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Math Help - Equicontinuity Question

  1. #1
    Member Haven's Avatar
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    Equicontinuity Question

    Let f be a continuous function on \mathbb{R}. Let f_n = f(nt) for n \in \mathbb{N} be equicontinuous on [0,1]. I.e., \forall \epsilon >0, \exists \delta >0 such that if |x-y| < \delta, then |f_n(x) - f_n(y)| < \epsilon for all n.

    What can we conclude about f?

    All I am able to get is that since each f_n is defined on a compact set, then each f_n is pointwise bounded. So, \{f_n \} is uniformly bounded. So, f is uniformly bounded.

    Is there something else that I can conclude?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Haven View Post
    Let f be a continuous function on \mathbb{R}. Let f_n = f(nt) for n \in \mathbb{N} be equicontinuous on [0,1]. I.e., \forall \epsilon >0, \exists \delta >0 such that if |x-y| < \delta, then |f_n(x) - f_n(y)| < \epsilon for all n.

    What can we conclude about f?

    All I am able to get is that since each f_n is defined on a compact set, then each f_n is pointwise bounded. So, \{f_n \} is uniformly bounded. So, f is uniformly bounded.

    Is there something else that I can conclude?
    You can show that f must be constant. Can you prove that?
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  3. #3
    Member Haven's Avatar
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    Quote Originally Posted by Drexel28 View Post
    You can show that f must be constant. Can you prove that?
    Actually, I am having a bit of difficult getting showing it is constant. I consider x \in \mathbb{R} and let f(x) = c. Now I suppose \exists y\in \real such that f(y) \neq c. Now let n be a natural number larger than y. It follows that f_n(y/n)= f(y) \neq c.

    However, I am having troubles with the \delta's in this case. for a given epsilon, Does the intermediate value theorem guarantee that I can find z such that x<z<y and z-x < \delta?
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  4. #4
    Super Member girdav's Avatar
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    You can write that |f(x)-f(y)| =|f_n\left(\dfrac xn\right)-f_n\left(\dfrac yn\right)|. Take a \varepsilon >0. We can find a n such that \left|\dfrac xn-\dfrac yn\right|\leq \delta hence |f(x)-f(y)|\leq\varepsilon. Now you can conclude.
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