Let $\{\alpha_{n}\}_{n\in\mathbb{N}}$ be positive sequence such that $$\sum_{n=1}^{\infty}\alpha_n<\infty.$$

Question: Is there any chance to evaluate the operator norm of the matrix operator $$C=\begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 & \dots\\ 0 & \alpha_2 & \alpha_3 & \dots\\ 0 & 0 & \alpha_3 & \dots\\ \vdots & \vdots & \vdots & \ddots\\ \end{pmatrix},$$ acting on $\ell^{2}(\mathbb{N})$, in terms of the sequence $\{\alpha_{n}\}_{n\in\mathbb{N}}$?

A remark: Note $C$ can be decomposed as $C=VD$ where $V_{i,j}=1$, if $i\leq j$, $V_{i,j}=0$, if $i>j$, and $D=\mbox{diag}(\alpha_1,\alpha_2,\alpha_3,\dots)$.

An additional comment concerning boundedness of $C$: As noticed by Miel Sharf below, the assumptions do not guarantee the boundedness of $C$. Operator $C$ is bounded, for instance, under the following additional assumption: $$\sup_{n}\frac{1}{\alpha_{n}}\sum_{k\geq n}\alpha_{k}^{2}<\infty.$$ This is, in fact, the case I am interested in.

  • $\begingroup$ Why is $C$ positive? $\endgroup$ – Suvrit Apr 7 '15 at 22:11
  • $\begingroup$ @Suvrit: it's not positive because it's not self-adjoint. $\endgroup$ – Nik Weaver Apr 8 '15 at 1:09
  • $\begingroup$ @Nik Weaver You are right. The remark is wrong. I edit it. $\endgroup$ – Twi Apr 8 '15 at 10:06
  • $\begingroup$ @NikWeaver: my question was a rhetoric question :-) sorry, I should have just asked it more directly (because it is the lack of that positivity that breaks this question). $\endgroup$ – Suvrit Apr 8 '15 at 13:28
  • $\begingroup$ @Suvrit: it is hard to read "tone" from printed text ... it always surprises me when people can't tell I'm being sarcastic, for example $\endgroup$ – Nik Weaver Apr 8 '15 at 15:11

I think that this operator might not be bounded.

Let $e_i$ denote the i-th vector in the standard orthonormal basis form $\ell^2(\mathbb{N})$. Then

$$ C\cdot e_i=\alpha_i\sum_{j=1}^i e_j $$

thus $||C||\ge\sup_{n}{(\sqrt{n}\alpha_n)}$. However, the rightmost amount might be infinite even if you assume that $\alpha_n$ are positive and that the sum $\sum \alpha_n$ converges.

Indeed, take the sequence defined by $\alpha_n = \frac{1}{log_2(n)^2}$ if $n$ is a power of 2 and $\alpha = 0$ otherwise. The sum $\sum \alpha_n$ becomes $\sum_{k=1}^\infty \frac{1}{k^2}$, which does converge. However, it's clear that

$$ \sup_{n}{(\sqrt{n}\alpha_n)} = \sup_{k} (2^{k/2}*\frac{1}{k^2})=\infty $$

if you want $\alpha_n$ to be strictly positive, then replace $\alpha_n=0$ by $\alpha_n=\frac{1}{2^n}$ (or any rapidly decaying sequence).

  • $\begingroup$ Thanks. Any idea for the norm $\|C\|$ in the case when $C$ is bounded? $\endgroup$ – Twi Apr 8 '15 at 10:09
  • $\begingroup$ Not really, but here is what I would try to do - look at operators of the same shape on finite dimensional spaces. You know that the operator norm there is just the highest singular value, so (hopefully) you can compute it. Now, the operator norm of the original operator is a supermum on all sequences in $\ell^2$ of some expression, which is the same as the supermum of the same expression on eventually zero sequences. For those, you already have the expression for the operator norm by the above calculation. $\endgroup$ – Miel Sharf Apr 8 '15 at 11:28

Let $\sum_{p\geq 1}p{\alpha_p}^2=\tau$ and assume that $\tau<\infty$. From $trace(C^TC)=\tau$, we deduce that $||C||=\sqrt{\max(spectrum(C^TC))}\leq\sqrt{\tau}$.

Moreover, if the $(\alpha_i)_i$ are $0$ except $\alpha_k$, then $||C||=\sqrt{\tau}$.

  • $\begingroup$ Thanks. Nevertheless this is only an upper bound for the norm in the case when $C^{T}C$ is a trace class operator. I assume $\alpha_k>0$ for all $k$. $\endgroup$ – Twi Apr 13 '15 at 15:13

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