hey guys and girls i've ran into a bit of a fix and i'm not sure how to go about this problem.
as the question title states im trying to find the values of P such that $f_n(x)=\frac{nx}{1+n^2x^p}$ converge uniformly to its pointwise limit (which unless im wrong is 0)
i've considered several ways of going about this but cant quite seem to get this right. ok so heres my current plan
in order to find p i'm using this theorem
Let $(f_n)$ be a sequence if real-valued differentiable function on $[a,b]$. suppose that $f_n(x)$ converges for at least one $x \in [a,b]$ and that $f'(x)$ converges uniformly on $[a,b]$. then
(i) $f_n$ converges uniformly on $[a,b]$
(ii) $f = \lim_{n \rightarrow \infty} f_n$ is differentiable on $[a,b]$, and for all $x \in [a,b]$, $f'(x) = \lim_{x \rightarrow \infty} f'_{n}(x)$
so heres the plan.
(1) find whether $f_n$ converges pointwise to a limit (it does it's zero)
(2) differentiate $f_n$ and determine whether that converges pointwise (again it does)
(3) determine whether $f'_n(x)$ converges uniformly by finding a maximum value of $f'_n(x)$ and hoping the limitations of p jumps out at me.
so here we go.
(1.) $$\lim_{n \rightarrow \infty}f_n = \lim_{n \rightarrow \infty} \frac{nx}{1+n^2x^p} = 0$$
(2.) $$f'(x) = \frac{d}{dx}\left(\frac{nx}{1+n^2x^p}\right)= n\frac{d}{dx}\left(\frac{x}{1+n^2x^p}\right)$$
$$n\left(\frac{\frac{d}{dx}(x)(1+n^2x^p) - \frac{d}{dx}(1+n^2x^p)x}{(1+n^2x^p)^2} \right)$$
$$n\left(\frac{1+n^2x^p(1-p)}{(1+n^2x^p)^2} \right) = \frac{n+n^3x^p(1-p)}{(1+n^2x^p)^2}$$
$$\lim_{n \rightarrow \infty}\frac{n+n^3x^p(1-p)}{(1+n^2x^p)^2} = 0 \text{ pointwise}$$
right in order to determine whether this converges uniformly
$f_n$ converges uniformly if $\forall \epsilon > 0 \exists N \in \mathbb{N} \text { such that when } n > N \Longrightarrow \sup_{x \in [0,1]} |f_n(x) - f| < \epsilon $
we need to find the largest value of $f'_n(x)$ can take (so long as x remains in the interval). in order to do this we differentiate $f'_n(x)$ and find it's critical point
i make $$f''_n(x) = \frac{n^3px^{p-1}(n^{2}(p-1)x^p-p-1)}{(1+n^2x^p)^3}$$ which means that $$f''_n(x) = 0 \text{ at } x=0 \text{ and } x^{p} = \frac{1+p}{n^2(p-1)}$$
with the latter being the maximum of $f'n$
so going from the definition of uniform convergence.
let $\epsilon > 0 \exists N \in \mathbb{N}$ such that when $ n> N$ implies
$$\sup_{x \in [0,1]}|f'_n(x)-f'(x)| = \sup_{x \in [0,1]}|f'_n(x)| \leq f'_n \left( \sqrt[p]\frac{1+p}{n^2(p-1)} \right)$$ which is equal to $$\frac{n+n^3\left( \frac{1+p}{n^2(p-1)} \right)(1-p)}{\left(1+n^2\left( \frac{1+p}{n^2(p-1)} \right)\right)^2} = \frac{n+n(1+p)}{{\left(1+\left( \frac{1+p}{p-1} \right)\right)^2}}=\frac{n+n(1+p)}{{\left(\frac{p-1+1+p}{p-1}\right)^2}}=\frac{(p-1)^2(n+n(1+p))}{4p^2}$$
annnnd this is where im stuck. i was hoping the values of p would become ovbious for this to converge. any suggestions? any help. i'd greatly appreciate it.
would it be smarter to, rather then try and show $f'_n$ converges uniformly. just running through the def of uniform convergence for $f_n$ using the maxima i find with $f'_n$?
HINT
I would appeal directly to the second statement you make that $f_n$ converges uniformly to $0$ if and only if $\sup_{x\in [0,1]}|f_n(x)| \to 0.$
So since $f_n\ge 0$ on $[0,1],$ you can calculate the maximum of $f_n$ on $[0,1]$ and take the limit as $n\to \infty$ and see for which $p$ the limit is zero.