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 $$ B = \sum_{k} k\rangle \langle k+1. $$
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$\begingroup$ 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 $\lim_\limits{n \to \infty} \left(1  \frac{a}{n} \right)^{ (np)} = e^a$, 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