Limited operator I am trying to prove that the given operator is bounded in the $L_2[0, 1]$ space: $$ (Ax)(t) = t^{r-1} \int\limits^{t}_{0} \frac{x(s)}{s^r} ds
$$
The way I'm trying to do it is via a Hölder's inequality, Fubini's theorem and also using Hardy's inequality for integrals. I know that $r$ must be less than $\frac{1}{2}$ in order for the operator above to be bounded. However, I am unable to prove that. Can anyone help me? Thank you.
 A: $$\frac{1}{t}\int^t_0\big(\tfrac{t}{s}\big)^rx(s)\,ds=\int^1_0\frac{x(ut)}{u^r}\,du$$
Then, using Minkowski's inequality (the generalized version rather)
\begin{align}
\Big(\int^1_0\Big|\int^1_0\frac{x(ut)}{u^r} \,du\Big|^2\,dt\Big)^{1/2}&\leq \int^1_0\frac{1}{u^r}\Big(\int^1_0|x(ut)|^2\,dt\Big)^{1/2}\,du\\
&=\int^1_0\frac{1}{u^r}\frac{1}{u^{1/2}}\Big(\int^u_0|x(t)|^2\,dt\Big)^{1/2}\,du\\
&\leq \|x\|_2\int^1_0\frac{1}{u^{r+1/2}}\,du
\end{align}
For the last integral to converge, it is necessary that $r+\tfrac12<1$.
A: This is to address a comment of the OP which is not part of the original posting.
Denote $A_r$ the operator $A_rx(t)=\int^1_0\frac{x(tu)}{u^r}\,du$ on $L_2(0,1)$.
Suppose $r=1/2$.
If $x_p(s)=\frac{1}{s^p}$ with $p<1/2$, then $\|x\|_2=\frac{1}{(1-2p)^{1/2}}<\infty$. Let $X_p(t)=(1-2p)^{1/2}x_p(t)$. Then,
$$A_{1/2}X_p(t):=\frac{\sqrt{1-2p}}{t^p}\int^1_0\frac{1}{u^{p+1/2}}\,du=\frac{1}{t^p}\frac{2}{\sqrt{1-2p}}=\frac{2}{1-2p}X_p(t)$$
Hence
$$\|A_{1/2}X_p\|_2=\frac{2}{1-2p}$$
Notice that $\|X_p\|_2=1$ for all $p<1/2$, and that $\lim_{p\rightarrow1/2-}\|AX_p\|_2=\infty$, that is $A_{1/2}$ is unbounded.
Similar examples can be used to show that $A_r$ is unbounded of $r>1/2$.
