TITLE: Is $1/\max(i,j)$ a bounded matrix on Hilbert spaces?
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: 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
$$\|A\|=\|CC^{*}\|=\|C\|^{2}\leq 4.$$