# Monotone and bounded sequence of self adjoint operators converges pointwise

It is easy to prove that a monotone and bounded sequence of self-adjoint operators $$\{T_n\}_{n\in\mathbb{N}}$$ in a Hilbert space $$H$$ converges pointwise to a self-adjoint operator $$T$$ and that $$\lVert T \rVert \leq \sup\{\lVert T_n \rVert:\; n\in \mathbb{N}\}$$.

The idea of the proof is to show that the sequence $$\{T_nx\}_{n\in\mathbb{N}}$$ is Cauchy and, therefore, convergent. Then we define $$Tx=\lim_n T_nx$$ (which is linear and self-adjoint). The latter also shows that $$\lVert T \rVert \leq \sup\{\lVert T_n \rVert:\; n\in \mathbb{N}\}$$. My reference book however, states that $$\lVert T \rVert = \sup\{\lVert T_n \rVert:\; n\in \mathbb{N}\}$$. While it's true that $$\langle Tx, x\rangle = \sup_{n\in \mathbb{N}}\langle T_nx, x\rangle$$ which in extent proves that $$T\geq T_n$$ for all $$n\in \mathbb{N}$$, I don't see how this would imply $$\lVert T \rVert \geq \lVert T_n \rVert$$ for all $$n \in \mathbb{N}$$ which is what we need for the claim to hold.

Any help is appreciated!

• I guess in your book the operators $T_n$ are positive and the sequence $T_n$ is nondecreasing. Otherwise every nonconstant nondecreasing sequence provides a counterexample: for example if $S_n\nearrow S,$ then $T_n=S_n-S\nearrow 0.$ Nov 29, 2022 at 6:28
• @LostStatistician18 Your deleted solution is fully correct provided that $T_n\ge 0.$ Moreover for a positive operator $\|T\|=\sup_{\|x\|=1}\langle Tx,x\rangle.$ I am pretty sure the textbook made the assumption $T_n\ge 0$, but the asker missed that. :) Nov 29, 2022 at 8:44
• @RyszardSzwarc No such assumption was made in my textbook, only that $T_n$ are self-adjoint and monotone. Nov 29, 2022 at 12:30
• So it is an obvious error, as the conclusion is not true for real (i.e. self-adjoint) numbers. Nov 29, 2022 at 14:02

Counterexample: $$T_n=-\frac1n{\rm Id}_H.$$