I am trying to understand the connection between the eigenspace of the continuous operator $$ H(x,y) = \frac{1}{x+y} $$ which is nothing but the square of the Laplace operator, and its discrete counterpart, i.e. the Hilbert matrix $$ H_{i,j} = \frac{1}{i+j}\,, i,j=1,\dots,N \,. $$ In particular, it can be proven that the generalized eigenfunctions of the continuous operator are $$ u_s(x) = \frac{1}{\sqrt{2\pi}} x^{-\frac{1}{2}+is}\,, \int_0^\infty dy \, H(x,y) u_s(y) = \vert \lambda_s \vert^2 u_s(x) $$ where $$ \lambda_s = \Gamma(\frac{1}{2}+is)\,, \quad \vert \lambda_s \vert^2 = \frac{\pi}{\cosh{\pi s}} \,. $$ Intuitively, the discretized spectrum and eigenvectors should converge to their continuous values, but I do not fully understand nor find in the literature a discussion about the two distinct effects of
- discretization, i.e. $i,j=1,2,\dots$, possibly up to $\infty$, and
- finite, but continuous extension (i.e. having $\int_0^X$ rather than $\int_0^\infty$).
I believe their effects need to be taken separately into account, since considering an infinite but not continuous amount of points leads to some contradictions, e.g. if we consider the trace of the continuous vs the discrete Hilbert operator we obtain $$ \mathrm{tr}{H} = \int_{-\infty}^\infty ds \, \vert \lambda_s \vert^2 = \pi < \sum_{n=1}^\infty \frac{1}{n+n} = \infty \,. $$
I was wondering whether there was a way to link finite and discrete (i.e. numerical) eigenvalues of the Hilbert matrix to their continuous operator counterpart. Thank you all!
EDIT
Additional question: since the spectrum of the continuous Hilbert operator $H(x,y) = \frac{1}{x+y}$, $x,y\in(0,\infty)$ is absolutely continuous in $[0,\pi]$ and the spectrum of the infinite Hilbert matrix $H_{ij}=\frac{1}{i+j}$, $i,j=1,2,\dots$ has the same property, what is the crucial difference between the two (besides the obvious discretization of indices)? If studying the infinite Hilbert matrix is not enough to recover the continuous Hilbert operator, what should complement the study of their spectrum, which is identical?
