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I have a problem I cannot solve. I have been working with problems with convergence of sequences of functions for some time now. But I can't seem to solve most of the problems. Anyway here is my problem:

Consider a continuous function [itex]f: [0, \infty) \rightarrow \mathbb{R} [/itex]. For each [itex]n [/itex] define [itex]f_n(x) = f(x^n) [/itex]. Show that the set of continuous function [itex]\{f_1, f_2, \ldots \}[/itex] is equicontinuous at [itex] x=1[/itex] if and only if [itex]f [/itex] is a constant function.

1. Assume [itex]f = k[/itex] is constant, then [itex] f_n(x) = f(x^n) = k [/itex] for all [itex] x[/itex].

[itex] |f_n(x) - f_n(y) | = |f(x^n) - f(y^n)| = |k-k| = 0 [/itex], so it is indeed equicontinuous everywhere, in particular at 1.

2. Assume [itex] f_n [/itex] is equicontinuous at [itex] x=1 [/itex]. By definition this means

[itex] \forall \epsilon > 0, \, \exists \delta > 0 [/itex] : [itex] |1 - y| < \delta \Rightarrow |f_n(1) - f_n(y)| < \epsilon [/itex] holds for all [itex]n [/itex].

Now [itex] f_n(1) = f(1^n) = f(1) [/itex] for all [itex]n [/itex], let's say [itex]f(1^n) = c [/itex], and [itex] f_n(y) = f(y^n) [/itex]. So we can write the above expression:

[itex] \forall \epsilon > 0, \, \exists \delta > 0 [/itex] : [itex] |1 - y| < \delta \Rightarrow |f(y^n) - c| < \epsilon [/itex].

Now from this I want to arrive at [itex] f [/itex] is constant. I think I need to use the fact that [itex]f [/itex] is continuous but I can't see how. It feels intuitively right that if [itex] |f(y^n) - c| < \epsilon[/itex] for every [itex]n [/itex], it is arbitrarily close to a constant function along a sequence which for [itex] y < 1 [/itex] converges to 0, and for [itex] y > 1 [/itex] diverges to [itex] \infty [/itex].

Thanks in advance!

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# Equicontinuity at a point if.f. continuous function constant

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