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.

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I changed the $L_1$ in your title to $L^1$. Hope you don't mind. –  NKS Feb 24 '12 at 0:17
Maybe I'm missing something, but an operator is compact if the image of the closed ball is pre-compact and by your argument that would imply $L^1(0,\infty)=L^\infty(0,\infty)$. –  azarel Feb 24 '12 at 0:19
The ball of $L^\infty$ is not norm dense in $L^1$. For instance, you cannot approximate the constant function 2 in $L^1$ norm by functions from the unit ball of $L^\infty$ (i.e. by functions bounded by 1). –  Nate Eldredge Feb 24 '12 at 2:43

Rademacher's functions are useful tools here. Let $\alpha, \beta \in \mathbb R$ be distinct real numbers. Define $f \in L^\infty(0,1)$ as $$f(x) = \begin{cases} \alpha & 0 < x \le 1/2\\ \beta & 1/2 < x < 1 \end{cases}$$ Then set $u_n(x) = f(2^nx \pmod 1).\$ You can compute directly that $(u_n)$ has no convergent subsequence in any $L^p$.
It’s even easier to see if you set $u_n(x)=f\left(2^nx\bmod 1\right)$. –  Brian M. Scott Feb 24 '12 at 0:28
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 $n\ne m$.