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Define $T: L^2(\mathbb{R})\to L^2(\mathbb{R})$ by $(Tf)(x)=\int_{\mathbb{R}}\frac{f(y)}{1+|x|+|y|}dy$.

Is this operator bounded? If it is, then is it also compact?

I got stuck in simply applying Cauchy-Schwarz inequality to show it is bounded.

Any hint on this problem? Thanks in advance!


Remark:

1, I have tried to set $f_n(y)=\frac{1}{\sqrt{n}}1_{[0,n]}$, then $\{||Tf_n||_2\} $ are uniformly bounded, so I guess $T$ is a bounded operator.

More details as following:

As above, take $f_n(y)=\frac{1}{\sqrt{n}}1_{[0,n]}$, then $(Tf_n)(x)=\frac{1}{\sqrt{n}}\int_0^n\frac{1}{1+|x|+y}dy$,

so $||Tf_n||_2^2=\int_{\mathbb{R}^1}\frac{1}{n}|\int_0^n\frac{1}{1+|x|+y}dy|^2dx=\int_{\mathbb{R}^1}\frac{1}{n}(ln(1+|x|+y)|_{y=0}^{y=n})^2dx=\int_{\mathbb{R}^1}\frac{1}{n}(ln(\frac{1+|x|+n}{1+|x|}))^2dx=\frac{2}{n}\int_0^{\infty}(ln(\frac{1+x+n}{1+x}))^2dx=\frac{2}{n}\int_1^{\infty}(ln(1+\frac{n}{y}))^2dy\overset{z=\frac{n}{y}}{=}2\int_{0}^n\frac{ln(1+z)^2}{z^2}dz\leq 2\int_{0}^{\infty}\frac{ln(1+z)^2}{z^2}dz$, while since $z\to 0, \frac{ln(1+z)^2}{z^2}\to 1; z\to\infty, \frac{ln(1+z)^2}{z^2}\leq \frac{1}{z^{1+\epsilon}}, \forall 0<\epsilon<1$, so value of this integral is finite.

2, Note that the kernel $K(x,y)=\frac{1}{1+|x|+|y|}$ appearing here does not belong to $L^2(\mathbb{R})$; otherwise, all the problems can be solved, for example, see Hilbert-Schmidt operator

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$\frac{f(x)}{1+|x|+|y|}\leq f(x)$ –  akkkk Dec 17 '12 at 16:41
    
@akkkk, but $f$ might not be in $L^1(\mathbb{R})$. –  ougao Dec 17 '12 at 16:56
    
Are you sure it is $f(x)$ and not $f(y)$ inside the integral defining $T$? –  Julián Aguirre Dec 17 '12 at 17:29
    
@Julián, sorry, it should be $f(y)$, I have fixed it. –  ougao Dec 17 '12 at 17:35
    
What did you find for $\lVert Tf_n\rVert$ in the first remark? Can you show your computations? –  Davide Giraudo Dec 18 '12 at 12:28
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1 Answer

up vote 7 down vote accepted

$T$ is bounded but not compact.


To see $T$ is bounded, given $f\in L^2(\mathbb{R})$, for the change of variables $y=xt$(when $x\ne 0$ is fixed), we have $$|Tf(x)|\le\int_{\mathbb{R}}\frac{|f(y)|}{|x|+|y|}dy=\int_{\mathbb{R}}\frac{|f(xt)|}{1+|t|}dt.\tag{1}$$ Applying Minkowski's integral inequality to $(1)$ and using the changes of variables $x=\frac{y}{t}$(when $t\ne 0$ is fixed) and $t=s^2$ successively, we have $$\|Tf\|_2\le\int_{\mathbb{R}}\frac{\big(\int_{\mathbb{R}}|f(xt)|^2dx\big)^{\frac{1}{2}}dt}{1+|t|}=\int_{\mathbb{R}}\frac{\big(\int_{\mathbb{R}}|f(y)|^2dy\big)^{\frac{1}{2}}dt}{|t|^\frac{1}{2}(1+|t|)}=\int_{\mathbb{R}}\frac{2\|f\|_2ds}{1+s^2}=2\pi\|f\|_2.$$ It follows that $\|T\|_2\le 2\pi$, i.e. $T$ is bounded.


To see $T$ is not compact, as in your question, let $f_n(y)=\frac{1}{\sqrt{n}}1_{[0,n]}$. Since $\|f_n\|_2=1$, it suffices to show that $\{Tf_n\}$ has no Cauchy subsequence in $L^2(\mathbb{R})$. By definition, $$Tf_n(x)=\frac{1}{\sqrt{n}}\log(1+\frac{n}{1+|x|}).$$ Given $n\in\mathbb{N}$, let $m\gg n$, say $m\ge 144n$. When $m\le |x|\le 2m$,

$$|Tf_m(x)|\ge\frac{1}{\sqrt{m}}\log(1+\frac{m}{1+2m})\ge\frac{1}{3\sqrt{m}},$$

and $$|Tf_n(x)|\le \frac{1}{\sqrt{n}}\log(1+\frac{n}{m})\le\frac{\sqrt{n}}{m}\le\frac{1}{12\sqrt{m}}.$$

It follows that when $m\le |x|\le 2m$, $|Tf_m(x)-Tf_n(x)|\ge\frac{1}{4\sqrt{m}}$, and hence

$$||Tf_m-Tf_n||_2\ge 2\big(\int_m^{2m}|Tf_m-Tf_n|^2dx\big)^{\frac{1}{2}}\ge \frac{1}{2}.\tag{2}$$ $(2)$ implies that for any subsequence $\{f_{n_k}\}$ of $\{f_n\}$, $\{Tf_{n_k}\}$ cannot be a Cauchy sequence in $L^2(\mathbb{R})$, i.e. $T$ is not compact.

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The integration is incorrect: $$\int_{\mathbb{R}} \frac{dy}{(1+|x|+|y|)^2}=\frac{2}{1+|x|}.$$ –  Julián Aguirre Dec 17 '12 at 17:45
    
@JuliánAguirre: Thank you. –  23rd Dec 17 '12 at 17:48
    
+1 Very nice... –  copper.hat Dec 19 '12 at 16:04
    
@copper.hat: Thank you! –  23rd Dec 19 '12 at 16:14
    
@richard, thank you !!! –  ougao Dec 21 '12 at 18:09
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