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Let $(x_k)_{k\geq 1}\in\ell_2$. Consider $\left(\sum\limits_{k=1}^\infty \dfrac{x_k}{j+k}\right)_{j\geq 1}$.

Now my question is that whether $\left(\sum\limits_{k=1}^{\infty}\frac{x_k}{j+k}\right)_{j\geq 1}$ belongs to $\ell_2$ or not.

The following is an idea, but I am not sure whether it's right.

Let $\varphi(t)=i(\pi-t)$ be a $2\pi$-period function. Then it's easy to see that $\widehat\varphi(n)=\frac{1}{n}$ for every nonzero $n\in\mathbb Z$ and $\widehat \varphi(0)=0$. Then we obtain $$\sum\limits_{1\leq j,k\leq N}\frac{a_jb_k}{j+k}=\frac{1}{2\pi} \int_0^{2\pi}\left(\sum\limits_{j=1}^N a_j e^{-ijt}\right)\left(\sum\limits_{k=1}^Nb_ke^{-ikt}\right)\varphi(t)dt$$ where $(a_1,\cdots,a_N),(b_1,\cdots,b_N)\in\mathbb C^N$.Then by Cauchy-Schwarz inequality, we have

$$\left|\sum\limits_{1\leq j,k\leq N}\frac{a_jb_k}{j+k}\right|\leq\|\varphi\|_{\infty}\left|\left(\sum\limits_{j=1}^N|a_j|^2\right)^{1/2}\left(\sum\limits_{k=1}^N|b_k|^2\right)^{1/2}\right|.$$ Now for any $x,y\in \ell_2$, we have $$\left|\langle u(x),y\rangle\right|\leq \|\varphi\|_{\infty}\|x\|_2\|y\|_2$$ where $u(x)=(\sum\limits_{k=1}^{\infty}\frac{x_k}{j+k})_{j\geq 1}$. Note that $\langle u(x),y\rangle$ exists, since $\lim\limits_{N\to\infty}\sum\limits_{1\leq j,k\leq N}\frac{|x_j||y_j|}{j+k}$ exists. Then $\langle u(x),\cdot\rangle$ is a linear functional on $\ell_2$ for fixed $x\in\ell_2$. By Riesz representation theorem, there exists $z\in \ell_2$ such that $\langle u(x),y\rangle=\langle z,y\rangle$ for all $y\in\ell_2$. Hence $z=u(x)$, i.e. $u(x)=z\in\ell_2$.

Maybe I have made some mistakes in the proof. And anyone know some other proofs?

Thank you for you help.

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3  
It looks good to me. You also seem to get that the $\ell^2$ norm is $\le\pi\|x_k\|_2$. –  robjohn May 11 '12 at 18:37
    
What you have done is called Toeplitz method. –  Norbert May 11 '12 at 19:26
    
@robjohn Yes, if we can show that $u(x)\in\ell_2$, then $u$ is a continuous linear map from $\ell_2$ to $\ell_2$. And the norm of $u$ is exactly $\pi$. –  molan May 12 '12 at 1:47
    
@Norbert In fact, I don't know that. Can you give me some reference? –  molan May 12 '12 at 1:47
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2 Answers 2

up vote 5 down vote accepted

There are a couple of similar approaches to your problem.

1) The first one you can find in The Cauchy-Schwarz Master Class on page 155. This is a direct proof using basic ideas of calculus. It is already mentioned in Byron Schmuland's answer.

2) The second approach is a bit more advanced. This is exactly what you have done in your answer. This approach is called Toeplitz method, and you can find it in the same book on page 165.

3) The third approach uses Schur test. You can find it in Halmos's problem book on Hilbert spaces. See problems 45 and 46.

4) Of course, your problem can be expressed as the question of boundedness of some particular operator on $\ell^2$, which is unitarily equivalent to an infinite dimensional analogue of Hilbert matrix. You can find several solutions using this operator approach, and much more, in this article. See problems V, VII and X.

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Nice,thank you so much. Although I don't know how to use Hilbert's Inequality to do it, Shur test and Hilbert matrix are really perfect. –  molan May 12 '12 at 11:29
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Reference: The inequality $$\sum_{m=1}^\infty \sum_{n=1}^\infty {a_m b_n\over m+n}\leq C \left(\sum_{m=1}^\infty a_m^2\right)^{1/2} \left(\sum_{n=1}^\infty b_n^2\right)^{1/2} $$ is called Hilbert's inequality and is (10.1) on page 155 of The Cauchy-Schwarz Master Class by J. Michael Steele. He also shows that the constant $C$ can be taken to be $\pi$, and that $C=\pi$ is optimal.

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That book should be mandatory reading for introductory analysis classes. –  Alex R. May 11 '12 at 17:05
    
I checked some books about Hilbert's inequality. However, I still don't know how to give another proof. Anyway, thank you very much. And the book Cauchy-Schwarz Master Class is nice. –  molan May 12 '12 at 2:58
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