Evaluate the simple convergence and uniform convergence over the interval $(0, \infty)$ for the sequence of functions:
For $n \geq 0$ and $\forall x \in (0, \infty),~~f_n(x) = \tan^{-1}\bigl( \frac{n+x}{1+nx} \bigr)$.
We have $f_n(0) = \tan^{-1}(n)$ which converges to $\pi/2$ as $n \rightarrow \infty$, so I can say that the sequence converges but what the convergence uniformly?
Thanks to the comment below, it seems that I can't consider $f_n(0)$. Consequently, can I use inequalities?
i.e to say $|f_n(x)| \le\frac\pi2\ $ and so converges ?
Best Answer
We have $f_n(x) \to f(x) = \tan^{-1} \left(\frac{1}{x}\right)$ pointwise for $x \in (0,\infty)$.
To examine uniform convergence, consider
$$\begin{align}\sup_{x \in (0,\infty)}|f_n(x) - f(x)| &= \sup_{x \in (0,\infty)}\left| \tan^{-1} \left(\frac{n+x}{1+nx}\right) - \tan^{-1} \left(\frac{1}{x}\right)\right| \\ &= \sup_{x \in (0,\infty)}\left| \tan^{-1}\left( \frac{\frac{n+x}{1+nx} - \frac{1}{x}}{1 +\frac{n+x}{1+nx}\frac{1}{x} } \right)\right| \\ &= \sup_{x \in (0,\infty)}\left| \tan^{-1}\left( \frac{x(n+x)- 1 - nx}{x(1+nx) + n+x} \right)\right| \\ &= \sup_{x \in (0,\infty)}\left| \tan^{-1}\left( \frac{x^2-1}{n(x^2+1) + 2x} \right)\right| \end{align}$$
Can you find the behavior of the RHS as $n \to \infty$? Recall that the arctangent is an increasing function.