# uniform convergence fn to f

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?

• 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 $1/x$.
– user683478
Commented Jun 30, 2021 at 21:08
• 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. Commented Jun 30, 2021 at 21:12
• $f_n(x)$ 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. Commented Jun 30, 2021 at 21:23

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

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

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.