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Problem:

Prove that the sequence of functions $\{ f_n \}$ defined by: $f_n(x)= n \sin (\sqrt{4\pi^2n^2+x^2})$ converges uniformly on $[ 0, \alpha]$ where $\alpha > 0$. Does $\{ f_n \}$ converge uniformly on $\mathbb{R}$?

Here is what I did: I proved that the pointwise limit of $\{ f_{n} \}$ is the function $f( x ) = \frac{x^2}{4\pi}$. Then, in order to prove the uniform convergence, I need to prove that $$ \sup_{x\in [ 0, \alpha] } \left \{ | f_n (x)- f(x) | \right \} \to 0 ,$$ as $n \to \infty $ and that's where I am stuck. In the book, there is a hint saying that for $x$ in the mentioned interval $x \in [ 0, \alpha]$, and using the inequality $\sin x \geqslant x-\frac{x^{3}}{3!}$, we get: $$ \left| n \sin \sqrt{4\pi^2 n^2+x^2} -\frac{x^2}{4\pi} \right| \leqslant \frac{a^2}{4\pi } \left ( 1-\frac{2}{\sqrt{1+\frac{a^2}{4\pi^2 n^2}}+1} \right ) + \frac{n}{3!} \frac{\alpha^6}{8n^3 \pi^3} .$$

I don't understand how the book got this inequality based on $\sin x \geqslant x - \frac{x^3}{3!}$ for $x \geqslant 0$. Can anyone give me a detailed proof how to get that inequality given by the book? From that point, I can easily prove the uniform convergence.

For the uniform convergence on $\mathbb R$, there is hint saying that I should use $| \sin x | \leqslant \left | x \right |$ to get: $$ \left| n \sin\sqrt{4\pi^2 n^2+x^2} \ -\frac{x^2}{4\pi}\right| \geqslant \frac{x^2}{4\pi} \left( 1-\frac{2}{\sqrt{1+\frac{x^2}{4\pi^2 n^2}}+1} \right) .$$

Any help please how can we derive the last inequality too? because from this inequality, I can easily prove the non-uniform convergence on $\mathbb R$.

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If any of you guys reading this post, please I am waiting your answers or comments. Thanks –  M.Krov Nov 16 '11 at 4:26

1 Answer 1

up vote 3 down vote accepted

Uniform convergence over $[0, \alpha]$. I presume you are trying the apply the inequality to bound the given expression directly. Though this is correct, this is not useful because the inequality $\sin \theta \geqslant \theta - \frac{\theta^3}{3!}$ ($\theta \geqslant 0$) is tight only for small $\theta$, whereas the expression inside the "sin" grows unbounded in our case. So we massage the function a little before employing the inequality.

$$ \begin{eqnarray*} \sin (\sqrt{4\pi^2 n^2 + x^2}) &=& \sin (\sqrt{4\pi^2 n^2 + x^2} - 2 n \pi) \\ &=& \sin \left(\frac{4\pi^2 n^2 + x^2 - (2 n \pi)^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} \right) \\ &=& \sin \left(\frac{x^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} \right). \end{eqnarray*} $$ Notice that $\frac{x^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi}$ is small, so we can hope to apply the inequality at this stage. Doing so gives $$ \begin{eqnarray*} \frac{x^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} &\geqslant& \sin (\sqrt{4\pi^2 n^2 + x^2}) \\ &\geqslant& \frac{x^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} - \frac{1}{3!} \left( \frac{x^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} \right)^3. \end{eqnarray*} $$

Try to take it from here.

EDIT: More steps added. The left hand side inequality implies that $$ \frac{x^2}{4\pi} - n\sin(\sqrt{4 \pi^2 n^2 + x^2}) \geqslant 0, $$ so we only need to upper bound the difference. For this, we use the right hand side inequality: $$ \begin{eqnarray*} \frac{x^2}{4 \pi} - n\sin(\sqrt{4 \pi^2 n^2 + x^2}) &\leqslant& \frac{x^2}{4 \pi} - \frac{n x^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} + \frac{n x^6}{6 (\color{Green}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi})^3} \\ &\leqslant& \frac{x^2}{4 \pi} - \frac{nx^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} + \frac{nx^6}{6 (\color{Green}{4n\pi})^3} \\ &\leqslant& \color{Red}{\frac{x^2}{4 \pi}} \left[ 1 - \frac{2}{\sqrt{1 + \frac{x^2}{4 \pi^2n^2}} + 1} \right] + \color{Red}{\frac{nx^6}{6 (4n\pi)^3}} \\ &\leqslant& \color{Red}{\frac{{\alpha}^2}{4 \pi}} \left[ \color{Blue}{ 1 - \frac{2}{\sqrt{1 + \frac{x^2}{4 \pi^2n^2}} + 1} } \right] + \color{Red}{\frac{n{\alpha}^6}{6 (4n\pi)^3}} \\ &\leqslant& \frac{\alpha^2}{4 \pi} \left[ \color{Blue}{ 1 - \frac{2}{\sqrt{1 + \frac{\alpha^2}{4 \pi^2n^2}} + 1}} \right] + \frac{n\alpha^6}{6 (4n\pi)^3}, \end{eqnarray*} $$ repeatedly using the fact that $0 \leqslant x \leqslant \alpha$.


Non-uniform convergence over $[0, \infty)$. We saw that $$ \begin{eqnarray*} n\sin (\sqrt{4\pi^2 n^2 + x^2}) &\leqslant& \frac{nx^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} \\ &=& \frac{x^2}{4 \pi} \frac{2}{\sqrt{1 + \frac{x^2}{4\pi^2n^2}} + 1} . \end{eqnarray*} $$ Can you get the inequality claimed by the hint from here?

EDIT: More hints on the non-uniform convergence. This problem is meant to highlight the (subtle at first glance) difference between pointwise and uniform convergence. For any fixed $x$, as $n \to \infty$, it is true that the sequence $f_n(x)$ converges to $f(x)$; but this is what pointwise convergence is about.

If we want to show uniform convergence, we need to show that the sequence $$ u_n := \sup \{ |f_n(x) - f(x)| \colon x \in \mathbb R \} $$ goes to $0$ as $n \to \infty$. The idea is to provide a uniform upper bound on the error term $|f_n(x) - f(x)|$ that is independent of $x \in \mathbb R$, such that the upper bound goes to $0$ as $n \to \infty$.

In our example, we are interested in the sequence $$ \left| n\sin (\sqrt{4\pi^2 n^2 + x^2}) - \frac{x^2}{4 \pi} \right| = \frac{x^2}{4 \pi} - n\sin (\sqrt{4\pi^2 n^2 + x^2}) . $$ To prove non-uniform convergence (over $[0, \infty)$), we need to lower bound the sequence $$ u_n := \sup_{x \in [0, \infty)} \left( \frac{x^2}{4 \pi} - n\sin (\sqrt{4\pi^2 n^2 + x^2}) \right) . $$ By the inequality given to you in the text-book hint, we have $$ \begin{eqnarray*} u_n &\geqslant& \sup_{x \in [0, \infty)} \frac{x^2}{4\pi} \left ( 1-\frac{2}{\sqrt{1+\frac{x^2}{4\pi^2n^2}}+1} \right) . \\ &\geqslant& \sup_{x \in [0, \infty)} \frac{x^2}{4\pi} \left ( 1-\frac{2}{\Big(\frac{|x|}{2 n \pi} \Big)} \right) . \\ &=& \sup_{x \in [0, \infty)} \left( \frac{x^2}{4\pi} - n|x| \right) . \end{eqnarray*} $$ Taking $x = 8 n \pi$, we get $$ \begin{eqnarray*} u_n \geqslant \frac{64 n^2 \pi^2}{4 \pi} - n \cdot 8 n \pi = 8 n^2 \pi \to \infty, \end{eqnarray*} $$ as $n \to \infty$. Therefore, clearly, $u_n$ cannot approach $0$ as $n \to \infty$. In fact, by choosing $x$ a little more carefully,* you can show that $u_n=\infty$ for each $n$.

*Note: Notice that our choice of $x$ varies as $n$ varies. This is unavoidable because we already know that for any fixed $x$, the above deviation term goes to $0$.

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I am still not able to reach the final answer based on your hint. Can you please elaborate more? –  M.Krov Nov 16 '11 at 14:41
    
I just need one more step to finish my proof: for non-uniform convergence on $ mathbb R$, using your provided hint, I could prove the second inequality given in the statement of the problem (the one given as a hint by the book to prove non-uniform converegence). I think that the idea is to prove that right hand side doesn't tend to zero as n tends to infinity. –  M.Krov Nov 16 '11 at 17:49
    
But, actually as n tends to infinity, the right hand side tends tends to zero, and thus the left hand side of the inequality is greater than or equal to 0 which is not sufficient to prove non uniform convergence, because I need it to be strictly greater than 0. Can you, help me out with this one? and Thanks for all the hints you gave so far. –  M.Krov Nov 16 '11 at 17:50
1  
@Zi2 Precisely. In fact, for any $n$, the right hand side is unbounded (so that, sup over all $x \in \mathbb R$ is infinity). [More explicitly, given $n$, take $x = 2 \pi n$ and see what happens to the difference.] –  Srivatsan Nov 16 '11 at 17:53
    
Indeed. The right hand side is unbounded which makes it clear now that the convergence cannot be uniform. Thanks again. –  M.Krov Nov 16 '11 at 18:00

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