Can the space of linear operators on a Hilbert space be made into a Hilbert space? 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 $\dim H\ge 2$.
 A: (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 $2^\mathfrak c$.
A: The nonseparable case:
This answer shows that for an uncountable set $I$, the space $\ell^\infty(I)$ cannot be embedded into a Hilbert space i.e. there is no injective bounded linear map $\ell^\infty(I) \to H$ where $H$ is a Hilbert space.
Now assume that $H$ is nonseparable and that $|\cdot|$ is a Hilbert norm on $\mathcal L(H)$ which is equivalent to the operator norm $\|\cdot\|$. Let $E$ be an (uncountable) orthonormal basis for $H$. Then we have an isometric embedding $(\ell^\infty(E), \|\cdot\|_\infty) \to (\mathcal{L}(H), \|\cdot\|)$ as diagonal operators. Compose this with the linear homeomorphism $(\mathcal{L}(H), \|\cdot\|) \to (\mathcal{L}(H), |\cdot|)$ and we get an embedding $(\ell^\infty(E), \|\cdot\|_\infty) \to (\mathcal{L}(H), |\cdot|)$ which is a contradiction.
