5
$\begingroup$

I'm given the following function sequence:

$$f_n = \frac{nx}{1+nx^2}, \forall x \in A = [0,\infty].$$

I show the following that:

$$\lim_{n \to \infty} \frac{nx}{1+nx^2} \le \frac{nx}{nx^2} \le \frac{1}{x}.$$

And thus my convergent function I compute is $f(x) = \frac{1}{x}.$

However the answer appears to be $f(x) = \frac{1}{2x}$ using A/G mean inequality.

This leads to my next question that if $f_n \to f$ converges EITHER point wise or uniformly to $f$, is $f$ unique?

$\endgroup$
3
  • 1
    $\begingroup$ To compute the pointwise limit, you shouldn't have those inequalities. Simply hold $x$ constant and compute the limit by diving the numerator and denominator by $n$ to get that the pointwise limit is the function $1/x$. $\endgroup$
    – user683478
    Commented Jun 30, 2021 at 21:08
  • $\begingroup$ It is easy to prove, from the definition of uniform or pointwise convergence (or any convergence on metric spaces, actually) that the limit function is unique in both cases. $\endgroup$
    – Albert
    Commented Jun 30, 2021 at 21:12
  • $\begingroup$ $f_n(x)$ converges pointwise to the function $f(x)=\frac{1}{x}\mathbb{1}_{(0,\infty)}(x)$ on $[0,\infty)$, that is, $f(0)=0$ and $f(x)=\frac{1}{x}$ for $x>0$. Convergence is not uniform however. $\endgroup$
    – Mittens
    Commented Jun 30, 2021 at 21:23

2 Answers 2

2
$\begingroup$

The question of uniform convergence is solved by considering the supremum of the fraction $$\sup\limits_{(0,+\infty)}\left|\frac{nx}{1+nx^2}-\frac{1}{x}\right| = \sup\limits_{(0,+\infty)}\frac{1}{x(1+nx^2)}=+\infty$$

$\endgroup$
2
  • $\begingroup$ Is there a missing factor of $x$ in the denominator? Either way, convergence is not uniform since the supremum is $\infty$. $\endgroup$
    – shoteyes
    Commented Jul 2, 2021 at 3:23
  • $\begingroup$ Thanks, @shoteyes. Fixed. $\endgroup$
    – zkutch
    Commented Jul 2, 2021 at 3:28
0
$\begingroup$

The pointwise convergent function $f$ is not $f = \frac{1}{x}$ on the domain given A. Please note the correct answer below for f:

$f = 0: x = 0$

$f = \frac{1}{x}: x > 0 $

Note in the sequence $f_n(0) = 0, \forall n \in \mathbb N$

Convergence is not uniform since f is not continuous.

If $f_n \to f$ converges pointwise f is unique and which can be shown by a contradiction taking $2\epsilon = |f_1 - f_2|$ where $f_1 \ and \ f_2$ are diffrent pointwise convergence functions and resulting in a $ 2\epsilon < 2\epsilon \ \forall \epsilon > 0$ which statement is a contradiction.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .