Is the norm of the Hilbert matrix equal to $\pi$?

Question

Let $A$ be a Hilbert matrix,

$$a_{ij}=\frac{1}{1+i+j}$$

We have the result $\| A \| \leq \pi$. I am using the subordinate norm of the Euclidean norm, i.e.,

$$\| A \| = \sup\{\langle Ax,y\rangle:\quad x,y\in\mathbb{R}^n,\quad\Vert x\Vert_2\leq 1,\quad\Vert y\Vert_2\leq 1\}$$

This inequality can be proved using Hilbert's Inequality. Look here.

Question: Do we have an equality? I found nothing on the Internet about such equality.

Answer 1

Certainly not (if you're talking about a finite matrix, rather than an operator on $\ell^2$). $\pi$ is transcendental, and the norm of a finite matrix with rational entries is an algebraic number.

Answer 2

According to this paper, the answer is yes. I'm not familiar with the theorem of Nehari mentioned there; if you assume that, though, you can check that the value of $g(n)$ does produce the Hilbert matrix as indicated and has $L^\infty$ norm $\pi$. (Here $\hat g$ is the Fourier transform of $g$). (The paper refers to the infinite-dimensional Hilbert matrix as an operator $A : \ell^2 \to \ell^2$, but it's not hard to show that a positive result in that case shows that $\pi$ is also the supremum of the corresponding Hilbert matrices $A_n : {\mathbb{R}}^n \to {\mathbb{R}}^n$, if you're dealing with the finite-dimensional case.)

Of course, if you're asking whether any particular $A: {\mathbb{R}}^n \to {\mathbb{R}}^n$ has norm exactly $\pi$, the answer is no; see the comment above.