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
Hint:
The reverse triangle inequality states that, if $x,y\in \mathbb{R}$ then: $$ \left||x|-|y|\right| \leq |x-y|. $$
Can you apply this to your question?