Thread: Real Analysis - Sequences

1. Real Analysis - Sequences

Hey, I'm pretty sure I already solved part a. of this question, but decided to put it up just in case. Any help would be greatly appreciated. Thanks.

Let fn(x) = nx / (1 + nx^2)

a. Find the pointwise limit of (fn) for all x Є (0, infinity).
b. Is the convergence uniform on (0, infinity)?
c. Is the convergence uniform on (0,1)?
d. Is the convergence uniform on (1, infinity)?

2. Originally Posted by ajj86
Hey, I'm pretty sure I already solved part a. of this question, but decided to put it up just in case. Any help would be greatly appreciated. Thanks.

Let fn(x) = nx / (1 + nx^2)

a. Find the pointwise limit of (fn) for all x Є (0, infinity).
b. Is the convergence uniform on (0, infinity)?
c. Is the convergence uniform on (0,1)?
d. Is the convergence uniform on (1, infinity)?
Note that for x≥1, f_n(x) = nx/(1+nx^2) < nx/(nx^2) = 1/x ≤ 1 so each of the $f_n$ is bounded above by $1$ on $[1, infinity)$ so since $f_n$ is continuous on $[0, 1]$ and $[0, 1]$ is compact we have by the extreme value theorem that $f_n(x)$ is bounded above on $[0, 1]$ taking the greater of $1$ and the upper bound for $f_n(x)$ on $[0, 1]$, upper bound for $f_n(x)$ on $[0, infinity]$. Therefore each one of the $f_n$ is bounded above on $[0, infinity]$.

to find the pointwise limit of the $f_n$, we see that for $x>0$

$f_n(x)$

$= \frac{nx}{1+nx^2}$

$= \frac{1}{(1/(nx) + x)}$

this approaches 1/x as n\forwardarrow\infinity now for every n, $f_n(0) = 0$, we have

n---->infinity lim f_n(x) = {0 if x=0, 1/x if x≠0}

this shows it is not bounded above on $[0, infinity]$. It follows that the $f_n$ cannot converge uniformly