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.

  • $\begingroup$ 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. $\endgroup$
    – dan93
    Apr 5, 2020 at 0:40
  • $\begingroup$ I have a strong feeling that you may need uniform boundedness principle. $\endgroup$ Apr 5, 2020 at 8:01
  • $\begingroup$ 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. $\endgroup$ Apr 5, 2020 at 8:17

1 Answer 1


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 $f\in L^2(0,1)$.

(1st inequality: triangular, 2nd inequality: Holder)

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 $[0,1]$.

A comment: This operator is special and has a name, it is called the Hardy operator. There are some more things known about it, as you can see here.

  • $\begingroup$ Thank you very much, you made my day. I will also see much more on Hardy's operator, thank you again, have a nice day. $\endgroup$
    – dan93
    Apr 5, 2020 at 14:22
  • $\begingroup$ @dan93 Glad to be of help :) $\endgroup$ Apr 5, 2020 at 14:25

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