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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 $a \otimes b \gt x$? How can we noncommutativize the following proof, in the commutative case: Let $F^2$ be a positive function on $X\times Y$. Define $f(x)=\sup F(x,y)$ and $g(y)=\sup F(x,y)$, then $F^2 \lt fg$.

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up vote 8 down vote accepted

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 $\| e \| \| f \| = \| x \|$.

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Thank you, But could you please explain why the last inequality hold? –  Ali Taghavi Jul 11 '13 at 11:03
    
That's because $e \otimes f = \sum_{m,n}2^{-(m+n)}e_m \otimes f_n$ is an unconditionally norm convergent series of positive elements. –  Narutaka OZAWA Jul 11 '13 at 23:38
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