For any two positive operator $a,b$ with norm $<1$ there is a $c$ such that $a,bleq c$ and $|c|<1$. Does this hold for $|a|=|b|=1$?

The answer is negative and here is a counter-example. Let us begin with the following:

Lemma. If $T$ and $S$ are positive operators on a Hilbert space, with $0leq Tleq Sleq 1$, and if $xi $ is any vector such that $Txi =xi $, then $Sxi =xi $, as well.

Proof. We have $$ |xi |^2 = langle xi , xi rangle = langle Txi , xi rangle leq langle Sxi , xi rangle leq |Sxi ||xi | leq |xi |^2, $$ so equality holds throughout. Cauchy-Scwartz inequality, namely the second inequality above, being an equality, we deduce that $Sxi =xi $, as desired. QED

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Let us introduce the subalgebra $$ Asubseteq C([0, 1])otimes M_2(mathbb C) = Cbig ([0, 1], M_2(mathbb C)big ) $$ formed by all continuous functions $$ f: [0, 1]to M_2(mathbb C), $$ such that $$ f(1) = pmatrix{z&0cr 0&0}, $$ for some $z$ in $mathbb C$. Consider the elements $a$ and $b$ in $A$ given, for every $t$ in $[0,1]$, by $$ a(t) = pmatrix{1&0cr 0&0}, $$ and $$ b(t) = pmatrix{t&sqrt{t-t^2}cr sqrt{t-t^2}&1-t}. $$

Notice that both $a$ and $b$ are projections, hence positive elements with norm 1.

Theorem. There exists a unique element $c$ in $C([0, 1])otimes M_2(mathbb C)$, such that $a, bleq c$, and $|c|leq 1$.

Proof. The existence is easily verified by taking $c=1$. Next suppose that $cin C([0, 1])otimes M_2(mathbb C)$ is such that $a, bleq cleq 1$. It then follows that $$ a(t), b(t)leq c(t)leq 1, $$ for every $tin [0, 1]$.

Setting $xi =(1,0)$, we have that $a(t)xi =xi $, so $c(t)xi =xi $, by the Lemma.

On the other hand, setting $eta _t=left(t,sqrt{t-t^2}right)$, we have that $b(t)eta _t=eta _t$, so again $c(t)eta _t=eta _t$.

Observing that ${xi , eta _t}$ spans $mathbb C^2$, for $tin (0,1)$, we deduce that $c(t)$ is the identity $2times 2$ matrix for every such $t$, and hence also for $t=0$ and $t=1$, by continuity. QED

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Observing that the element $c$ referred to above is not in $A$, we see that there is no element $c$ in $A$ doing the job!