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 ?

  • $\begingroup$ You wrote $f_n(0)$ but $f_n$ is not defined at $0.$ $\endgroup$ – zhw. Feb 5 at 19:59
  • $\begingroup$ @zhw Yes you are right. So, I need to consider the $\sup$ ? $\endgroup$ – Jean Feb 5 at 20:03

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.

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  • $\begingroup$ Great help. Unfortunately, I am not sure about what I found. It is positive for some values of $x$ and negative for the rest and I am not familiar with such cases. I would be grateful if you can give more details $\endgroup$ – Jean Feb 6 at 16:30
  • $\begingroup$ @Jean: The arctangent is an increasing odd function. For $x > 1$ we have $\frac{x^2 - 1}{n(x^2+1) + 2x} < \frac{x^2}{nx^2} = \frac{1}{n}$ and $\tan^{1} \left( \frac{x^2 - 1}{n(x^2+1) + 2x}\right) <\tan^{-1}(1/n) \to 0$ as $n \to \infty$. For $0 < x \leqslant 1$ we have $\left|\tan^{1} \left( \frac{x^2 - 1}{n(x^2+1) + 2x}\right) \right| = \tan^{1} \left( \frac{1- x^2}{n(x^2+1) + 2x}\right) < \tan^{-1}(1/n) \to 0$. $\endgroup$ – RRL Feb 6 at 18:24

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