Test for pointwise and uniform convergence
Test for pointwise and uniform convergence of $u_n(x)=\arctan(nx)$ on $(0,\infty)$.
Explanation of Method
Since $nx\to\infty$ for every fixed $x>0$, the pointwise limit is the constant $\pi/2$. But as $x\to0^+$ (for any fixed $n$), $\arctan(nx)\to0$, so the error $\pi/2-\arctan(nx)$ approaches its worst possible value $\pi/2$ arbitrarily closely — ruling out uniform convergence on the full half-line. Restricting to $[a,\infty)$ for $a>0$ removes this problem, since the error is then bounded by its value at the left endpoint, which does shrink with $n$.
- Find the pointwise limit $u(x)=\pi/2$.
- Show $\sup_{(0,\infty)}|u_n-u|=\pi/2$ for every $n$, ruling out uniform convergence there.
- Show uniform convergence holds on $[a,\infty)$ for any $a>0$.
Step 1 · Pointwise limit
For fixed $x>0$: $nx\to\infty$, so $\arctan(nx)\to\pi/2$. Thus $u(x)\equiv\pi/2$ on $(0,\infty)$.
Step 2 · No uniform convergence on $(0,\infty)$
$|u_n(x)-u(x)|=\pi/2-\arctan(nx)$, a decreasing function of $x$. As $x\to0^+$: $\arctan(nx)\to0$, so this difference $\to\pi/2$. Hence $\sup_{x>0}|u_n(x)-u(x)|=\pi/2$ for every $n$, so $\lim_n\sup|u_n-u|=\pi/2\ne0$: not uniform.
Step 3 · Uniform convergence on $[a,\infty)$, $a>0$
Since $\pi/2-\arctan(nx)$ is decreasing in $x$, its supremum on $[a,\infty)$ occurs at $x=a$: $\sup_{x\ge a}|u_n(x)-u(x)|=\pi/2-\arctan(na)\to0$ as $n\to\infty$ (since $na\to\infty$). Uniform convergence holds.
Summary Table
| Interval | Convergence |
|---|---|
| $(0,\infty)$ | Pointwise only |
| $[a,\infty)$, $a>0$ | Uniform |
Additional Notes
- The mechanism here is the mirror image of Problem 176's example: there, the failure occurred near the right endpoint $x=1$; here, it occurs as $x\to0^+$, the left boundary of the domain that is excluded from the interval.
- Since $u(x)\equiv\pi/2$ is itself continuous, this example shows that the continuity of the limit function is not sufficient for uniform convergence — the same lesson as Problem 177's moving-bump example, but arising here from an unbounded domain rather than a shifting peak.
- This is a standard illustration of the general principle that convergence which is uniform on every closed sub-interval $[a,b]$ of an open (or half-open) domain need not be uniform on the domain itself — "local uniform convergence" does not imply "global uniform convergence."
Result
Pointwise on $(0,\infty)$; not uniform there; uniform on $[a,\infty)$ for $a>0$: