Strangely defined ball compact in $L^p(I)$ or not?

Question

Let $I = (0, 1)$ and $1 \le p \le \infty$. Set$$B_p = \{u \in W^{1, p}(I) : \|u\|_{L^p(I)} + \|u'\|_{L^p(I)} \le 1\}.$$When $1 < p \le \infty$, does it necessarily follow that $B_p$ is compact in $L^p(I)$?

Answer 1

Note that

$B_p$ compact in $L^p$ $\Rightarrow$ $B_p$ closed in $L^p$ $\Rightarrow$ $B_p$ complete in $L^p$ $\Rightarrow$ $W^{1,p}(I)$ complete w.r.t. $\|\cdot\|_{L^p}$

So, $B_p$ isn't compact in $L^p$.

However, $B_p$ is relatively compact in $L^p$.