How can I show that $\{ x^{-a} : a \in [0,1) \}$ isn't compact in $L^1(0,1)$, but $\{ x^{-a} : 0 \leq a \leq 1-\delta \}$ is compact for $\delta \in (0,1)$? It probably has something to do with $x^{-1}$ not being in $L^1(0,1)$, but how can I use that? Can I somehow argue for the first set that $x^{a_k}$ with $a_k = \frac{k}{k+1}$ has no convergent subsequence? Or can I somehow use that $||x^{-a_k}||_1$ goes against $\infty$? What does convergence mean in the $L^1$ norm? ($L^1$ is the $L^p$ space with $p=1$.)

# Compact set in $L^1(0,1)$

compactnessfunctional-analysislebesgue-integrallp-spacesmeasure-theory

## Best Answer

If $\delta\in(0,1)$, the the map$$\begin{array}{ccc}[0,1-\delta]&\longrightarrow&L^1(0,1)\\x&\mapsto&x^{-a}\end{array}$$is continuous. Since $[0,1-\delta]$ is compact, its range is compact.

On the other hand, if $f_n(x)=x^{-1+1/n}$, then $\|f_n-f_m\|_1=|n-m|$. So, the sequence $(f_n)_{n\in\Bbb N}$ has no Cauchy subsequence, and therefore it has no convergent subsequence. So, $\{x^{-a}\mid a\in[0,1)\}$ is not compact.