Monthly Archives: May 2015

Some related problems on L^2 bounded operator

Problem 1. Suppose w is a measurable function on \mathbb{R}^{d} with 0<w\left(x\right)<\infty for a.e. x, and K is a measurable function on \mathbb{R}^{2d} that satisfies:

  • \int_{\mathbb{R}^{d}}\left|K\left(x,y\right)\right|w\left(y\right)dy\le Aw\left(x\right) for almost every x\in\mathbb{R}^{d}, and
  • \int_{\mathbb{R}^{d}}\left|K\left(x,y\right)\right|w\left(x\right)dx\le Aw\left(y\right) for almost every y\in\mathbb{R}^{d}.

Prove that the integral operator defined by

    \[ Tf\left(x\right)=\int_{\mathbb{R}^{d}}K\left(x,y\right)f\left(y\right)dy,\quad x\in\mathbb{R}^{d} \]

is bounded on L^{2}\left(\mathbb{R}^{d}\right) with \left\Vert T\right\Vert \le A.

Note as a special case that if \int\left|K\left(x,y\right)\right|dy\le A for all x, and \int\left|K\left(x,y\right)\right|dx\le A for all y, then \left\Vert T\right\Vert \le A.

Proof. If f\in L^{2}\left(\mathbb{R}^{d}\right), note that by Cauchy-Schwarz inequality we have

    \begin{align*} & \mathrel{\phantom{=}}\int_{\mathbb{R}^{d}}\left|K\left(x,y\right)\right|\left|f\left(y\right)\right|dy\\ & =\int_{\mathbb{R}^{d}}\left|K\left(x,y\right)\right|^{\frac{1}{2}}w\left(y\right)^{\frac{1}{2}}\left|K\left(x,y\right)\right|^{\frac{1}{2}}\left|f\left(y\right)\right|w\left(y\right)^{-\frac{1}{2}}dy\\ & \le\left(\int_{\mathbb{R}^{d}}\left|K\left(x,y\right)\right|w\left(y\right)dy\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^{d}}\left|K\left(x,y\right)\right|\left|f\left(y\right)\right|^{2}w\left(y\right)^{-1}dy\right)^{\frac{1}{2}}\\ & \le A^{\frac{1}{2}}w\left(x\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^{d}}\left|K\left(x,y\right)\right|\left|f\left(y\right)\right|^{2}w\left(y\right)^{-1}dy\right)^{\frac{1}{2}}. \end{align*}

For any f\in L^{2}\left(\mathbb{R}^{d}\right), note that by Fubini’s theorem, we have

    \begin{align*} & \mathrel{\phantom{=}}\left\Vert Tf\right\Vert _{L^{2}\left(\mathbb{R}^{d}\right)}^{2}\\ & =\left(\int_{\mathbb{R}^{d}}\left|Tf\left(x\right)\right|^{2}dx\right)\\ & \le\int_{\mathbb{R}^{d}}Aw\left(x\right)\left(\int_{\mathbb{R}^{d}}\left|K\left(x,y\right)\right|\left|f\left(y\right)\right|^{2}w\left(y\right)^{-1}dy\right)dx\\ & =\int_{\mathbb{R}^{d}}\left|f\left(y\right)\right|^{2}w\left(y\right)^{-1}A\left(\int_{\mathbb{R}^{d}}Aw\left(x\right)\left|K\left(x,y\right)\right|dx\right)dy\\ & \le\int_{\mathbb{R}^{d}}\left|f\left(y\right)\right|^{2}A^{2}dy\\ & =A^{2}\left\Vert f\right\Vert _{L^{2}\left(\mathbb{R}^{d}\right)}^{2}. \end{align*}

So we get \left\Vert Tf\right\Vert _{L^{2}\left(\mathbb{R}^{d}\right)}\le A\left\Vert f\right\Vert _{L^{2}\left(\mathbb{R}^{d}\right)} for all f\in L^{2}\left(\mathbb{R}^{d}\right). So by definition of operator norm, we get

    \[ \left\Vert T\right\Vert \le A. \]

An application of the above problem is the following:

Example 2. The operator T defined on L^2 (0,\infty)

    \[ Tf(x) = \int_{0}^{\infty} \frac{f(y)}{x+y} dy\]

is bounded with \Vert T \Vert \leq 1.

Solution. By previous problem, it suffices to find a function w so that

    \[ \frac{1}{\pi}\int_{0}^{\infty}\frac{1}{x+y}w\left(y\right)dy\le w\left(x\right). \]

Define w\left(x\right)=\frac{1}{\sqrt{x}}. Then this is our desired function. To use the change of variable, set \sqrt{y}=u to get

    \begin{align*} \frac{1}{\pi}\int_{0}^{\infty}\frac{1}{x+y}\cdot\frac{1}{\sqrt{y}}dy & =\frac{1}{\pi}\int_{0}^{\infty}\frac{1}{x+u^{2}}2du\\ & =\frac{2}{\pi}\left[\frac{1}{\sqrt{x}}\arctan\left(\frac{u}{\sqrt{x}}\right)\right]_{0}^{\infty}\\ & =\frac{1}{\sqrt{x}}. \end{align*}

Similary, we get

    \[ \frac{1}{\pi}\int_{0}^{\infty}\frac{1}{x+y}\cdot\frac{1}{\sqrt{x}}dx=\frac{1}{\sqrt{y}}. \]

Apply the previous theorem to yield \Vert T \Vert| \leq 1.