# Completeness of Lowner order in separable Hilbert space

Consider a separable Hilbert space $\mathcal{H}$, we can first define $T(\mathcal{H})$ the trace class of $\mathcal{H}$, then $D(\mathcal{H})$ to denote the interesection of positive operator and trace class with trace no more than 1.

For $A,B\in D(\mathcal{H})$, we say $A\leq B$ if $A-B$ is positive, or (A,B) satisfies the Lowner order.

Now suppose we have increasing sequence $$A_0,A_1,\cdots,A_n,\cdots$$ with $(A_i,A_{i+1})$ satisfies the Lowner order, and $A_i\in D(\mathcal{H})$.

Does this sequence has least upper bound? In other words, does this class has limit?

• You need some condition to exclude a sequence line $A_n = nA$ where $A$ is some fixed nonzero positive trace-class operator. – Nate Eldredge Jul 10 '18 at 2:27
• A_i are in trace class. – gondolf Jul 10 '18 at 5:50
• @gondolf That doesn't affect Nate's counterexample. If $A$ is a trace-class operator, so is $nA$ for any $n \in \mathbb{N}$, and its trace is $n$ times the trace of $A$. You need to put an upper bound on the sequence. – Robert Furber Jul 10 '18 at 6:20
• @gondolf: restricting to "trace no more than 1" is unnecessary. You just need boundedness in operator norm. – Nik Weaver Jul 10 '18 at 22:11

This isn't research level, but if the sequence is increasing and bounded then yes, it has a least upper bound. For each $v \in H$ the sequence $\langle A_iv,v\rangle$ is bounded and increasing and therefore converges to its least upper bound in $\mathbb{R}$. By polarization it follows that for all $v,w \in H$ the sequence $\langle A_iv,w\rangle$ converges. Thus the sequence $(A_i)$ converges in the weak operator topology. (In fact, it also converges strong operator.)