# Thread: Uniform convergence and differentiation

1. ## Uniform convergence and differentiation

(1)

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

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

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

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

(b) Now show that $\displaystyle (g'_n)$ converges on [0,1]. Is the convergence uniform? Set $\displaystyle h = lim$ $\displaystyle g'_n$ and compare $\displaystyle h$ and $\displaystyle g'$. 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 $\displaystyle h_n(x)=\frac{sin(nx)}{n}.$ Show that $\displaystyle h_n \rightarrow 0$ uniformly on $\displaystyle \mathbb{R}.$ At which points does the sequence of derivatives $\displaystyle h'_n$ converge?
Okay, I will give this a shot.

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

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

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

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

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

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

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

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