Peano curves in function algebras
Peano curves in function algebras
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函数代数中的皮亚诺曲线
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发表时间：
1972
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通讯作者：
L. Eifler
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作者：
L. Eifler
我们对Pelczyiiski得到的结果作一个简短的证明。如果X是一个不可数的紧度量空间，如果A是X上的函数代数，那么在A中存在f使得f(X)在平面内。设X是紧度量空间，a是X上的函数代数。如果限制a中每个f的“大小”，那么空间X也同样受到限制。例如，Rudin[4]证明了f'(X)对于A中的每个f是可数的当且仅当X是可数的。当然，f'(X)对于A中的每个f都是可数的，这意味着A=C(X)。Pelczyn'ski[2]证明了X是不可数的当且仅当A包含一个闭子空间M且X包含一个完全的闭子集K，使得f-*f K是M到C(K)上的等距。我们的目的是用一种不同的方法来解读佩尔琴斯基的一些作品。如果X是平面C的紧子集，则P(X)表示多项式在C(X)中的一致闭包。我们从下面的结果开始。定理。设X是C的不可数紧子集。存在X的不可数闭子集K，使得K是P(X)的峰集，且P(X)|K= C(K)。证明。我们可以假设X是多项式凸的。根据Wermer对P(X)的湮灭子的表征[6]，在X的边界aX上存在一个非负测度，使得vp (X)和v在aX上被支持，这意味着v相对于它是绝对连续的。因此，根据Bishop对Rudin-Carleson定理的推广[1]，我们只需要找到Kc AX使u，(K)=O，且K是闭不可数的。这样的K是存在的。也就是说，由于AX是不可数的，所以在AX上存在一个连续映射g到[0,1]x[0,1]上。则u（g-1({x} x[0,1])）对于除可数个数外的所有x([0,1]) =0。推论。假设Y是一个紧度量空间，a是Y上的一个函数代数。如果(Y)在C中对于a中的每一个f都没有内部，那么Y是可数的，因此a = C Y)。编辑于1971年6月15日收到，1971年9月13日修改。AMS 1970学科分类。主要46 j10。
We give a short proof of the following result which was obtained by Pelczyiiski. If X is an uncountable, compact metric space and if A is a function algebra on X, then there exists f in A such that f(X) has interior in the plane. Let X be a compact metric space and let A be a function algebra on X. If one restricts the "size" off(X) for eachf in A, then the space X is likewise restricted. For example, Rudin [4] has shown thatf'(X) is countable for eachf in A if and only if X is countable. Of course, f'(X) is countable for eachf in A implies A=C(X). Pelczyn'ski [2] has shown that X is uncountable if and only if A contains a closed subspace M and X contains a perfect, closed subset K such thatf-*f K is an isometry of M onto C(K). Our purpose is to give a different approach to some of Pelczyn'ski's work. If X is a compact subset of the plane C, then P(X) denotes the uniform closure in C(X) of the polynomials. We begin with the following result. THEOREM. Sulppose X is an uncountable, compact subset of C. There is a closed, uncountable subset K of X such that K is a peak set for P(X) and P(X)|K= C(K). PROOF. We may assume that X is polynomially convex. By Wermer's characterization [6] of the annihilator of P(X), there is a nonnegative measure It on aX, the boundary of X, such that v P(X) and v is supported on AX implies that v is absolutely continuous with respect to It. Hence, by Bishop's generalization [1] of the Rudin-Carleson theorem, we only need to find Kc AX such that u,(K)=O and K is closed and uncountable. Such a K exists. Namely, since AX is uncountable, there is a continuous map g on AX onto [0, 1] x [0, 1]. Then u(g-1({x} x [0, 1]))=0 for all but countably many x in [0, 1]. COROLLARY. Suppose Y is a conmpact metric space and A is a function algebra on Y. Iff( Y) has no interior in Cfor eachf in A, then Y is countable and hence A= C Y). Received by the editors June 15, 1971 and, in revised form, September 13, 1971. AMS 1970 subject classifications. Primary 46J10.