The inclusion $j:L^{\infty}(0,1)\to L^1(0,1)$ is continuous but not compact. I'm stuck on this problem, namely I cannot find a bounded subset in $L^\infty(0,1)$ such that it is not mapped by the canonical inclusion $$j: L^\infty(0,1)\to L^1(0,1)$$ onto a relatively compact subset in $L^1(0,1)$. Can anybody provide me an example? Really I don't see the point.
My thoughts are wondering on the fact that the ball of $L^\infty(0,1)$ is norm dense in $L^1(0,1)$ so the inclusion cannot be compact, however, as i said, no practical examples come to my mind.
Thank you very much in advance. 

This is actually just a variant of a special case of NKS’s example, but it may be especially easy to visualize with this description.
For $n\in\mathbb{Z}^+$ and $x\in(0,1)$ let $f_n(x)$ be the $n$-th bit in the unique non-terminating binary expansion of $x$. Then $\|f_n\|_\infty=1$, but $\|f_n-f_m\|_1=\frac12$ whenever $n\ne m$.
