Is this Hardy-like operator $f \mapsto x^{-\frac{1}{2}} \int_{x}^{2x} f(y) dy$ bounded from $L^2$ to $L^2$?

Question

I'm interested in knowing whether

$$Tf(x) = x^{-\frac{1}{2}}\int_{x}^{2x}f(y)dy = \int_{\mathbb R} K(x,y)f(y)dy, \quad x > 0$$

where $K(x,y) = x^{-\frac{1}{2}} 1_{0 < y \leq x \leq 2y}(x,y)$, defines an operator which is bounded from $L^2((0,\infty))$ to $L^2((0,\infty))$.

Using Minkowski inequality, one can prove the boundedness from $L^1((0,\infty))$ to $L^2((0,\infty))$ :

\begin{align*} ||Tf||_{L^2} &= \left( \int_{\mathbb R} \left| \int_{\mathbb R} K(x,y)f(y)dy \right|^2 dx \right)^{\frac{1}{2}} \leq \int_{\mathbb R} \left( \int_{\mathbb R} |K(x,y)f(y)|^2dx \right)^{\frac{1}{2}} dy \\ &= \int_{0}^{+\infty} \left( \int_{\frac{1}{2}y}^{y} x^{-1}|f(y)|^2dx \right)^{\frac{1}{2}} dy \lesssim ||f||_{L^1} \end{align*}

but I was not able to prove or disprove the $L^2-L^2$ boundedness. This looks related to Hardy's inequality.

Answer 1

Your $T$ is not well-defined in $L^2(0,\infty)$.

Let $f(t)=\frac1t\,1_{[1,\infty)}$. Then $f\in L^2(0,\infty)$ and $$ \|Tf\|_2^2 \geq\int_{1}^\infty \frac1x\,\Big|\int_x^{2x}\frac1t\,dt\Big|^2\,dx =\log^22\,\int_{1}^\infty\frac1x\,dx=\infty. $$ Hence $Tf\not\in L^2(0,\infty)$.