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

Question

I mean, for two positive operator $a,b$ with norm $1$, is there still always a positive $c$ such that $a,b\leq c$ and $\|c\|=1$? Suppose it is in a non-unital C*algebra.

And $a\leq b$ means $b-a$ is positive.

Answer 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 $0\leq T\leq S\leq 1$, and if $\xi $ is any vector such that $T\xi =\xi $, then $S\xi =\xi $, as well.

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

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

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

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 $t\in (0,1)$, we deduce that $c(t)$ is the identity $2\times 2$ 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!