Let $f_n(x) := x^n$ for $x \in [0, 1]$, for $n \in \mathbb N$ . Show that the sequence $(f_n)$ is convergent in $(C[0, 1],\|\|_1 )$ whereas it is not convergent in $(C[0,1],\|\|_\infty)$.
Here's my attempt on the second part : Suppose $f_n(x)$ is convergent in $(C[0,1],\|\|_\infty)$ and let its limit point be $g(x)$, So for $\epsilon > 0$, $\exists$ $n_0$ $\in \mathbb{N}$ such that $\|f_n - g\|_\infty$ < $\epsilon$ for $n$ $\geq n_0$ i.e $\|x^n - g(x)\|_\infty$ < $\epsilon$ for $n$ $\geq n_0$ or $sup_{(0\leq x\leq1)}|x^n - g(x)|$ $< \epsilon$ for $n$ $\geq n_0$, Now if $(i)$ $g(x) = 1$ for some $ x\in [0,1]$ then $sup_{(0\leq x\leq1)}|x^n - g(x)|$ $\geq 1$ so by taking $\epsilon = 1/2$ implies that $f_n(x)$ is not convergent. $(ii)$ $g(x)\ne1$ for $x \in[0,1]$ so $|1-g(t)|>0$ for some $t\in[0,1]$ then $sup_{(0\leq x\leq1)}|x^n - g(x)|\geq |1-g(t)| $, so by taking $\epsilon = |1-g(x)|/2$ implies that $f_n(x)$ is not convergent, therefore in either case it is not convergent in $(C[0,1], \|\|_\infty)$
If $f_n \to f$ in $\|\cdot\|_\infty$ then $f_n(x)\to f(x)$ for all $x \in [0,1]$. So there is only one candidate limit if the sequence converges.
The pointwise limit $(f_n)$ is $$f(x)=\begin{cases} 0 & 0 \le x < 1\\ 1 & x=1\end{cases}$$
which is not continuous so is not in $C([0,1])$.
In $\|\cdot \|_1$ the sequence converges to $g(x)\equiv 0, g \in C([0,1])$, as $$\|f_n - g\|_1 = \int_0^1 |f_n(x)-g(x)|\,dx = \int_0^1 x^n dx = \frac{1}{n+1} \to 0$$