0
$\begingroup$

I am not familiar with functional analysis. Could you tell me please, how to prove the following statement (if it is true)? $$ \lim_\limits{M \to \infty} \|T_A - T_b \| = 0, $$ here operator norm defined as $\|A\| = \sup_\limits{\| x \| = 1}\| Ax \|$, $x \in \ell^2$. $T_A$ and $T_B$ are operators defined as $$ T_A = \exp\{\alpha B\}, $$ $$ T_B = \left(E - \frac{\alpha}{M}B \right)^{-(M - p)}, $$ here $\alpha \in \mathbb{R}$, $M \in \mathbb{N}$, $p \in \mathbb{N}$ , and $B$ is the following linear operator $$ B = \sum_{k} |k\rangle \langle k+1|. $$

$\endgroup$
  • $\begingroup$ Is $E$ the identity operator? $\endgroup$ – Nate Eldredge Jun 23 at 20:10
  • $\begingroup$ @Nate Eldredge, Yes E - identity operator. $\endgroup$ – MightyPower Jun 23 at 20:27
  • $\begingroup$ @Nate Eldredge, Why? I compare this with a similar limit for functions and $\lim_\limits{n \to \infty} \left(1 - \frac{a}{n} \right)^{- (n-p)} = e^a$, not $e^{pa}$. $\endgroup$ – MightyPower Jun 23 at 20:30
  • $\begingroup$ Oh you're right, never mind. $\endgroup$ – Nate Eldredge Jun 23 at 20:32
  • 1
    $\begingroup$ This should have the same proof as the scalar calculus identity that you mention since all the operators involved here ($1,B$) commute, so the fact that we're dealing with operators rather than numbers can't really make itself felt. $\endgroup$ – Christian Remling Jun 23 at 21:46

Your Answer

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

Browse other questions tagged or ask your own question.