Uniform Convergence on an Equicontinuous Set

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SUMMARY

The discussion centers on proving that an equicontinuous sequence of functions \( (f_n) \) in \( C(X, \mathbb{R}) \) converges uniformly to the zero function, given that \( (f_n(x)) \to 0 \) for every fixed \( x \) in a compact metric space \( (X,d) \). The proof utilizes the properties of compactness and equicontinuity, establishing that the sequence is Cauchy with respect to the uniform norm. Key steps include the use of a delta-net and the elimination of dependence on \( x \) through compactness, confirming uniform convergence.

PREREQUISITES
  • Understanding of compact metric spaces
  • Familiarity with equicontinuity in function sequences
  • Knowledge of uniform convergence and Cauchy sequences
  • Proficiency in the properties of continuous functions in \( C(X, \mathbb{R}) \)
NEXT STEPS
  • Study the concept of delta-nets in metric spaces
  • Learn about the properties of equicontinuous function sequences
  • Explore the implications of compactness in functional analysis
  • Investigate uniform convergence and its applications in analysis
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Mathematicians, particularly those focused on functional analysis, students studying real analysis, and anyone interested in the properties of continuous functions in compact spaces.

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Homework Statement


Let (X,d) be a compact metric space, (f_n) be an equicontinuous sequences of functions in C(X, [itex]\mathbb{R}[/itex])such that, for every fixed x in X, [itex](f_n(x)) \to 0[/itex].

Show that (f_n) converges uniformly to the zero function

Homework Equations





The Attempt at a Solution


First things first, I started by noting that since X is compact, then C(X,R) (the space of all continuous functions from X to R) is equivalent to C_b(X,R), the set of all continuous bounded functions from X to R. Now R complete implies that F_b(X,R) is complete (set of bounded functions), and C_b(X,R) is a closed subset of F_b(X,R) and so is also complete. Thus it suffices to show that (f_n) is Cauchy with respect to the distance induced by the uniform norm. That is

[tex]\rho(f,g) = \sup \{ |f(x) - g(x) | | x \in X \}[/tex]

Let [itex]\epsilon > 0[/itex]. Then (f_n) uniformly continuous implies [itex]\exists \delta >0[/itex] such that [itex]\forall x,y \text{ satisfying } d(x,y) < \delta, |f(x)-f(y)| < \epsilon[/itex]. Since [itex](f_n(x)) \to 0, \exists n_x[/itex] for each fixed x, such that [itex]|f_n(x)| < \epsilon[/itex].

Since X is compact, let [itex]x_1, \ldots, x_p[/itex] be a delta-net for X. Then [itex]\forall x \in X, \exists x_j \text{ such that } d(x,x_j) < \delta[/itex] and so [itex]|f_n(x) - f_n(x_j)| < \epsilon[/itex] for all n.

I'm not sure if I ended up doing some of this for no reason, but I can't quite figure out where to go from here.
 
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You know f_n converges to zero. That means that for every e>0 there is a N(x) such that for all n>N(x) |f_n(x)|<e. You only have to show that it's uniform. If f_n is continuous that means that there is a d_n(x) such that if |x-x'|<d_n(x) then |f_n(x)-f_n(x')|<e. Equicontinuity let's you eliminate the dependence of d on n. Now you want to eliminate the dependence on x. Use compactness.
 

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