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The resulting metric topology corresponding to the norm given by: $\|f\| = \sup\limits_{x\in X} | f(x) |$ on $C^*(X)$ is called the uniform norm topology on $C^*(X)$. Show that, in uniform norm topology is uniform convergence of the functions.

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    What does the $C^*$ indicate, as opposed to $C$? Bounded functions? I think your last sentence is missing some words too. It would be nice to add a whole new sentence or two saying what you've tried. I mean, take a convergent sequence $f_n \to f$ for $\|\,\|$ and try to show that the $f_n$ converge uniformly to $f$. Where do you get stuck?2012-07-25
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    @Dylan: yes. Bounded and continuous real-valued functions. See also [this previous question](http://math.stackexchange.com/q/173589/5363).2012-07-25
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    Standard notation for this is $C_b(X)$.2012-07-25
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    @ncmathsadist Both notations are standard, the $C^*(X)$ notation is common in the litterature on Stone-Cech compactification and Stone Spaces, for instance in _Stone spaces_ by Peter T.Johnstone.2012-07-25
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    Is this approach correct: Let {Fn}, (n = 1, 2, 3, …) be a sequence of functions converging pointwise on C*(X) to a function F. Let Є > 0. Now for each f in C*(X), lim Fn(f) = F(f) or equivalently for a chosen f in C*(X) there is a natural number k such that | Fn(f) – F(f)| < Є whenever n ≥ k. We are to show that, {Fn}, (n = 1, 2, 3, …) converges uniformly on C*(X) to F i.e. to show that for a chosen Є > 0 there is a natural number k such that | Fn(f) – F(f)| < Є whenever n ≥ k (for all f in C*(X)).2012-07-25
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    @SugataAdhya please learn to use MathJax for the mathematics. When one is accustomed to reading things like "$\lvert F_n(f) - F(f)\rvert < \varepsilon$", seeing it rendered as "| Fn(f) – F(f)| < Є" is slightly confusing. It's also a simple fact that people take questions with properly formatted mathematics (and spelling and grammar—not an issue for you) a lot more seriously.2012-11-17

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As I understand it, you are trying to show that if $C^*(X)$ is given the uniform norm topology, then if a sequence of functions on $C^*(X)$ converges, then it converges uniformly. So you started out taking a sequence $F_1, F_2,...$ of elements of $C^*(X)$, but these are functions of $X$, so you should not have written $F(f)$ where $f$ is in $C^*(X)$. But if you fix that mistake, then your answer is right. This question is really just a matter of getting the definitions right.

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    I've taken Fn:C∗(X) --> R2012-07-25
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    But you're starting with a set of points of $C^*(X)$ that converge, right? And then trying to show that those functions converge uniformly?2012-07-25
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    I didn't start with points of C*(X). Do I've to do that?2012-07-25
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    I thought I need to choose a sequence Fn: C*(X) ---> R that converges uniformly to a F: C*(X) ---> R and then I've to show the convergence is uniform. That's what I thought.2012-07-25
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    I don't think so. You're trying to prove something about convergence of functions in $C^*(X)$. By that we mean a sequence of functions that converge in the uniform norm topology. You're trying to show that convergence in the uniform norm topology is the same as uniform convergence. Convergence of what in the uniform norm topology? Well, $C^*(X)$ is given the uniform norm topology. So convergence refers to this space. What does it mean for a sequence of elements of $C^*(X)$ to converge?2012-07-25
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    I think what I've to prove is this: Let f_n converges to f in Uniform norm topology than f_n converges to f uniformly and conversely if f_n converges to f uniformly on X then f_n converges to f in the topology of C*(X)2012-07-26
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    Yes, that's right.2012-07-26