As I was going over the definitions of pointwise and uniform convergence I came to the following problem: since the canonical example for continuous functions on $[0,1)$ which are pointwise but bot uniform convergent(wrt the constant function $f=0$) is sequence of functions $f_n(x) = x^n$ I ask myself is there such sequence of functions for the interval $[0,1]$. So far I couldn't think find any example and I am stating to believe that the answer might be negative. So what do think, is there such sequence and if not can it be proven that such sequence does not exist?
[Math] Pointwise but not uniform convergence of continuous functions on $[0,1]$
uniform-convergence
Related Solutions
Comparison
Pointwise convergence means at every point the sequence of functions has its own speed of convergence (that can be very fast at some points and very very very very slow at others).
Imagine how slow that sequence tends to zero at more and more outer points:
$$\frac{1}{n}x^2\to 0$$
Uniform convergence means there is an overall speed of convergence.
In the above example no matter which speed you consider there will be always a point far outside at which your sequence has slower speed of convergence, that is it doesn't converge uniformly.
Another Approach
One can check uniform convergence by considering the "infimum of speeds over all points". If it doesn't vanish then it is uniformly convergent. And that gives another characterization as the ones with nonvanishing overall speed of convergence.
For pointwise convergence, you fix $x\in [0,1] $ and you compute $\lim_ {n\to+\infty}f(x) $. There will be pointwise convergence in the set containing $x $ for which the limit exists $(\in\mathbb R) $. in your example,
for $x=0$, the limit is zero. for $x=1$, it is zero. for $0 <x <1$, write $(1-x)^n=e^{n\ln (1-x)} $ and you will find zero since exponential is faster than polynomial.
thus all the limits are zero in $[0,1] $ . $(f_n) $ converges (pointwise) to function $0$ at $[0,1] $
For uniform convergence
find the maximum of $|f_n (x)-0|=f_n (x)$ at $[0,1] $.
$$f'_n (x)=n^2 (1-x)^{n-1}(1-x-nx) $$
it is attained at $$x_n=\frac {1}{n+1} $$ $$f (x_n)=n (\frac {n}{n+1})^{n+1}$$
$$\lim_{n\to+\infty}f_n (x_n)=+\infty$$
the convergence is not uniform at $[0,1] $.
Best Answer
Certainly. take for example $f_n(x) = nxe^{-nx}$ which converges pointwise but not uniformly to $0$ on $[0,1]$. (And as pointed out in the comment, your own example works as well.)