For which values of the parameter $r$ the operator is compact?

Question

I have an operator $$A : L_{2}[0,1] \to L_{2}[0,1]$$ which is $$(Ax)(t)={t^{r-1}}\int_{0}^{t}\frac{x(s)}{s^r}ds$$

I already proved that operator A is bounded in $L_{p}[0,1]$ space when $p\in(1,\infty)$. Since I have $p=2$ an operator is bounded.

Then to show it is not a compact operator, usually we construct a bounded sequence $f_n$ where $Af_n$ does not have a convergent subsequence. But what we should do if we have parameter? I guess that given operator isn't compact at all values of the parameter.

Any help would be great!

Answer 1

The assumption $r<{1\over 2}$ is needed for the boundedness of $A.$ For $x_a(t)=t^a,$ where $a> -{1\over 2},$ the function belongs to $L^2(0,1).$ We get $$(Ax_a)(t)= {1\over a-r+1}t^a={1\over a-r+1}x_a(t)$$ Hence the spectrum of $A$ contains the interval $ \left [0,{({1\over 2}-r})^{-1}\right].$ Therefore the operator cannot be compact.