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!

  • $\begingroup$ 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 $T_n=S_n-S\nearrow 0.$ $\endgroup$ Nov 29, 2022 at 6:28
  • $\begingroup$ @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. :) $\endgroup$ Nov 29, 2022 at 8:44
  • $\begingroup$ @RyszardSzwarc No such assumption was made in my textbook, only that $T_n$ are self-adjoint and monotone. $\endgroup$ Nov 29, 2022 at 12:30
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    $\begingroup$ So it is an obvious error, as the conclusion is not true for real (i.e. self-adjoint) numbers. $\endgroup$ Nov 29, 2022 at 14:02

1 Answer 1


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

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    $\begingroup$ Wow. It was that simple wasn't it? :) $\endgroup$ Nov 29, 2022 at 0:43

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