Let $H$ be a Hilbert space, and $\mathcal L(H)$ be the space of linear operators on $H$. Can we find an inner product on $\mathcal L(H)$ that induces an equivalent norm on $\mathcal L(H)$, i.e., a norm that is equivalent to the operator norm?
For finite-dimensional spaces, the Frobenius inner product answers the question in the affirmative. So any counter-examples necessarily is infinite-dimensional.
I am aware of the negative result, that there is no inner product that induces the operator norm if $\dim H\ge 2$.
Best Answer
(generalisation for unseparable case is thanks to MaoWao in comment)
Let $H'$ be a closed separable subspace of $H$.
Any subspace of space with topology induced by inner product also has topology induced by inner product. Any space with topology induced by inner product is isomorphic to its dual. If $H$ has topology induced by inner product, then $\mathcal L(H')$ is isomorphic to a closed subspace of $\mathcal L(H)$ (take orthogonal complement to $H'$ and define operator as $0$ on it to continue it from $H'$ to entire $H$).
Thus, if $H$ has topology induced by inner product, then $\mathcal L(H')$ is isomorphic to its dual. Let's prove that $\mathcal L(H')$ has cardinality less than $\mathcal L^*(H')$ and thus not isomorphic to it.
Let $x_i$ be basis of $H'$, then $A$ from $\mathcal L(H')$ is completely defined by countable number of real numbers $\langle Ax_i, x_j\rangle$, so $\mathcal L(H')$ has cardinality continuum.
But set of all diagonal operators on $H'$ is isomorphic to $l_\infty$, thus $\mathcal L^*(H')$ contains at least as many elements as $l_\infty^*$, which has cardinality of at least $2^\mathfrak c$.