A question on the Diagonal Theorem of Functional Analysis

Could you please help me understand why from the stated conditions it follows that each row in the matrix $x_{q_i q_j}$ converges to zero, as the proof of the theorem claims?

This is given for the columns but it's apparently a lot more subtle for the rows. Thank you in advance.

• Note that since the sum in (b) must eventually be finite (for sufficiently large $i$) then the sum has a tail which vanishes. Without loss of generality we can pass to a subsequence so that the sum is finite for all $i$. Then, convergence of the sum implies $x_{q_i q_j}\to 0$ as $j\to \infty$. – Matt Apr 20 '17 at 14:04
• @Matt Hi Matt, thanks for your comment. If you would like to write it with more detail as an answer, I can mark it as the correct one. – JohnK Apr 20 '17 at 14:06

1 Answer

Since $\lim_{i\to \infty} \sum_{j=1}^\infty x_{q_i q_j} = 0$ then the sum is eventually finite. Thus we may pass to an additional subsequence $(q_i)$ (abusing notation) so that $\sum_{j=1}^\infty x_{q_i q_j} <\infty$ for all $i$. An infinite sum is finite only if the terms converge to zero, so $x_{q_i q_j} \to 0$ as $j\to \infty$ for each $i$. On the other hand (a) implies that $x_{q_i q_j} \to 0$ as $i\to \infty$ for each $j$. Thus each matrix row and column converges to zero as claimed.

• But what about the sum of terms such as $-1, 1, -1, 1, \ldots$. Isn't that a finite sum? – JohnK Apr 20 '17 at 14:31
• No this is not finite. A sum $\sum x_i$ is finite only if its partial sums $S_n = \sum_{i=1}^n x_i$ converge: $S_n \to S$. In the alternating sequence you give you have $S_1 = -1$, $S_2 = 0$, $S_3 = -1$, and so on. This does not converge and so the sum is not finite. – Matt Apr 20 '17 at 14:35
• Aha, thanks a lot. – JohnK Apr 20 '17 at 14:36
• You're welcome. Good luck! – Matt Apr 20 '17 at 14:39