# How can I show that the closed unit ball in $L^1([0,1])$ is not compact?

Consider the following subset of $L^1([0,1])$, $S=\left\{f\in L^1([0,1]):{\|f\|}_1\leq1\right\}$. Prove that $S$ is not compact.

Should I start with an open cover and prove that it has no finite subcover or find a convergent sequence that has no convergent sub-sequence? I am quite confused.

• In general, convergent subsequences are much easier to work with. Mar 22 '17 at 5:04
• True in any infinite-dimensional normed space: math.stackexchange.com/questions/287360/…. Mar 27 '17 at 17:11

Take $f_n(x)$ to be the $n$th binary digit in the expansion of $x \in (0,1)$. Then $\|f_n\|_1 = { 1\over 2}$ and $\|f_n-f_m\|_1 = {1 \over 2}$ for all $m \neq n$.

Let $U_n = B(f_n,{1 \over 4})$, this is an open cover with no finite sub cover.

Addendum: To illustrate $f_n$, suppose $x= {1 \over 2} + {1 \over 8} = 0.101\bar{0}$, then $f_1(x) = 1$, $f_2(x) =0$, $f_3(x) = 1$ and$f_k (x) = 0$ for all $k > 3$. In general ($x \in (0,1)$) $f_n(x) = \lfloor 2^n x \rfloor\pmod 2$.

• Thanks a lot. But I don't quite get what does it mean by "the nth digit in the expansion of x". Thanks again. Mar 22 '17 at 12:24
• @emma_s $f_n(x)=\big\lfloor 10(10^{n-1}x - \lfloor10^{n-1}x \rfloor \big\rfloor$ or in other words if $x=0.a_1a_2\ldots$ then $f_n(x)=a_n$. Mar 26 '17 at 21:59
• @freakish: I presume you meant $10$ rather than $2$? Mar 26 '17 at 22:24
• @copper.hat $2$? The sequence $(a_1, a_2, \ldots)$ is a decimal expansion of $x$, $a_i$ is a digit. Written as digits $x=0.a_1a_2a_3a_4\ldots$. Mar 26 '17 at 22:27
• @freakish: The $f_n$ in the answer are the binary digits of $x$. Mar 26 '17 at 22:28

Consider $f_n:=2^n\mathbf{1}_{[0,1/2^n]}$.