# Thread: Uniform convergence and differentiation

1. ## Uniform convergence and differentiation

(1)

(a) Let $h_n(x)=\frac{sin(nx)}{n}.$ Show that $h_n \rightarrow 0$ uniformly on $\mathbb{R}.$ At which points does the sequence of derivatives $h'_n$ converge?

(b) Modify this example to show that it is possible for a sequence $(f_n)$ to converge uniformly but for $(f'_n)$ to be unbounded.

(2) Consider the sequence of functions defined by $g_n(x)= \frac{x^n}{n}$.

(a) Show $(g_n)$ converges uniformly on $[0,1]$ and find $g=lim$ $g_n$. Show that $g$ is differentiable and compute $g'(x)$ for all $x \in [0,1].$

(b) Now show that $(g'_n)$ converges on [0,1]. Is the convergence uniform? Set $h = lim$ $g'_n$ and compare $h$ and $g'$. Are they the same?

Thanks for any and all help!

2. Show what you did. You asked several questions on convergence of series of functions. You surly had to learn something. Post what you did.

(1)

(a) Let $h_n(x)=\frac{sin(nx)}{n}.$ Show that $h_n \rightarrow 0$ uniformly on $\mathbb{R}.$ At which points does the sequence of derivatives $h'_n$ converge?
Okay, I will give this a shot.

Show that $h_n \rightarrow 0$ uniformly on $\mathbb{R}$
$|h_n(x)|< 1/n \rightarrow 0$
$h_1 = sinx$
$h_2 = \frac{sin(2x)}{2}$

Show $\forall \epsilon > 0 \exists N_{\epsilon} : n>N \Rightarrow |\frac{sin(nx)}{n} - 0| < \epsilon$ for all x's in $\mathbb{R}$

$|\frac{sin(nx)}{n}| \leq 1/n$

So we need to make $1/n$ $< \epsilon$ $\Rightarrow$ $n > \frac{1}{\epsilon}$

(b) Not exactly sure on this part but..

$h'_n(x) = \frac{1}{n}n$ $cos(nx)$
$h'_n(\frac{\pi}{2}) = cos(\frac{n\pi}{2}) = 0$
$h'_n(\frac{\pi}{4}) = cos(n\cdot\frac{\pi}{4})$

Please let me know how I can improve on this. Thanks!

$h'_n(x) = \frac{1}{n}n$ $cos(nx)$
$h'_n(\frac{\pi}{2}) = cos(\frac{n\pi}{2}) = 0$
$h'_n(\frac{\pi}{4}) = cos(n\cdot\frac{\pi}{4})$
$h'_n(x) = \sin (nx)$ while $h(x) = 0 \implies h'(x) = 0$. The problem is asking to find all $x$ such that $\lim h_n'(x)$ exists and is equal to $h'(x)$, i.e. $\lim \sin (nx) = 0$. If $x \not = 2\pi k$ then $\sin (nx) \not = 0$ and so $\{ \sin (nx) \}$ will oscillate. Which means there can be no limit. It seems that $x=2\pi k$ are the points where the converge works.