# 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 $1/x$. Jun 30 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.
– Hal
Jun 30 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. Jun 30 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 $\infty$. Jul 2 at 3:23
• Thanks, @shoteyes. Fixed. Jul 2 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.