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Let $f_n(x)=\sin \sqrt{x+4n^2x^2}$ on $[0,\infty)$.

(1) Prove that $f_n$ is equi-continuous on $[0,\infty)$.
(2) $f_n$ is uniformly bounded.
(3) $f_n \to 0$ pointwise on $[0,\infty)$
(4) There is no subsequence of $f_n$ that converges to $0$ uniformly
(5) Compare with the Arzela-Ascoli Theorem.

Below is the definition of equi-continuous.

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Does this problem mean equi-continuous at a point or uniformly?

Could someone give me some hint? Thanks!

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    $\begingroup$ You have a problem at $0:$ $f_n(1/n)-f_n(0)$ does not converge to $0.$ $\endgroup$ – zhw. Apr 20 '15 at 16:56
  • $\begingroup$ @Why $f_n(1/n)-f_n(0)$ does not converge to 0? I thought it is convergent to 0. Could you please tell me the reason why it is not convergent? $\endgroup$ – Sherry Apr 21 '15 at 4:23
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Note the last sentence of the first paragraph you quote: "The family is equicontinuous if it is equicontinuous at every point of $X$." Here $X = [0,\infty)$. So you would need to show that for every $x_0 \in [0,\infty)$ and every $\epsilon > 0$, there exists $\delta > 0$ such that for every $n$ and every $x$ with $|x-x_0| < \delta$, we have $|f_n(x_0) - f_n(x)| < \epsilon$.

However, zhw's comment shows that this does not hold when $x_0 = 0$, so in fact the family is not equicontinuous on $[0,\infty)$. Since $f_n(x) = \sin\left(\sqrt{x + 4 n^2 x^2}\right)$, we have $f_n(1/n) = \sin\left(\sqrt{\frac{1}{n} + 4}\right)$ and $f_n(0) = 0$. As $n \to \infty$, by continuity of $\sin$ and $\sqrt{\cdot}$ we get $|f_n(1/n) - f_n(0)| \to \sin(2) \ne 0$. So choosing some $\epsilon < \sin(2)$, for every $\delta > 0$ we can choose $n$ so large that $1/n < \delta$ and so that $|f_n(1/n)| > \epsilon$, and taking $x = 1/n < \delta$ we have $|f_n(x) - f_n(0)| > \epsilon$.

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  • $\begingroup$ Why $f_n(1/n)-f_n(0)$ does not converge to 0? I thought it is convergent to 0. Could you please tell me the reason why it is not convergent? $\endgroup$ – Sherry Apr 21 '15 at 4:24
  • $\begingroup$ @Sherry: I added an explanation. Is it possible you have copied the question incorrectly? $\endgroup$ – Nate Eldredge Apr 21 '15 at 5:32
  • $\begingroup$ Although I copied it incorrectly, your explanation give me hint. Thanks! $\endgroup$ – Sherry Apr 21 '15 at 5:53

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