The problem:

I've hit what might be a dead-end. If it is true, I would like to show that for $c \in (0,1)$ and $1 \leq k \leq J$, the sum $$ \sum_{j=k}^J jc^{1-j-k} = \sum_{n=1}^{J+1-k} (k-n-1) c^{n} $$ has an upper bound that is independent of $k$. I tried the lazy approach of letting $J \to \infty$, but that doesn't seem to work. Any input here would be appreciated.

The context: This is for a PDE course where we are dabbling in functional analysis. I am trying to show that the spectrum of the right-shift operator on $\ell^2$ is $\{\lambda:|\lambda| \leq 1\}$, and this summation comes from the $|\lambda|>1$ case. We have not established the notion of an operator norm so I would prefer to avoid using it, though I am well aware it makes quick work of this part of the problem.

  • $\begingroup$ Is $J > k{}{}$? Is $k > 0$? $\endgroup$ – Antonio Vargas Apr 16 '14 at 1:09
  • $\begingroup$ @AntonioVargas yes, $1 \leq k \leq J$ $\endgroup$ – Ben Grossmann Apr 16 '14 at 1:15

If $0 < c < 1$ then $1/c > 1$, so if $1 \leq k \leq J$ we have

$$ \begin{align} \sum_{j=k}^J jc^{k - j - 1} &= \sum_{j=k}^J j \left(\frac{1}{c}\right)^{1+j-k} \\ &\leq \sum_{j=k}^J j \left(\frac{1}{c}\right)^{1+j-1} \\ &\leq \sum_{j=1}^J j \left(\frac{1}{c}\right)^{j}. \end{align} $$

The first inequality holds holds because the map $x \mapsto \left(\frac{1}{c}\right)^x$ is increasing.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your answer! Unfortunately, this doesn't happen to suit my needs since I need my upper bound to be finite as $J \to \infty$, sorry for not having that in the question statement. $\endgroup$ – Ben Grossmann Apr 16 '14 at 4:12
  • $\begingroup$ @Omnomnomnom, your quantity doesn't have a finite upper bound that holds for all $J,k$ with $1 \leq k \leq J$. The inequalities in my argument are tantamount to taking $k=1$. $\endgroup$ – Antonio Vargas Apr 16 '14 at 4:14
  • $\begingroup$ My apologies! The sign of the exponent was wrong. This certainly should change things $\endgroup$ – Ben Grossmann Apr 16 '14 at 4:20
  • $\begingroup$ Now take $k = J$ in the new version; it's still unbounded as $J \to \infty$. $\endgroup$ – Antonio Vargas Apr 16 '14 at 4:26
  • $\begingroup$ Well, that takes care of that then. Thank you. $\endgroup$ – Ben Grossmann Apr 16 '14 at 4:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.