Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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 $\mathbb R$.

share|cite|improve this question
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
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 $0$.

share|cite|improve this answer
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
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.