Full Text
UDC 517.432
MATHEMATICS
A. Ya. HELEMSKII
ON AN ANALYTIC CONDITION ON THE RADICAL OF A COMMUTATIVE BANACH ALGEBRA AND RELATED QUESTIONS OF DECOMPOSABILITY
(Presented by Academician Yu. V. Linnik on 30 VI 1965)
The purpose of the present work is to establish a condition on the radical, “analytic” in its character, which ensures the decomposability of certain classes of commutative Banach algebras (c.b.a.) with such a radical. Let \(\mathfrak A\) be a c.b.a.; \(\mathfrak R\) its radical; \(A\) the quotient algebra \(\mathfrak A / \mathfrak R\); and \(P_{\mathfrak A}\) and \(P_A\) the sets of idempotents of the algebras \(\mathfrak A\) and \(A\). The natural epimorphism \(\pi:\mathfrak A \to A\) induces a mapping \(P_\pi: P_{\mathfrak A} \to P_A\), which, as shown in \((^1)\), establishes a one-to-one correspondence between these two sets.
Bade and Curtis showed \((^{2,3})\) that questions of decomposability of a number of c.b.a.’s reduce to the following: will the preimage (under the mapping \(P_\pi\)) of every bounded subset of \(P_A\) be bounded in \(\mathfrak A\)? They also proved in \((^3)\) that the latter is always the case if the radical \(\mathfrak R\) is algebraically nilpotent, i.e. \(\mathfrak R^n = 0\) for some natural \(n\).
Definition. By the radical sequence of a c.b.a. \(\mathfrak A\) with radical \(\mathfrak R\) we shall mean the sequence
\[ f(\mathfrak A,n)=\sup_{\substack{r\in\mathfrak R\\ \|r\|=1}}\|r^n\|. \]
Lemma 1. Let \(\mathfrak R_1 \subseteq \mathfrak R\) be a closed ideal in \(\mathfrak A\). Then
\[ f(\mathfrak A/\mathfrak R_1,n)\leq f(\mathfrak A,n). \]
The proof follows immediately from the fact that the radical of the algebra \(\mathfrak A/\mathfrak R_1\) is the image of the radical \(\mathfrak R\) under the natural epimorphism of \(\mathfrak A\) onto \(\mathfrak A/\mathfrak R_1\), and from the definition of the norm in the quotient algebra.
Lemma 2. Let \(f(n)>0,\ n=1,2,\ldots,\) be a sequence such that
\[ \lim_{n\to\infty}\sqrt[n^2]{f(n)}=0; \]
let \(K\) be a constant. Then the set
\[ \{M>0:\ M-M^n f(n)\leq K^n;\ n=1,2,\ldots\} \]
is bounded.
Put \(F(n)=f(n)^{-1/n}\); since
\[ \lim_{n\to\infty}\sqrt[n]{F(n)}=\infty, \]
there is, evidently, a natural number \(N_0\) such that \(F(n)>K^{n+1}+1\) for all \(n\geq N_0\). Take \(M>F(N_0)\); for some natural \(N\geq N_0+1\),
\[ F(N-1)<M\leq F(N). \]
We have
\[ M-M^N f(N)>F(N-1)-F(N)^N f(N). \]
But \(F(N)^N f(N)=1\), while \(F(N-1)>K^N+1\) in view of \(N-1\geq N_0\). Therefore \(M-M^N f(N)>K^N\). Consequently, the inequality \(M-M^n f(n)\leq K^n\) for \(M>F(N_0)\) is not satisfied for all \(n\), which proves the lemma.
The main result of the paper is the following.
Theorem 1. Let \(\mathfrak A\) be a c.b.a., \(\mathfrak R\) its radical, and suppose that the radical sequence satisfies the condition
\[ \lim_{n\to\infty}\sqrt[n^2]{f(\mathfrak A,n)}=0. \]
Let, further, a set \(Q\subseteq P_{\mathfrak A}\) be such that its image \(\pi Q\) is bounded in \(A=\mathfrak A/\mathfrak R\). Then \(Q\) is bounded in \(\mathfrak A\).
For each \(p\in P\mathfrak A\) put \(\mathfrak R_p^0=\{r\in\mathfrak R:\ pr=0\}\) and \(\mathfrak R_p^1=\{r\in\mathfrak R:\ pr=r\}\). The \(\mathfrak R_p^0\) and \(\mathfrak R_p^1\) are closed ideals in \(\mathfrak A\), \(\mathfrak R_p^0\mathfrak R_p^1=0\) and \(\mathfrak R=\mathfrak R_p^0\oplus\mathfrak R_p^1\).
We introduce the following notation: \(\hat{\mathfrak A}_p^0=\mathfrak A/\mathfrak R_p^0\); \(\hat{\mathfrak A}_p^1=\mathfrak A/\mathfrak R_p^1\); \(\hat{\mathfrak R}_p^0\) and \(\hat{\mathfrak R}_p^1\) are the radicals in \(\hat{\mathfrak A}_p^0\) and \(\hat{\mathfrak A}_p^1\); \(\|\cdot\|_p^0\) and \(\|\cdot\|_p^1\) are the norms in \(\hat{\mathfrak A}_p^0\) and \(\hat{\mathfrak A}_p^1\); \(\pi_p^0:\mathfrak A\to\mathfrak A/\mathfrak R_p^0\); \(\pi_p^1:\mathfrak A\to\mathfrak A/\mathfrak R_p^1\); \(\hat\pi_p^0:\hat{\mathfrak A}_p^0\to\hat{\mathfrak A}_p^0/\hat{\mathfrak R}_p^0\) and \(\hat\pi_p^1:\hat{\mathfrak A}_p^1\to\hat{\mathfrak A}_p^1/\hat{\mathfrak R}_p^1\) are the corresponding natural epimorphisms; finally, for convenience put \(p_0=\pi_p^0p\), \(p_1=\pi_p^1p\).
The following relations are obvious: \(\hat{\mathfrak R}_p^0=\mathfrak R/\mathfrak R_p^0\); \(\hat{\mathfrak R}_p^1=\mathfrak R/\mathfrak R_p^1\); \(A=\mathfrak A/\mathfrak R=\hat{\mathfrak A}_p^0/\hat{\mathfrak R}_p^0=\hat{\mathfrak A}_p^1/\hat{\mathfrak R}_p^1\) and \(\pi=\hat\pi_p^0\pi_p^0=\hat\pi_p^1\pi_p^1\). In addition, note that \(p_1\hat{\mathfrak R}_p^1=0\) (since \(p\mathfrak R\subset\mathfrak R_p^1\)).
We shall carry out the proof of the theorem in three stages.
\(1^0\). The norms \(\|p_1\|_p^1,\ p\in Q\), are bounded in the aggregate.
The set \(\pi Q\), by the condition of the theorem, is bounded in \(A\), i.e., for some constant \(K_1\), \(\|\pi p\|<K_1\) for \(p\in Q\). But \(\pi=\hat\pi_p^1\pi_p^1\); therefore from the definition of the norm in a quotient algebra it follows that for each \(p\in Q\) there exist \(\hat p_1\in\hat{\mathfrak A}_p^1\) and \(r_p^1\in\hat{\mathfrak R}_p^1\) such that \(\hat p_1=p_1+r_p^1\), with \(\|\hat p_1\|_p^1<K_1\). Clearly, in order to prove assertion \(1^0\), it suffices, in view of the triangle inequality, to establish the boundedness of the set \(\{\|r_p^1\|_p^1;\ p\in Q\}\).
According to the multiplicative condition, \(\|\hat p_1^{\,n}\|_p^1\le(\|\hat p_1\|_p^1)^n<K_1^n\). On the other hand, in view of \(r_p^1\in\hat{\mathfrak R}_p^1\) and \(p_1\hat{\mathfrak R}_p^1=0\), and also \(p_1^n=p_1\), \(\hat p_1^{\,n}=(p_1+r_p^1)^n=p_1+(r_p^1)^n=p_1+r_p^1+(r_p^1)^n-r_p^1=\hat p_1+(r_p^1)^n-r_p^1\). Applying the triangle inequality, we obtain \(\|\hat p_1^{\,n}\|_p^1=\|r_p^1-(r_p^1)^n-\hat p_1\|_p^1\ge\|r_p^1\|_p^1-\|(r_p^1)^n\|_p^1-\|\hat p_1\|_p^1\), whence \(\|r_p^1\|_p^1-\|(r_p^1)^n\|_p^1\le K_1^n+K_1<(K_1+1)^n\). But from the definition of the radical sequence it follows that \(\|(r_p^1)^n\|_p^1\le(\|r_p^1\|_p^1)^n f(\hat{\mathfrak A}_p^1,n)\), and by Lemma 1, \(f(\hat{\mathfrak A}_p^1,n)\le f(\mathfrak A,n)\). Thus,
\[ \|r_p^1\|_p^1-\bigl(\|r_p^1\|_p^1\bigr)^n f(\mathfrak A,n)\le (K_1+1)^n \]
simultaneously for all \(p\in Q\) and \(n=1,2,\ldots\). Consequently (Lemma 2), the norms \(\|r_p^1\|_p^1,\ p\in Q\), are bounded in the aggregate, which, by the observation made above, proves assertion \(1^0\).
\(2^0\). The norms \(\|p_0\|_p^0,\ p\in Q\), are bounded in the aggregate. Let \(K_2\) be a constant (which exists, as shown in \(1^0\)) such that \(\|p_1\|_p^1<K_2\) for \(p\in Q\). Put \(\bar p=e-p\) (\(e\) is the identity of the algebra \(\mathfrak A\)); obviously, without loss of generality one may assume that the set \(Q\), together with each \(p\), also contains \(\bar p\) (the latter does not affect the boundedness of the set \(\pi Q\)). As is easily seen, \(\mathfrak R_{\bar p}^0=\mathfrak R_p^1\) and \(\mathfrak R_{\bar p}^1=\mathfrak R_p^0\). Therefore
\[ \|p_0\|_p^0=\|\pi_p^0p\|_p^0=\|\pi_{\bar p}^{\,1}p\|_{\bar p}^{\,1} =\|\pi_{\bar p}^{\,1}(e-\bar p)\|_{\bar p}^{\,1}\le \]
\[ \le \|\pi_{\bar p}^{\,1}e\|_{\bar p}^{\,1}+\|\pi_{\bar p}^{\,1}\bar p\|_{\bar p}^{\,1} \le \|e\|+\|\bar p_1\|_{\bar p}^{\,1}<\|e\|+K_2, \]
which proves assertion \(2^0\).
\(3^0\). Completion of the proof. Let \(K_3=\|e\|+K_2\); in assertion \(2^0\) it was shown that \(\|p_0\|_p^0<K_3\) for \(p\in Q\). Consequently, for each \(p\in Q\) there exists \(r_p^0\in\mathfrak R_p^0\) such that \(\|p+r_p^0\|<K_3\). But \(p\mathfrak R_p^0=0\), whence also \(pr_p^0=0\). Therefore, quite analogously to assertion \(1^0\), one can prove that
\[ \|r_p^0\|-\|r_p^0\|^n f(\mathfrak A,n)\le (K_3+1)^n,\qquad n=1,2,\ldots \]
Applying Lemma \(2^0\) once more, we obtain the boundedness of the set \(\{\|r_p^0\|:\ p\in Q\}\), and with it, in view of the triangle inequality, of the set \(Q\) itself. Thus, the theorem is proved.
Remark. The conditions of Theorem 1 are satisfied, in particular, by the algebraically nilpotent radicals considered in \((^3)\); in this case the radical sequence is, obviously, finite.
Combining Theorem 1 with the general results of \((^3)\), we shall apply it to questions of decomposability of certain commutative Banach algebras. Recall that a commutative Banach algebra \(\mathfrak A\) is called strongly decomposable if \(\mathfrak A = B \oplus \mathfrak R\), where \(B\) is a closed subalgebra and \(\mathfrak R\) is the radical of the algebra.
Theorem 2. Let \(\mathfrak A\) be a commutative Banach algebra with radical \(\mathfrak R\) such that
\[ \lim_{n\to\infty}\sqrt[n^2]{f(\mathfrak A,n)}=0. \]
Suppose further that \(A=\mathfrak A/\mathfrak R=C(\Omega)\), where \(\Omega\) is a totally disconnected compactum. Then \(\mathfrak A\) is strongly decomposable.
Since \(A=C(\Omega)\), the set \(P_A\) is bounded; consequently, by Theorem 1, the set \(P_{\mathfrak A}\) in \(\mathfrak A\) is also bounded. Applying Theorem 2.4 of \((^3)\) (see also \((^4)\), Corollary 2 to Theorem 3.10), we obtain the required result.
Theorem 3. Let \(\mathfrak A\) be a commutative Banach algebra with radical \(\mathfrak R\) such that
\[ \lim_{n\to\infty}\sqrt[n^2]{f(\mathfrak A,n)}=0. \]
Suppose further that \(A=\mathfrak A/R=l_1\) (the algebra of absolutely convergent series with coordinatewise multiplication). Then \(\mathfrak A\) is strongly decomposable.
Denote by \(Q\) the set of irreducible idempotents of the algebra \(\mathfrak A\). Since \(\pi Q\), the set of irreducible idempotents of the algebra \(l_1\), is bounded, it follows, by Theorem 1, that \(Q\) is also bounded in \(\mathfrak A\). To complete the proof it remains to apply Theorem 5.6 of \((^3)\).
In conclusion, the author expresses his deep gratitude to M. A. Naimark for his attention to the work.
Received
24 VI 1965
REFERENCES
- C. E. Rickart, General Theory of Banach Algebras, Princeton, New Jersey, 1960.
- W. G. Bade, P. C. Curtis, Am. J. Math., 82, No. 3, 589 (1960).
- W. G. Bade, P. C. Curtis, Am. J. Math., 82, No. 4, 851 (1960).
- J. E. Moyal, The Theory of Spectral and Scalar Algebras. Australian National University (at present Argonne National Laboratory), Preprint.