sequence of functions

**skybluesea2010** · November 15th 2011, 06:26 PM

**Problem:**

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

Here is what I did: I proved that the pointwise limit of $\left \{ f_{n} \right \}$ is the function $f\left ( x \right )=\frac{x^{2}}{4\pi }$ . Then, in order to prove the uniform convergence, I need to prove that $\sup_{x\in\left [ 0, \alpha \right ] } \left \{ \left | f_{n}\left ( x \right )-f\left ( x \right ) \right | \right \}n \to 0$ 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 $x\in\left [ 0, \alpha \right ]$ , and using the inequality $\sin\left ( x \right )\geq x-\frac{x^{3}}{3!}$ ,
we get:
$\left| n \sin\sqrt{4\pi ^{2}n^{2}+x^{2}} \right| -\frac{x^{2}}{4\pi }|\leqslant \frac{a^{2}}{4\pi }\left ( 1-\frac{2}{\sqrt{1+\frac{a^{2}}{4\pi ^{2}n^{2}}}+1} \right )+\frac{n}{3!}\frac{\alpha ^{6}}{8n^{3}\pi ^{3}}$

I don't understand how the book got this inequality based on $\sin\left ( x \right )\geq x-\frac{x^{3}}{3!}$ . 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 R, there is hint saying that I should use $\left | \sin\left ( x \right ) \right | \leqslant \left | x \right |$ to get:
$\left| n \sin\sqrt{4\pi ^{2}n^{2}+x^{2}} \ -\frac{x^{2}}{4\pi }\right|\geq \frac{x^{2}}{4\pi }\left ( 1-\frac{2}{\sqrt{1+\frac{x^{2}}{4\pi ^{2}n^{2}}}+1} \right)$

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