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In PMA, Rudin's book, there is the following theorem (7.29):

Let $B$ be the uniform closure of an algebra $A$ of bounded functions. (Here, an algebra means a family of function satisfying that it is closed under addition, multiplication, and scalar multiplication.)

Then, $B$ is a uniformly closed algebra.

I ask whether the condition 'bounded' is actually the 'uniform' bounded. I think, if it means just the pointwise bounded, then we cannot show that product sequence of two sequences converges uniformly.

Do you think so?

Jeff
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  • What is bounded here is every individual function. $A$ need not be pointwise bounded, but each $f\in A$ must be bounded on all of $X$ (calling the domain of the functions $X$). – Daniel Fischer Apr 23 '14 at 18:55

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No, it is fine if $A$ is an algebra of bounded functions. In Rudin's proof of this theorem, he says it is easy to show that $f_ng_n \rightarrow fg$ where $f_n \rightarrow f$ and $g_n \rightarrow g$ uniformly. Well, you can see from this question that every uniformly convergent sequence of bounded functions is uniformly bounded.

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Jeff
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