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

compactnessfunctional-analysislebesgue-integrallp-spacesmeasure-theory

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 $$p=1$$.)

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.