adam512
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Hello,
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 f: [0, \infty) \rightarrow \mathbb{R}. For each n define f_n(x) = f(x^n). Show that the set of continuous function \{f_1, f_2, \ldots \} is equicontinuous at x=1 if and only if f is a constant function.
1. Assume f = k is constant, then f_n(x) = f(x^n) = k for all x.
|f_n(x) - f_n(y) | = |f(x^n) - f(y^n)| = |k-k| = 0, so it is indeed equicontinuous everywhere, in particular at 1.
2. Assume f_n is equicontinuous at x=1. By definition this means
\forall \epsilon > 0, \, \exists \delta > 0 : |1 - y| < \delta \Rightarrow |f_n(1) - f_n(y)| < \epsilon holds for all n.
Now f_n(1) = f(1^n) = f(1) for all n, let's say f(1^n) = c, and f_n(y) = f(y^n). So we can write the above expression:
\forall \epsilon > 0, \, \exists \delta > 0 : |1 - y| < \delta \Rightarrow |f(y^n) - c| < \epsilon.
Now from this I want to arrive at f is constant. I think I need to use the fact that f is continuous but I can't see how. It feels intuitively right that if |f(y^n) - c| < \epsilon for every n, it is arbitrarily close to a constant function along a sequence which for y < 1 converges to 0, and for y > 1 diverges to \infty.
Thanks in advance!
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 f: [0, \infty) \rightarrow \mathbb{R}. For each n define f_n(x) = f(x^n). Show that the set of continuous function \{f_1, f_2, \ldots \} is equicontinuous at x=1 if and only if f is a constant function.
1. Assume f = k is constant, then f_n(x) = f(x^n) = k for all x.
|f_n(x) - f_n(y) | = |f(x^n) - f(y^n)| = |k-k| = 0, so it is indeed equicontinuous everywhere, in particular at 1.
2. Assume f_n is equicontinuous at x=1. By definition this means
\forall \epsilon > 0, \, \exists \delta > 0 : |1 - y| < \delta \Rightarrow |f_n(1) - f_n(y)| < \epsilon holds for all n.
Now f_n(1) = f(1^n) = f(1) for all n, let's say f(1^n) = c, and f_n(y) = f(y^n). So we can write the above expression:
\forall \epsilon > 0, \, \exists \delta > 0 : |1 - y| < \delta \Rightarrow |f(y^n) - c| < \epsilon.
Now from this I want to arrive at f is constant. I think I need to use the fact that f is continuous but I can't see how. It feels intuitively right that if |f(y^n) - c| < \epsilon for every n, it is arbitrarily close to a constant function along a sequence which for y < 1 converges to 0, and for y > 1 diverges to \infty.
Thanks in advance!