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 $\|a\|=\|b\|=1$? 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.
 A: 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

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

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!
