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Suppose that $X = L_1([0,1]):= \{f:[0,1]\mapsto [0,\infty): \int_0^1 |f(x)|~dx <\infty\}$. Equip $X$ with the standard metric $d(f,g) = \int_0^1 |f(x)-g(x)|~dx$. Now, define: $$C:= \{f\in X: f(x) \in [0,1]~\forall x \in [0,1] \}~.$$ In other words, $C$ is the set of all non-negative, integrable functions on $[0,1]$, bounded above by $1$. My question is, is $C$ a compact subset of $X$?

I have seen examples showing that the unit $L_1$ ball is non-compact (for example Unit sphere in $L^p([0,1])$ is not compact.), but all these examples are constructing a sequence of unbounded functions on $[0,1]$. In other words, what I want, is a sequence of non-negative, bounded functions in $X$ that does not have a convergent subsequence. Any help will be greatly appreciated!

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I recall seeing this in a Real Analysis course: Define the functions: $f_n:[0,1]\to{}\{0,1\}$ as $f_n(x)=$ the $n$th bit of the binary expression for $x$. You then have that $\int_0^1|f_n-f_m|d\mu=\frac{1}{2}$ (since $f_n-f_m=1$ precisely when the $m$th bit is $0$ and the $n$th is $1$ or vice versa) and thus no subsequence converges.

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  • $\begingroup$ Nice construction, thanks, $\endgroup$ – Usermath Apr 16 '19 at 17:30
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I believe you can also look at the functions generated by the mapping of Alspach (``A fixed point free nonexpansive map,'' Dale Alspach, Proc. of the AMS, Vol. 82, No. 3 (Jul., 1981), pp. 423-424), which is an adaptation of the Baker transform (https://en.wikipedia.org/wiki/Baker%27s_map).

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