sup_{u\in\mathcal{U}(\mathcal{A})}\|\phi(u)x_{0}\|$. Thus, $\omega(1)=\|\omega\|$ and hence $\omega$ is a positive linear functional.

By contradiction, suppose that there are $\epsilon>0$ and a sequence $(a_{k})_{k\in\mathbb{N}}$ in $\mathcal{A}$ such that $\|a_{k}\|\leq 1$ and $\|\phi(a_{k})x_{0}\|\geq\epsilon$ for all $k\in\mathbb{N}$ and $\omega(a_{k}^{*}a_{k})\to 0$ as $k\to\infty$. We have

[TABLE]

for all $k\in\mathbb{N}$. For $k\in\mathbb{N}$, let $b_{k}=\frac{a_{k}}{\|a_{k}\|}$. We have $\|b_{k}\|=1$ and $\|\phi(b_{k})x_{0}\|\geq\epsilon$ for all $k\in\mathbb{N}$ and $\omega(b_{k}^{*}b_{k})\to 0$ as $k\to\infty$.

Since $\|x_{n}^{*}\|=1$,

[TABLE]

Thus,

[TABLE]

But

[TABLE]

and $\|\phi(v_{n})x_{0}\|\leq\omega(1)$ for all $n\in\mathbb{N}$. Take

[TABLE]

in Lemma 2.7 and note that $\omega(|b_{k}|)\leq\omega(b_{k}^{*}b_{k})^{\frac{1}{2}}\omega(1)^{\frac{1}{2}}\to 0$ as $k\to\infty$. We have

[TABLE]

Thus,

[TABLE]

So $\|\phi(|b_{k}|)x_{0}\|\to 0$ as $k\to\infty$. Since $b_{k}=b_{k}(|b_{k}|+\frac{1}{k})^{-1}(|b_{k}|+\frac{1}{k})$ and $\|b_{k}(|b_{k}|+\frac{1}{k})^{-1}\|\leq 1$, it follows that $\|\phi(b_{k})x_{0}\|\to 0$ as $k\to\infty$ which contradicts with $\|\phi(b_{k})x_{0}\|\geq\epsilon$. ∎

Proof of Theorem 2.1(i).

Without loss generality, we may assume that $\mathcal{A}$ is unital by extending $\phi$ to a homomorphism from the unitization of $\mathcal{A}$ into $B(l^{p}(J))$. We may also assume that $\phi$ is unital since $\phi(1)$ is an idempotent on $l^{p}(J)$ and the range of every idempotent on $l^{p}(J)$ is isomorphic to $l^{p}(J_{0})$ for some set $J_{0}$ [7], [3].

Let $x_{0}\in l^{p}$. Let $(a_{k})_{k\in\mathbb{N}}$ be a sequence in $\mathcal{A}$ such that $\|a_{k}\|\leq\frac{1}{2}$ for all $k\in\mathbb{N}$. We need to show that $(\phi(a_{k})x_{0})_{k\in\mathbb{N}}$ has a norm convergent subsequence.

Case 1: $p>2$

Passing to a subsequence, we may assume that $(\phi(a_{k})x_{0})_{k\in\mathbb{N}}$ converges weakly to an element of $l^{p}(J)$. Thus, $\phi(a_{k_{1}}-a_{k_{2}})x_{0}\to 0$ weakly as $k_{1},k_{2}\to\infty$.

By Lemma 2.5, we have $\lim_{k_{1},k_{2}\to\infty}\omega((a_{k_{1}}-a_{k_{2}})^{*}(a_{k_{1}}-a_{k_{2}}))=0$ for every positive linear functional $\omega\colon\mathcal{A}\to\mathbb{C}$ of the form $\omega(a)=y_{0}^{*}(\phi(a)x_{0})$ for $a\in\mathcal{A}$. By Lemma 2.9, we have $\lim_{k_{1},k_{2}\to\infty}\|\phi(a_{k_{1}}-a_{k_{2}})x_{0}\|=0$. So $(\phi(a_{k})x_{0})_{k\in\mathbb{N}}$ is norm convergent.

Case 2: $p<2$

Passing to a subsequence, we may assume that $(y_{0}^{*}\circ\phi(a_{k}^{*}))_{k\in\mathbb{N}}$ converges weakly to an element of $(l^{p}(J))^{*}$. Thus, $y^{*}\circ\phi(a_{k_{1}}^{*}-a_{k_{2}}^{*})\to 0$ weakly as $k_{1},k_{2}\to\infty$ for every $y^{*}\in(l^{p}(J))^{*}$.

By Lemma 2.6, we have $\lim_{k\to\infty}\omega((a_{k_{1}}^{*}-a_{k_{2}}^{*})(a_{k_{1}}^{*}-a_{k_{2}}^{*})^{*})=0$ for every positive linear functional $\omega\colon\mathcal{A}\to\mathbb{C}$ of the form $\omega(a)=y_{0}^{*}(\phi(a)x_{0})$ for $a\in\mathcal{A}$. By Lemma 2.9, we have $\lim_{k_{1},k_{2}\to\infty}\|\phi(a_{k_{1}}-a_{k_{2}})x_{0}\|=0$. So $(\phi(a_{k})x_{0})_{k\in\mathbb{N}}$ is norm convergent. ∎

Lemma 2.10** ([4], Theorem 2.24).**

Let $G$ be a group. Let $\mathcal{H}$ be a Hilbert space. Let $\varphi\colon G\to B(\mathcal{H})$ be a unital homomorphism such that $\varphi(g)$ is unitary for all $g\in G$. If $\{\varphi(g)x\colon g\in G\}$ is norm precompact in $\mathcal{H}$ for all $x\in\mathcal{H}$, then $\mathcal{H}$ is the direct sum of some finite dimensional subspaces $\mathcal{H}_{\alpha}$, for $\alpha\in\Lambda$, such that $\mathcal{H}_{\alpha}$ is invariant under $\varphi(g)$ for all $\alpha\in\Lambda$ and $g\in G$.

Proof of Theorem 2.1(ii).

As in the proof Theorem 2.1(i), we may assume that $\mathcal{A}$ is unital and $\phi$ is unital. We may also assume that $\text{ker }\phi=\{0\}$. Let $a_{0}\neq 0$. There exists $x_{0}\in l^{p}(J)$ such that $\phi(a_{0})x_{0}\neq 0$. By Lemma 2.9, there exists $y_{0}^{*}\in(l^{p}(J))^{*}$ such that $\omega\colon\mathcal{A}\to\mathbb{C}$,

[TABLE]

for $a\in\mathcal{A}$, defines a positive linear functional and $\omega(a_{0}^{*}a_{0})\neq 0$.

Equip $\mathcal{A}$ with the positive semidefinite sesquilinear form

[TABLE]

for $a,b\in\mathcal{A}$. Consider the ideal $\mathcal{A}_{0}=\{a\in\mathcal{A}\colon\langle a,a\rangle=0\}$ of $\mathcal{A}$. Let $\mathcal{H}$ be the completion of the quotient space $\mathcal{A}/\mathcal{A}_{0}$. Then $\mathcal{H}$ is a Hilbert space. For each $a\in\mathcal{A}$, we can define a bounded linear operator on $\mathcal{H}$ by sending $b+\mathcal{A}_{0}$ to $ab+\mathcal{A}_{0}$ for $b\in\mathcal{A}$. So $\eta\colon\mathcal{A}\to B(\mathcal{H})$,

[TABLE]

for $a,b\in\mathcal{A}$, defines a unital $*$-homomorphism. We have

[TABLE]

for all $a_{1},a_{2},b\in\mathcal{A}$. By Theorem 2.1(i), we have that $\{\phi(a)x_{0}\colon a\in\mathcal{A},\,\|a\|\leq 1\}$ is norm precompact so $\{\eta(a)(b+\mathcal{A}_{0})\colon a\in\mathcal{A},\,\|a\|\leq 1\}$ is norm precompact for all $b\in\mathcal{A}$. Let $\mathcal{U}(\mathcal{A})$ be the set of all unitary elements of $\mathcal{A}$. By Lemma 2.10, we have that $\mathcal{H}$ is the direct sum of some finite dimensional subspaces $\mathcal{H}_{\alpha}$, for $\alpha\in\Lambda$, such that $\mathcal{H}_{\alpha}$ is invariant under $\eta(u)$ for all $\alpha\in\Lambda$ and $u\in\mathcal{U}(\mathcal{A})$. Note that $\mathcal{H}_{\alpha}$ is thus invariant under $\eta(a)$ for all $a\in\mathcal{A}$.

Since $\omega(a_{0}^{*}a_{0})\neq 0$, we have $\eta(a_{0})\neq 0$. So $\eta(a_{0})\neq 0$ on $\mathcal{H}_{\alpha_{0}}$ for some $\alpha_{0}\in\Lambda$. Thus, $\mathcal{A}$ is residually finite dimensional. ∎

Proof of Corollary 2.2.

One direction follows from Theorem 2.1. For the other direction, suppose that $\mathcal{A}$ is a residually finite dimensional $C^{*}$-algebra. Then there is a collection $(\phi_{\alpha})_{\alpha\in\Lambda}$ of $*$-representations of $\mathcal{A}$ on finite dimensional Hilbert spaces $\mathcal{H}_{\alpha}$ such that $\|a\|=\sup_{\alpha\in\Lambda}\|\phi_{\alpha}(a)\|$ for all $a\in\mathcal{A}$. Define $\phi\colon\mathcal{A}\to B((\oplus_{\alpha\in\Lambda}\mathcal{H}_{\alpha})_{l^{p}})$ by $\phi=\oplus_{\alpha\in\Lambda}\phi_{\alpha}$. Thus $\phi$ is a norm preserving homomorphism. However, it is a classical result of Pełczyński [7] that for $1<p<\infty$, the $l^{p}$ direct sum of finite dimensional Hilbert spaces is isomorphic to $l^{p}(J)$ for some set $J$. Therefore, $\mathcal{A}$ is isomorphic to a subalgebra of $B(l^{p}(J))$, via the map $a\mapsto S\phi(a)S^{-1}$ where $S\colon(\oplus_{\alpha\in\Lambda}\mathcal{H}_{\alpha})_{l^{p}}\to l^{p}(J)$ is any invertible operator. ∎

Acknowledgements: The author is grateful to the referee for some suggestions that improved the exposition. The author is supported by NSF DMS-1856221.