Given the operator $T:L^2((0,1);\mathbb{R} )\rightarrow L^2((0,1);\mathbb{R} )$ defined by $Tf(x)=\dfrac{1}{x}\displaystyle\int_{0}^{x}f(y)\,\mathrm dy$, say if it is well defined and discuss its boundedness.
This is part of an exercise which asked the same thing for operators of the form $T_{\alpha}f(x)=\dfrac{1}{x^\alpha}\displaystyle\int_{0}^{x}f(y)\,\mathrm dy$ with $\alpha >0$. We find that $T_\alpha$ is bounded for $\alpha \in (0,1)$ and not even well defined for $\alpha >1$.
For $\alpha=1$ we tried to exhibit a sequence $f_n\in L^2$ for which $\|Tf_n\|^2_2/\|f_n\|^2_2$ diverges, but we find that truncated funtions $h(x)\chi(x)_{(1/n,1)}$ with $h(x)=x^\beta$ or $\dfrac{\ln(x)}{x}$ don't make the trick. Another attempt was to write $f_n(x)=\Sigma_kf_{nk}(x)\chi(x)_{(s(k),s(k+1))}$ for some "partitioning function" $s$ and hope to balance the growth of $f_{nk}$ with the speed of interval $(0,1)$ subdivision. However calculations are very heavy, and led us nowhere.
Another thought was that $T$ could maybe by bounded over some dense subspace, which would answer to the question in a certain sense.
Thank you for reading, hope the best.