Test for pointwise and uniform convergence
Test for pointwise and uniform convergence of $u_n(x)=n(1-x)x^{n-1}$ on $[0,1]$.
Explanation of Method
Although the pointwise limit is identically $0$ (the factor $x^{n-1}$ eventually beats the linear growth of $n$ at every fixed $x<1$), the function's graph develops a "moving bump" whose peak height does not decay: locating the maximum via calculus shows it occurs at $x=1-\frac1n$ and stabilizes at height $\to1/e$, so the supremum of the error never tends to $0$.
- Find the pointwise limit at each fixed $x\in[0,1]$.
- Locate the maximum of $u_n$ via calculus.
- Evaluate the limiting height of that maximum to conclude non-uniform convergence.
Step 1 · Pointwise limit
At $x=0$ and $x=1$: $u_n=0$ for every $n$. For $0<x<1$: writing $u_n(x)=(1-x)\cdot nx^{n-1}$, and since $nx^{n-1}\to0$ (exponential decay of $x^{n-1}$ beats the linear factor $n$), $u_n(x)\to0$. So $u(x)\equiv0$ on $[0,1]$.
Step 2 · Locate the maximum
$u_n'(x)=nx^{n-2}\left[(n-1)-nx\right]=0$ at $x=\dfrac{n-1}n=1-\dfrac1n$. Evaluating: $$ u_n\!\left(1-\tfrac1n\right) = n\cdot\tfrac1n\cdot\left(1-\tfrac1n\right)^{n-1} = \left(1-\tfrac1n\right)^{n-1}. $$
Step 3 · Limiting height
$\left(1-\dfrac1n\right)^{n-1}\to e^{-1}$ as $n\to\infty$ (a standard limit), so $$ \lim_{n\to\infty}\sup_{x\in[0,1]}|u_n(x)-0| = \frac1e \ne 0. $$ Not uniform.
Summary Table
| Quantity | Value |
|---|---|
| Pointwise limit | $u(x)\equiv0$ |
| Location of max of $u_n$ | $x=1-1/n$ |
| $\sup_{[0,1]}u_n$ | $(1-1/n)^{n-1}\to1/e$ |
Additional Notes
- This is the classical "moving bump" (or "gliding hump") example: the sequence converges pointwise to a limit that is uniformly small nowhere, since the region of non-negligible value slides toward $x=1$ as $n$ grows, always carrying a fixed amount of "mass" (here, a peak height near $1/e$) with it.
- The same gliding-hump mechanism, applied to infinite matrices rather than functions of a real variable, underlies the necessity direction of Problem 161's theorem about matrices mapping bounded sequences to null sequences.
- This example contrasts sharply with Problem 176: there, the failure of uniform convergence traced to a discontinuity in the limit function; here, the limit function $u\equiv0$ is perfectly continuous, yet uniform convergence still fails, because a bump of fixed height simply relocates rather than flattening.
Result
Converges pointwise to $0$; not uniformly, since the peak height stabilizes at $1/e$: