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Apr 9, 2015 at 20:17 history edited GH from MO CC BY-SA 3.0
fixed spelling
Apr 9, 2015 at 19:35 answer added Twi timeline score: 6
Jul 20, 2014 at 22:26 comment added user52733 A good reference for the weighted Schur Test, as per @fedja's response, is this paper of Dym and Katsnelson regarding the work of Issai Schur. The "weighted" test and it's application to the Hardy series is on page 17-18.
May 21, 2014 at 23:32 comment added Suvrit The operator norm of the first matrix is bounded above by $4$; the second one has an unbounded trace, and think its operator norm also goes to $\infty$ (the operator norm seems to be $O(n)$ for an $n \times n$ matrix)
May 21, 2014 at 21:14 comment added fedja Same as with weight $1$: $Aw\le Cw$ entry-wise. You can use two weights as well: $Aw\le C'v, A^*v\le C''w$ but for self-adjoint operators there is no difference.
May 21, 2014 at 20:40 comment added Bazin @fedja I should have said that I did not understand your reference to a Schur test with weight, while the raised question was without weight.
May 21, 2014 at 19:43 comment added fedja @Bazin I wrote "the Schur test with the weight $w(j)=j^{-1/2}$", not "the Schur test with the weight $w(j)=1$ (which, indeed, does not work here as you noted 100% correctly)
May 21, 2014 at 19:37 comment added Bazin @fedja The Schur test does not work: you would have to check $$\sup_{i\ge 1}\sum_{j\ge 1}\frac{1}{i+j}$$ which are all infinite. I hope that the explanations below could clarify the situation and qualify the question for MO.
May 21, 2014 at 19:36 answer added Bazin timeline score: 2
May 21, 2014 at 17:27 answer added Igor Rivin timeline score: 2
May 21, 2014 at 13:22 comment added fedja Run the Schur test with the weight $w(j)=j^{-1/2}$ for the first matrix. Note that there are huge square blocks consisting of positive numbers exceeding $\frac 12$ in the second. Keep in mind that this question is borderline for MO and more suitable for MSE.
May 21, 2014 at 12:24 history asked django CC BY-SA 3.0