Let x be a positive element in the spatial tensor product of two non unital C\* algebras
A and B. Is there a single element $a \otimes b \geq x$?
How can we noncommutativize the following proof, in the commutative case:
Let $F^2$ be a positive function on $X\times Y$. Define $f(x)=\sup\_{y\in Y} F(x,y)$ and $g(y)=\sup\_{x\in X} F(x,y)$,
then $F^2 \leq fg$.



positive elements in tensor product
Yes. For a self-adjoint element $y$, denote $s(y)=\sup{\rm Sp}(y)$. Then for $\gamma \geq s(y)$, one has
$$\inf \lbrace s(y - \gamma(e\otimes f)) : 0\le e\le 1,\ 0\le f\le 1\rbrace \le 0.$$
Indeed, if $e\_n$ and $f\_n$ are approximate units, then so is $g\_n:=e\_n\otimes f\_n$ and 
$y - \gamma g\_n \le y-g\_n^{1/2} y g\_n^{1/2} \to 0$.
Now let $y\_0 := x \le 1$ and find $0 \le e\_n \le 1$ and $0 \le f\_n \le 1$ recursively so that the elements
$y\_{n+1} := y\_n - 4^{-n}(e\_n \otimes f\_n)$ satisfy $s(y\_{n}) \le 4^{-n}$ for all $n$.
Let $e = \sum\_{n=0}^\infty 2^{-n}e\_n$ and likewise for $f$. Then, one has
$$x = y\_0 \le \sum\_n 4^{-n}e\_n\otimes f\_n \le e\otimes f.$$
If $A$ and $B$ have strictly positive elements, one can arrange $(e\otimes f) - x$ is strictly positive. 


I think with more efforts one can find $e$ and $f$ such that $\| e \| \| f \| = \| x \|$.


