Timeline for Is $1/\max(i,j)$ a bounded matrix on Hilbert spaces?
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| May 23, 2014 at 18:10 | comment | added | Igor Rivin | @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)? | |
| May 23, 2014 at 16:42 | comment | added | Bazin | 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. | |
| May 23, 2014 at 11:56 | comment | added | Igor Rivin | @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 $2$)? | |
| May 23, 2014 at 10:52 | comment | added | Bazin | 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 ? | |
| May 21, 2014 at 17:27 | history | answered | Igor Rivin | CC BY-SA 3.0 |