# The limit of the operator norm in a Hilbert space

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 $$B = \sum_{k} |k\rangle \langle k+1|.$$

• Is $E$ the identity operator? – Nate Eldredge Jun 23 at 20:10
• @Nate Eldredge, Yes E - identity operator. – MightyPower Jun 23 at 20:27
• @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 $e^{pa}$. – MightyPower Jun 23 at 20:30
• Oh you're right, never mind. – Nate Eldredge Jun 23 at 20:32
• 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. – Christian Remling Jun 23 at 21:46