# Pointwise convergence and uniform convergence

1. Mar 29, 2010

### complexnumber

1. The problem statement, all variables and given/known data

Define the sequence $$\displaystyle f_n : [0,\infty) \to \left[0,\frac{\pi}{2}\right)$$ by $$f_n(x) := \tan^{-1}(nx), x \geq 0$$.

2. Relevant equations

Prove that $$f_n$$ converges pointwise, but not uniformly on
$$[0,\infty)$$.

Prove that $$f_n$$ converges uniformly on $$[t, \infty)$$ for $$t > 0$$.

3. The attempt at a solution

$$\displaystyle \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \tan^{-1} (nx) = 0$$ for $$x = 0$$.

$$\displaystyle \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \tan^{-1} (nx) = \frac{\pi}{2}$$ for $$x \in (0, \infty)$$.

Let $$\displaystyle f : [0, \infty) \to \left[0, \frac{\pi}{2} \right)$$ be defined by
\begin{align*} f(x) = \left\{ \begin{array}{ll} 0 & \text{ if } x = 0 \\ \dfrac{\pi}{2} & \text{ if } x > 0 \end{array} \right. \end{align*}

Therefore $$f_n$$ converges pointwise to $$f$$.

Is function $$f$$ correct? How can I prove the rest?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 29, 2010

### HallsofIvy

Yes, that is correct. Now, just observe that each function $tan^{-1}(nx)$ is continuous for all x while f(x) is not continuous at x= 0. If convergence were uniform, the limit function would also be continuous.

3. Mar 29, 2010

### complexnumber

Thanks. I found a theorem that says if $$f_n$$ converges uniformly to $$f$$ then $$f$$ must be continuous. Is it enough to say that since $$f$$ is
discontinuous at $$0$$, $$f_n$$ cannot converge uniformly to $$f$$ on $$[0, \infty)$$?

Is the following correct for the second question?

For any $$\varepsilon > 0$$ there must be an $$N \in \mathbb{N}$$ such
that for all $$n > N$$ and for all $$x \in [t, \infty)$$
\begin{align*} \abs{\tan^{-1}(nx) - \frac{\pi}{2}} <& \varepsilon \\ \frac{\pi}{2} - \tan^{-1}(nx) < & \varepsilon \\ \frac{\pi}{2} - \varepsilon < & \tan^{-1}(nx) \\ \tan \left(\frac{\pi}{2} - \varepsilon \right) < & nx \\ \frac{\tan (\frac{\pi}{2} - \varepsilon)}{x} < & n \end{align*}

Therefore choose $$\displaystyle N = \frac{\tan (\frac{\pi}{2} - \varepsilon)}{t}$$. Then for any $$x \in [t, \infty)$$,
\begin{align*} \abs{f_n(x) - f(x)} = \abs{\tan^{-1}(nx) - \frac{\pi}{2}} \leq \varepsilon \end{align*}
whenever $$n > N$$. Hence $$f_n$$ converges uniformly to $$f$$ in $$[t, \infty)$$.

4. Mar 29, 2010

### VeeEight

Yes, this is sometimes called the Uniform Limit Theorem and is fine to use (The above poster mentioned it).