6
$\begingroup$

I'm having trouble with the following problem:

Prove that the operator $$Tf(x) = \frac{1}{\pi} \int_0^\infty \frac{f(y)}{x+y} dy$$ is bounded on $L^2(0,\infty)$ with norm $||T|| \leq 1$.

I have no idea where to start for this problem. I guess I can write $Tf(x) = \frac{1}{\pi} \int_0^\infty K(x,y)f(y) dy$ with $K(x,y) = \frac{1}{x+y}$ but I don't see how I can proceed from here. Can anyone help me out? Thanks!

$\endgroup$
1
  • $\begingroup$ Are you sure that it is $L^2(0,\infty)$? $\endgroup$
    – sinbadh
    Jan 27, 2016 at 5:41

1 Answer 1

2
$\begingroup$

Let $x\in (0,\infty)$ and $f\in L^2(0,\infty)$. Via the transformation $y\mapsto xy$,

$$Tf(x) = \frac{1}{\pi}\int_0^\infty \frac{f(xy)}{x + xy}\, x\, dy = \frac{1}{\pi} \int_0^\infty \frac{f(xy)}{1 + y}\, dy.$$

Use Minkowski's integral inequality and the transformation $x\mapsto x/y$ to get the estimate

$$\|Tf\|_{L^2} \le \frac{1}{\pi}\int_0^\infty \frac{\|f\|_{L^2}}{y^{1/2}\,(1 + y)}\, dy.$$

Via the transformation $y\mapsto y^2$,

$$\frac{1}{\pi}\int_0^\infty \frac{\|f\|_{L^2}}{y^{1/2}(1 + y)}\, dy = \left(\frac{2}{\pi}\int_0^\infty \frac{1}{1 + y^2}\, dy\right)\|f\|_{L^2} = \|f\|_{L^2}.$$

Hence $\|Tf\|_{L^2} \le \|f\|_{L^2}$. Since $f$ was arbitrary, $\|T\| \le 1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .