Is $1/\max(i,j)$ a bounded matrix on Hilbert spaces? 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.
 A: 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
$$\|A\|=\|CC^{*}\|=\|C\|^{2}\leq 4.$$
A: 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.
A: 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.
