Uniform convergence of $f_n(x) = \arctan(nx)$ in the interval $(0,+\infty)$

Question

Let's see if you can help me resolve this contradiction ...

Proved the convergence pointwise of $f_n$ to the function $f = \frac{\pi}{2}$ in the interval $(0,+\infty)$.

I have seen this demonstration of YES uniform convergence in the interval $(0,+\infty)$ :

$$ \sup_{x\in(0,\infty)} \, | f_n(x) - f(x) | = \sup_{x\in(0,\infty)} \, | \arctan (nx) - \frac{\pi}{2} | = \sup_{x\in(0,\infty)} \, \arctan \left( \frac{1}{nx} \right) \to 0 $$

using the trigonometric property $ x> 0 $: $$ \arctan (x) + \arctan (\frac {1}{x}) = \frac {\pi}{2} $$

There really is no uniform convergence in $(0,+\infty)$ !! ... What reason makes this reasoning wrong?? I do not understand.

Thank you very much for your help

Carlos

Answer 1

The mistake you made is to say that $$\sup_{x>0} \arctan(\frac{1}{nx})\to 0$$ All you can say is $$\arctan(\frac{1}{nx})\to 0$$

but

$$\sup_{x>0}\arctan(\frac{1}{nx})\ge \arctan(\frac{1}{n\frac 1n})$$

$$=\frac{\pi}{4}$$

the convergence is not uniform at $ (0,+\infty)$.

It is uniform at $ [A,+\infty) $ for any $ A>0$.