Is $Tf(x)=\frac{1}{x}\int_{0}^{x}f(y)dy$ bounded as operator on $L^2((0,1);\mathbb{R} )$?

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.

• Maybe i should be more precise, the question is say if $Tf \in L^2 \forall f\in L^2$ and, if the case, say if $T$ is bounded or not, thanks. Apr 5, 2020 at 0:40
• I have a strong feeling that you may need uniform boundedness principle. Apr 5, 2020 at 8:01
• If you can prove that $\int_0^1 |Tf(x)\,g(x)| \,dx\in\mathbb{R}$ for each $g\in L^2$, then $Tf\in L^2$ by uniform boundedness principle. Apr 5, 2020 at 8:17

The trick is to appear a factor of the form $$x^\alpha$$, for a suitable $$\alpha$$, small enough to work things out. I use $$\alpha=1/4$$. We have the following calculation: Let $$f\in L^2(0,1)$$.
Then $$|T(f)(x)|\leq\frac{1}{x}\int_0^x|f(y)|dy=\frac{1}{x}\int_0^x|f(y)|y^{-1/4}y^{1/4}dy\leq$$ $$\frac{1}{x}\bigg(\int_0^x|f(y)|^2y^{1/2}dy\bigg)^{1/2}\cdot\bigg(\int_0^xy^{-1/2}dy\bigg)^{1/2}=\frac{1}{x}\bigg(\int_0^x|f(y)|^2y^{1/2}dy\bigg)^{1/2}\cdot\sqrt{2}x^{1/4}=\frac{\sqrt{2}}{x^{3/4}}\bigg(\int_0^x|f(y)|^2y^{1/2}dy\bigg)^{1/2}$$
So $$|T(f)(x)|^2\leq\frac{2}{x^{3/2}}\int_0^x|f(y)|^2y^{1/2}dy.$$ Integrating and using Tonelli's theorem, $$\|T(f)\|_2^2\leq2\int_0^1\frac{1}{x^{3/2}}\int_0^x|f(y)|^2y^{1/2}dydx=2\int_0^1\int_y^1\frac{1}{x^{3/4}}|f(y)|^2y^{1/2}dxdy=$$ $$=2\int_0^1|f(y)|^2y^{1/2}(4-4y^{1/4})dy\leq 2M\|f\|_2^2,$$ where $$M$$ is the maximum value of $$y\mapsto y^{1/2}(4-4y^{1/4})$$ on $$[0,1]$$.