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Family of equicontinuous functions on compact set

  1. Nov 19, 2013 #1
    The problem statement, all variables and given/known data.

    Let ##X## be a compact metric space. Prove that if ##\mathcal F \subset X## is a family of equicontinuous functions ##f:X \to Y \implies \mathcal F## is uniformly equicontinuous.

    The attempt at a solution.

    What I want to prove is that given ##\epsilon>0## there exists ##\delta>0##: if ##d_X(x,y)<\delta \implies \forall f \in \mathcal F d_Y(f(x),f(y))<\epsilon##. I know that for an arbitrary ##x_0 \in X## and a given ##\epsilon## I can find ##\delta##. I also know that ##X## is compact. I think I should write ##X## as a union of open covers involving something with the ##\delta## that works for each point ##x_0##, then extract a finite subcover (I would have finite ##\delta##'s) and take the minimum of those deltas. I got stuck trying to find the proper union of open covers.
     
  2. jcsd
  3. Nov 19, 2013 #2

    pasmith

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    Homework Helper

    Think about what you need your covering balls to do.

    Let [itex]x \in X[/itex] and [itex]y \in X[/itex] be such that [itex]d_X(x,y) < \delta[/itex]. You want to show [itex]d_Y(f(x),f(y)) < \epsilon[/itex]. You have a finite subcover by open balls so there must exist some [itex]z \in X[/itex] such that [itex]x \in B(z,r(z))[/itex] for some ball [itex]B(z,r(z))[/itex] in the subcover, and the triangle inequality will then give you a bound for [itex]d_X(y,z)[/itex]. You can then, by appropriate choice of [itex]\delta[/itex] and [itex]r(z)[/itex], ensure that you have bounds on [itex]d_Y(f(x),f(z))[/itex] and [itex]d_Y(f(y),f(z))[/itex], and a further application of the triangle inequality will give a bound on [itex]d_Y(f(x),f(y))[/itex].
     
  4. Nov 19, 2013 #3
    Let ##\epsilon>0##, I know that for each ##x \in X##, there exists ##\delta_x##: if ##y \in B_X(x,\delta_x) \implies \forall f \in \mathcal F d_Y(f(x),f(y))<\frac {\epsilon} {2}##. If ##y,z \in B(x,\delta_x) \implies d_Y(f(y),f(z)<\epsilon##. I can write ##X## as ##X=\bigcup_{x \in X} B_X(x,\delta_x)##, which is an open cover of ##X##. By hypothesis, there exists ##\delta>0## a Lebesgue number for this open cover. This means that for any ##y,z## such that ##d(y,z)<\delta \implies y,z \in B_X(x,\delta_x)## for some ##x \in X \implies d_Y(f(y),f(z))<\epsilon##, this proves ##\mathcal F## is uniformly equicontinuous.

    I know you've suggested me to get a finite subcover but I've tried to to that before and I didn't know how to get the appropiate ##\delta##, I am still thinking how could I solve it without using the Lebesgue number.
     
  5. Nov 20, 2013 #4

    pasmith

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    So far so good.

    In my earlier post, I suggested that you look at a finite subcover obtained from the open cover [itex]\{ B(x, r(x)) : x \in X\}[/itex]. Thus there exists a finite [itex]Z \subset X[/itex] such that [itex]\{ B(z, r(z)) : z \in Z \}[/itex] is an open cover of [itex]X[/itex].

    The entire point of the construction is that we obtain a finite set of radii of covering balls, which allows us to take
    [tex]
    \delta = \min \{ r(z) : z \in Z \} > 0
    [/tex]
    and it remains to choose [itex]r(z) > 0[/itex].

    If [itex]x \in B(z,r(z))[/itex] and [itex]d_X(x,y) < \delta \leq r(z)[/itex] then the triangle inequality will give a bound on [itex]d_X(y,z)[/itex] in terms of [itex]r(z)[/itex]. Your aim is to ensure that [itex]d_X(y,z) < \delta_z[/itex].
     
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