For $p \geq 1$, let $\mathcal{C}_{p}^0 [0,1]$ be the set of continuous functions in $[0,1]$ with the norm:
$\lVert f \rVert_p = \left( \int_0^1 |f(x)|^p dx\right)^{\frac{1}{p}}$
Show that the unit sphere $S = \{ f \in \mathcal{C}_{p}^0 [0,1] : \lVert f \rVert_p = 1\}$ is not compact.
My attempt
I know how to prove it for $p=1$ by defining a sequence of functions $(f_k)$ of $S$ such that for every $k \in \mathbb{N}$, $f_k$ is a triangle of height $2k(k+1)$ in $[\frac{1}{k+1}, \frac{1}{k}]$ and zero everywhere else, so that $\lVert f_k \rVert_1 = 1$. It is easy to prove that the $(f_k)$ does not have a convergent subsequence. But when I am struggling to prove it for any $p \geq 1$. There is a similar question at: Unit sphere in $L^p([0,1])$ is not compact. But the sequence given there does not consist of continuous functions.