Family of equicontinuous functions on compact set

In summary, we want to prove that if a family of equicontinuous functions is contained in a compact metric space, then it will also be uniformly equicontinuous. To do this, we need to show that for any given epsilon, there exists a delta such that for any x and y in the space, if the distance between them is less than delta, then the distance between the images of the functions at those points will be less than epsilon. One approach to this is to use the fact that compact spaces can be covered by a finite number of open balls, and then use the triangle inequality to find a bound for delta based on the radii of these balls. By choosing the minimum of these radii, we can ensure
  • #1
mahler1
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Homework Statement .

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.
 
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  • #2
mahler1 said:
Homework Statement .

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.

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].
 
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  • #3
pasmith said:
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].

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.
 
  • #4
mahler1 said:
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##.

So far so good.

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.

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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Related to Family of equicontinuous functions on compact set

1. What is the definition of a family of equicontinuous functions?

A family of equicontinuous functions is a set of functions that are all continuous and have the same rate of change at every point on a given compact set. In other words, the functions in the family do not vary greatly in their rates of change, making them "equi" or equal in continuity.

2. How is a compact set defined?

A compact set is a subset of a larger set that is both closed and bounded. This means that all the points in the set are contained within a finite distance from each other, and the set contains all of its limit points.

3. What is the significance of a family of equicontinuous functions on a compact set?

A family of equicontinuous functions on a compact set is significant because it allows for a stronger guarantee of uniform continuity. This means that the functions in the family will have consistent rates of change, making them easier to analyze and work with mathematically.

4. How does equicontinuity differ from pointwise continuity?

Equicontinuity refers to the continuity of a set of functions, while pointwise continuity refers to the continuity of a single function at each individual point. A family of equicontinuous functions is pointwise continuous, but the converse is not always true.

5. Can a family of equicontinuous functions on a compact set have different rates of change at different points?

No, a family of equicontinuous functions on a compact set must have the same rate of change at every point on the set. This is what distinguishes it from a family of continuous functions, which may have varying rates of change at different points.

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