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I would like to know if the infinite matrix $[\frac{1}{\max(i,j)}]_{i,j\geq 1}$ represents a bounded operator on $\ell^2(\mathbb{N}^\star)$. It would be sufficient to know if the Lehmer matrix $[\frac{\min(i,j)}{\max(i,j)}]_{i,j\geq 1}$ is bounded on $\ell^2(\mathbb{N}^\star)$.

Thanks.

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    $\begingroup$ Run the Schur test with the weight $w(j)=j^{-1/2}$ for the first matrix. Note that there are huge square blocks consisting of positive numbers exceeding $\frac 12$ in the second. Keep in mind that this question is borderline for MO and more suitable for MSE. $\endgroup$
    – fedja
    Commented May 21, 2014 at 13:22
  • $\begingroup$ @fedja The Schur test does not work: you would have to check $$\sup_{i\ge 1}\sum_{j\ge 1}\frac{1}{i+j}$$ which are all infinite. I hope that the explanations below could clarify the situation and qualify the question for MO. $\endgroup$
    – Bazin
    Commented May 21, 2014 at 19:37
  • $\begingroup$ @Bazin I wrote "the Schur test with the weight $w(j)=j^{-1/2}$", not "the Schur test with the weight $w(j)=1$ (which, indeed, does not work here as you noted 100% correctly) $\endgroup$
    – fedja
    Commented May 21, 2014 at 19:43
  • $\begingroup$ @fedja I should have said that I did not understand your reference to a Schur test with weight, while the raised question was without weight. $\endgroup$
    – Bazin
    Commented May 21, 2014 at 20:40
  • $\begingroup$ Same as with weight $1$: $Aw\le Cw$ entry-wise. You can use two weights as well: $Aw\le C'v, A^*v\le C''w$ but for self-adjoint operators there is no difference. $\endgroup$
    – fedja
    Commented May 21, 2014 at 21:14

3 Answers 3

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The operator you introduced, say $A$, is bounded indeed. There is a simple proof for this:

First, note $A$ can be written as $A=CC^{*}$ where $$C_{i,j}=\begin{cases} \frac{1}{i}, & i\geq j,\\ 0, & i<j. \end{cases}$$ Next, observe that $$(I-C)(I-C)^{*}=\mbox{diag}\left(0,\frac{1}{2},\frac{2}{3},\frac{3}{4},\dots\right)$$ and hence $$\|I-C\|^{2}=\|(I-C)(I-C)^{*}\|=\sup_{n\in\mathbb{N}}\frac{n}{n+1}=1.$$ Consequently, we get $$\|C\|\leq\|I-C\|+\|I\|=2$$ and so $$\|A\|=\|CC^{*}\|=\|C\|^{2}\leq 4.$$

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    $\begingroup$ This is very nice! $\endgroup$
    – GH from MO
    Commented Apr 9, 2015 at 20:12
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I could be very confused (likely), but notice that your matrix the sum of $N$ and $N^t,$ where $N$ is the upper triangular matrix where $N_{ij}= 1/i,$ when $i<j$ and $0$ otherwise.

It seems that by Denis Serre's answer to this question, the answer to yours is YES.

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  • $\begingroup$ Are you saying that you can prove the $L^2$ boundedness of the discrete Hilbert transform (matrix $(i-j)^-1)$) or of the Hardy operator (matrix $(i+j)^-1)$) that way ? $\endgroup$
    – Bazin
    Commented May 23, 2014 at 10:52
  • $\begingroup$ @Bazin I did not say that, but what's wrong with the argument (other than the fact that I should divide the diagonal elements of $N$ by $2$)? $\endgroup$
    – Igor Rivin
    Commented May 23, 2014 at 11:56
  • $\begingroup$ I do not see the connection with D. Serre's result, which is comparing the numerical radius to the norm. The (discrete) Hardy operator is hard stuff to handle, in particular not trace class. To get its $\ell^2$ boundedness, I used the Hilbert transform, and to get the boundedness of the latter, Fourier transform, which is bounded as a sign function. $\endgroup$
    – Bazin
    Commented May 23, 2014 at 16:42
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    $\begingroup$ @Bazin I have no idea about the Hardy operator and the Hilbert transform (that was not the subject of this question), but Serre is pointing out that the norm of $M$ is bounded (above and below) by a (linear) function of the eigenvalues of $N.$ In this case, $N$ is upper (or lower, if you prefer) triangular, with distinct diagonal entries, so we know EXACTLY what its eigenvalues are. What exactly am I missing here (without reference to Hilbert or Hardy)? $\endgroup$
    – Igor Rivin
    Commented May 23, 2014 at 18:10
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On $\ell^p(\mathbb N^*)$, $1<p<+\infty$, the matrix $$ A=\left(\frac{1}{i+j}\right)_{1\le i,j},\tag 1 $$ is a bounded operator and since the entries are positive and "equivalent" to your matrix, the latter enjoys as well the same properties. Now, proving the above fact is not so easy.

A good way to start is to look at the "continuous" version, i.e. the operator $\mathcal H$ on $L^p(\mathbb R)$ with kernel $$ \kappa(x,y)=\frac{Y(x)Y(y)}{π(x+y)},\quad\text{where $Y$ is the Heaviside function.} $$ You have for $u\in L^2(\mathbb R)$, $$(\mathcal H u)(x)=Y(x)\int_{\mathbb R} \kappa(x,y) Y(y)u(y) dy =Y(x)\int_{\mathbb R} \frac{1}{π(x-y)} Y(-y)u(-y) dy,$$ so that $ \mathcal H= Y\mathcal H_0 CY,\quad (Cu)(x)=u(-x). $ Since $\mathcal H_0$ is the Hilbert transform and $C, Y$ have norm 1, you get the boundedness result. The nice and not-so-trivial thing is that you have also $$ \Vert{\mathcal H}\Vert_{L^2\rightarrow L^2}=1. $$

To handle the discrete case, use the discrete Hilbert transform and the factorization above. It seems that you can prove as well that the $\mathcal B(\ell^2(\mathbb N^*))$ norm of $A$ is $\pi$.

There are more general approaches linked to Calder\'on-Zygmund theory of singular integrals.

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