# Inner product of linear bounded operators between Hilbert spaces

Let $X$ and $Y$ be Hilbert spaces, and let $L(X,Y)$ be the set of bounded linear operators between Hilbert spaces.

Can we equip $L(X,Y)$ with a natural inner product? I think it should look like

$\langle S, T \rangle = \sup_{x \in X} \dfrac{ \langle S x, T x \rangle_Y }{ \|x\|^2_X }$

where $S$ and $T$ and are from $L(X,Y)$. I have not found such a construct in standard text books on Hilbert spaces, therefore I would like to learn whether this is the way to do it.

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You probably meant $\Vert x\Vert^2$ in the denominator, so that the supremum exists. But even with this correction, the supremum will ruin the linearity required for an inner product. The supremum of the sum of two functions is usually not the sum of their suprema. – Andreas Blass Mar 10 '12 at 1:53
Corrected. Thank you very much. -- However, a similar formula does exist when we regard the Hilbert space $L(X,\mathbb R)$ of bounded linear functionals. Whence this is not necessarily the last word. – shuhalo Mar 10 '12 at 2:01
One approach could be to look what the natural norms of $L(X,Y)$ are (e.g. operator norm). Then use that a norm belongs to a scalar product iff it satisfies the parallelogramm identity. – Ralph Mar 10 '12 at 2:15
Depending on what you mean by "natural", and which possible $Y$ you are interested in, my answer is "no, you cannot do this" – Yemon Choi Mar 10 '12 at 4:14
Taking a sup as you wrote destroys the bilinear form. A bilinear pairing does exist between bounded operators on $X=Y$, $<T,S>:=\mathrm{tr}(T^*S)$, but it is not defined for all pairs: we need either $T$ bounded and $S$ nuclear, or both Hilbert-Schmidt, or more generally in two conjugate trace classes (for diagonal operators wrto a given Hilbert bases, this restricts the duality $\ell _ p$ spaces). – Pietro Majer Mar 10 '12 at 7:38

If $X$ and $Y$ are both non separable, then there is no continuous one-to-one linear mapping of $L(X,Y)$ into a Hilbert space. This is because in this case $L(X,Y)$ contains a subspace isomorphic to $c_0(\omega_1)$, and an argument of Olagunju (A Banach space that cannot be made into a BIP space, Math. Proc. Cambridge Philos. Soc., 63 (1967) pp 949-950) shows that there is no continuous one-to-one linear mapping of $c_0(\omega_1)$ into a Hilbert space.
Whether or not this rules out any "natural" inner product for you presumably depends on what you are looking for; in any case, what I've written above rules out various possibilities in the non separable setting (in particular, the identity mapping on $L(X,Y)$ being operator-norm-to-inner-product continuous).
As for the separable setting (i.e., $X$ and $Y$ both separable), $L(X,Y)$ is isomorphic to a subspace of $L_\infty([0,1])$ via the Hahn-Banach theorem, which admits a continuous one-to-one linear mapping into the Hilbert space $L_2([0,1])$ (i.e., the formal inclusion operator), hence there is a continuous one-to-one linear mapping of $L(X,Y)$ into the separable Hilbert space $L_2([0,1])$. But whether or not a "natural" such mapping exists, I don't know.
Any Banach space $X$ is a quotient of $L(X,X)$. – Jan Veselý Mar 13 '12 at 20:24