For u_n(x)=x^n (n=1,2,...): (1) show the sequence
For $u_n(x)=x^n$ ($n=1,2,\ldots$): (1) show the sequence converges on $[0,1]$ and find its limit; (2) show it does not converge uniformly on $[0,1]$; (3) does it converge uniformly on $[0,a]$ for $a<1$?
Explanation of Method
The pointwise limit is discontinuous at $x=1$ (jumping from $0$ to $1$), and since each $u_n$ is continuous while uniform limits of continuous functions must be continuous, uniform convergence on $[0,1]$ is impossible. Restricting to $[0,a]$ with $a<1$ removes the troublesome endpoint, and the supremum of the error is then exactly $a^n\to0$.
- Find the pointwise limit $u(x)=\lim_nx^n$ on $[0,1]$.
- Show $\sup_{[0,1]}|u_n-u|=1$ for every $n$, ruling out uniform convergence.
- Show $\sup_{[0,a]}|u_n-u|=a^n\to0$, confirming uniform convergence on $[0,a]$.
Step 1 · Pointwise limit on $[0,1]$
For $x\in[0,1)$: $x^n\to0$. For $x=1$: $1^n=1\to1$. So $$ u(x)=\lim_{n\to\infty}x^n = \begin{cases}0,&x\in[0,1)\\1,&x=1.\end{cases} $$
Step 2 · No uniform convergence on $[0,1]$
$|u_n(x)-u(x)|=x^n$ for $x\in[0,1)$ and $=0$ at $x=1$. Since $x^n\to1$ as $x\to1^-$, $\sup_{x\in[0,1]}|u_n(x)-u(x)|=1$ for every $n$ (the supremum is approached but not attained in $[0,1)$). So $\lim_n\sup|u_n-u|=1\ne0$: not uniform. (Equivalently: each $u_n$ is continuous but $u$ is not, and uniform limits of continuous functions are continuous.)
Step 3 · Uniform convergence on $[0,a]$, $a<1$
Here $u\equiv0$ on $[0,a]$, and $x^n$ is increasing in $x$, so $\sup_{x\in[0,a]}|u_n(x)-0|=a^n$. Since $0\le a<1$, $a^n\to0$ — uniform convergence holds on $[0,a]$.
Summary Table
| Interval | Convergence |
|---|---|
| $[0,1]$ | Pointwise only; limit discontinuous at $x=1$ |
| $[0,a]$, $a<1$ | Uniform, sup-error $=a^n\to0$ |
Additional Notes
- This is the canonical textbook example illustrating that pointwise convergence does not preserve continuity, while uniform convergence does — the discontinuity of the limit at $x=1$ is itself a proof that the convergence cannot be uniform on any interval containing that point.
- The phenomenon here (uniform convergence on every proper closed sub-interval, but not on the full interval including the problematic endpoint) is extremely common and reappears in Problem 177's more elaborate "moving bump" example.
- The rate $a^n\to0$ on $[0,a]$ is geometric, so convergence there is not just uniform but exponentially fast — a useful contrast with slower polynomial rates like $O(1/n)$ seen in Problems 178–180.
Result
Pointwise on $[0,1]$; not uniform there; uniform on every $[0,a]$, $a<1$: