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] $.
Your question has been answered in comments, and the conclusion is that the text doesn't make any sense. Quotation: "With respect to the supremum-norm the sequence converges pointwise".
- Yes, it converges pointwise.
- Convergence with respect to the supremum-norm trivially implies pointwise convergence.
- Pointwise convergence is much weaker than norm-convergence.
- The statement "With respect to the supremum-norm the sequence converges pointwise" is complete rubbish.
Edited:
Convergence in $C([0,1])$ with respect to the supremum-norm $\lVert - \rVert_\infty$ is nothing else than uniform convergence. The (uniform) limit of a sequence $(f_n)$, if it exists, is again a function in $C([0,1])$. You see that $(x^n)$ has a pointwise limit which is not continuous at the point $x = 1$, hence the convergence cannot be uniform.
Consider the definition
$$||g||_{\infty} := \sup_{x\in[0,1]}{|g(x)|}$$
for $g \in C([0,1])$. It does not make any sense to write
"The sequence $(f_n)_{n\in\mathbb{N}_0}$ converges pointwise to the limit function $f$, given by, $f(x) = 0$ for $x \in [0,1)$ and $f(x) = 1$ for $x=1$, as it holds:
$$||f_n(x)-f(x)||_{\infty} = \sup_{x\in[0,1)}{|f_n(x)-f(x)|} = |x^n|<\epsilon$$
for all $x \in [0,1)$, and $||f_n(1)-1||_{\infty} = 0$."
The notation $||f_n(x)-f(x)||_{\infty}$ is meaningless. Moreover, it is not true that $\sup_{x\in[0,1)}{|f_n(x)-f(x)|} <\epsilon$ unless $\epsilon > 1$. In fact, for each $n$ you have $\sup_{x\in[0,1)}{|f_n(x)-f(x)|} = \sup_{x\in[0,1)}{|x^n|} = 1$.
The correct approach is to replace $||f_n(x)-f(x)||_{\infty}$ by $|f_n(x)-f(x)|$. Then for $x \in [0,1)$ you get correctly $|f_n(x)-f(x)| = |x^n| < \epsilon$ for $n \ge n_0$. For $x = 1$ you get $|f_n(1)-f(1)| = 0$ for all $n
$.
Instead of considering real-valued functions you may consider the set $C([0,1],V)$ of continuous functions $f : [0,1] \to V$, where $V$ is a normed linear space with norm $\lVert - \rVert$. This induces again a supremum norm $\lVert - \rVert_\infty$ on $C([0,1],V)$. Pointwise convergence of a sequence $(f_n)$ means that $\lVert f_n(x) - f(x) \rVert \to 0$ for each $x$, but it is does not make sense to write $\lVert f_n(x) - f(x) \rVert_\infty$.
Best Answer
It does converge pointwise to $0$, for the reason you said. Basically, $f_n(x) = 0$ for sufficiently large $n$ (where "sufficiently large" depends on $x$).
To prove it formally, suppose $x \in [0, 1]$. If $x = 0$, then by definition, $f_n(0) = 0$ for all $n$, so $f_n(0) \to 0$ as $n \to \infty$.
Otherwise $x > 0$. We can then use the fact that $\frac{1}{n} \to 0$ from above as $n \to \infty$. We can find some $N$ such that $$n \ge N \implies 0 < \frac{1}{n} < x.$$ For such $n$, we have $x \notin (0, \frac{1}{n}]$, and hence $$n \ge N \implies f_n(x) = 0.$$ For any $\varepsilon > 0$, we can set this same $N$ to see that $$n \ge N \implies |f_n(x) - 0| < \varepsilon,$$ i.e. $f_n(x) \to 0$ as $n \to \infty$.