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For each $n \in \mathbb{N}$ consider the function $f_n : [0,+\infty) \to \mathbb{R}$ given by $$f_n(x) := \sin\left(\sqrt{4\pi^2n^2+x}\right), \ \ \ \ \forall x \ge 0.$$

Prove that

  1. $f_n$ converges uniformly on each interval $[0,a]$ with $a > 0$;
  2. $f_n$ does not converge uniformly on $[0,+\infty)$.

Original image at http://i.stack.imgur.com/WNTyz.png

My attempt:

I showed how to prove (2), just let $x_n=\pi^2(1/4+2n)$ then for any $n$, $f_n(x_n)=1$, so $f_n(x_n)$ cannot converge uniformly on $[0,\infty)$.

I have question about (1), If I use Arzela-Ascoli theorem, I can only show there is a subsequence converge uniformly on $[0,a]$. How to conclude that $f_n$ actually converge uniformly?

Could someone kindly help? Thanks!

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  • $\begingroup$ Rewrite as $f_n(x) = sin\big(n\sqrt{4\pi+\frac{x}{n^2}}\big)$, but now $x$ is bounded above. $\endgroup$ – Xiao Jul 25 '15 at 4:26
  • $\begingroup$ In your answer to (2), you should write $x_n,$ not $x.$ I don't see why you think $f_n(x_n)=1$ with your choice of $x_n.$ $\endgroup$ – zhw. Jul 25 '15 at 5:26
  • $\begingroup$ ^For his choice of $x_n = \pi^2(1/4+2n)$, we have $f_n(x_n)$ $= \sin(\sqrt{4\pi^2n^2+\pi^2(1/4+2n)})$ $= \sin(\sqrt{\pi^2(2n+1/2)^2})$ $= \sin((2n+1/2)\pi)$ $= \sin(\pi/2) = 1$. $\endgroup$ – JimmyK4542 Jul 25 '15 at 6:20
  • $\begingroup$ @JimmyK4542 Thanks. $\endgroup$ – zhw. Jul 25 '15 at 7:07
  • $\begingroup$ I found it a little easier to understand with $x_n = (25\pi^2/4)n^2 - 4\pi^2n^2$ $\endgroup$ – zhw. Jul 25 '15 at 18:36
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Hint: For any $n \in \mathbb{N}$ and any $x \in [0,a]$, we have:

$|f_n(x)| = \left|\sin\left(\sqrt{4\pi^2n^2+x}\right)\right|$ $= \left|\sin\left(\sqrt{4\pi^2n^2+x} - 2\pi n\right)\right|$ $\le \left|\sqrt{4\pi^2n^2+x}-2\pi n\right|$

$= \sqrt{4\pi^2n^2+x}-2\pi n$ $= \dfrac{x}{\sqrt{4\pi^2n^2+x}+2\pi n}$.

Can you continue the above work to get a uniform bound on $|f_n(x)|$ over all $x \in [0,a]$?

EDIT Spoiler:

Continuing from above: $|f_n(x)| \le \dfrac{x}{\sqrt{4\pi^2n^2+x}+2\pi n} \le \dfrac{a}{\sqrt{4\pi^2n^2}+2\pi n} = \dfrac{a}{4\pi n}$ for all $n \in \mathbb{N}$ and all $x \in [0,a]$. So for any $\epsilon > 0$, choose an integer $N > \dfrac{a}{4\pi\epsilon}$. Then for all $n \ge N$ and all $x \in [0,a]$, you have $|f_n(x)| \le \dfrac{a}{4\pi n} < \epsilon$. Hence, $f_n \to 0$ uniformly on $[0,a]$.

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  • $\begingroup$ +1. Why the two downvotes, though?! $\endgroup$ – triple_sec Jul 25 '15 at 6:45
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I think the Arzelà–Ascoli theorem is an overkill here. Let me show you a proof based on more elementary methods.


Claim 1: Define, for each $n\in\mathbb N$, $g_n:[0,a]\to\mathbb R$ as $$g_n(x)\equiv\sqrt{4\pi^2 n^2+x}-\sqrt{4\pi^2 n^2}\quad\text{for every $x\in[0,a]$}.$$ Then, $(g_n)_{n\in\mathbb N}$ converges to $0$ uniformly on $[0,a]$.

Proof: First, let me show pointwise convergence. For any $x\in[0,a]$, \begin{align*} g_n(x)=&\,\frac{\left(\sqrt{4\pi^2 n^2+x}-\sqrt{4\pi^2 n^2}\right)\left(\sqrt{4\pi^2 n^2+x}+\sqrt{4\pi^2 n^2}\right)}{\sqrt{4\pi^2 n^2+x}+\sqrt{4\pi^2 n^2}}\\ =&\,\frac{(4\pi^2n^2+x)-(4\pi^2 n^2)}{\sqrt{4\pi^2 n^2+x}+\sqrt{4\pi^2 n^2}}\\ =&\,\frac{x}{\sqrt{4\pi^2 n^2+x}+\sqrt{4\pi^2 n^2}}\to0\quad\text{as $n\to\infty$,} \end{align*} given that the denominator diverges. Uniform converges follows from the fact that $0\leq g_n(x)\leq g_n(a)$ for each $x\in[0,a]$. $\blacksquare$


Claim 2: The sequence $(\sin\circ g_n)_{n\in\mathbb N}$ converges to $0$ uniformly on $[0,a]$.

Proof: Fix $\varepsilon>0$. Since $\sin$ is continuous at $0$, there exists some $\delta>0$ such that $y\in(-\delta,\delta)$ implies $|\!\sin(y)|<\varepsilon$. Pick $N\in\mathbb N$ so large that for all $n\in\mathbb N$ such that $n>\mathbb N$, one has $|g_n(x)|<\delta$ for each $x\in[0,a]$. Then, it follows that $|(\sin\circ g_n)(x)|<\varepsilon$ whenever $n\in\mathbb N$, $n>N$, and $x\in[0,a]$. $\blacksquare$


The final step is to simply observe that for each $n\in\mathbb N$ and $x\in[0,a]$, one has $$f_n(x)=\sin\left(\sqrt{4\pi^2n^2+x}\right)=\sin(g_n(x)+2\pi n)=\sin(g_n(x)),$$ given that $\sin$ has period $2\pi$.

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