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Let $H$ be a Hilbert space and $B(H)$ be the set of bounded linear mapping on $H$. If $\lambda \in B(H)^*$, is it true that $\exists x,y \in H$ s.t. $\lambda(T)=(Tx,y), \forall T \in B(H)$?

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 not at all. The set of functionals of that form is not even a linear space. – Pietro Majer Oct 3 2011 at 15:16
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 And if you take the norm-closure of the set of all linear combinations of functionals of that form, you just get the class of the trace operators, the pre-dual of $B(H)$, still far from the whole $B(H)^*$. – Pietro Majer Oct 3 2011 at 15:22
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 Also, this is already false for $\dim H>1$ finite. – a-fortiori Oct 3 2011 at 15:23
This is a well-formed question, but it seems to be the wrong level for MathOverflow. You might want to ask at one of the sites listed in the FAQ. As a-fortiori mentioned, you can find counterexamples by counting dimensions. – S. Carnahan♦ Oct 4 2011 at 6:07

closed as too localized by Qiaochu Yuan, Andrew Stacey, Bill Johnson, Yemon Choi, S. Carnahan♦ Oct 4 2011 at 6:07

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