# 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 $$\dim H\ge 2$$.

• Any Hilbert space is reflexive. For a reflexive Banach space every closed subspace is reflexive. But $L(H)$ for infinite dimensional $H$ is not reflexive as it contains closed nonreflexive subspaces, for example the subspace of compact operators, which contains in turn a closed subspace isomorphic to $c_0(\mathbb{N})$, the sequences convergent to $0.$ Nov 2, 2022 at 19:52
• There is a natural inner product on operators on a Hilbert space, but it is only well-defined on the Hilbert-Schmidt operators. (And, it is not equivalent to the operator norm.) This is the natural generalization of the Frobenius inner product to the infinite-dimensional case. Nov 3, 2022 at 1:40

## 2 Answers

(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 $$2^\mathfrak c$$.

• Note that being isomorphic to a Hilbert space passes to closed subspaces, and $B(H)$ for non-separable $H$ contains $B(\ell^2)$ as a closed subspace. So you're answer actually settles the question in full generality. Nov 2, 2022 at 13:14
• @MaoWao could you please explain your argument in more detail ? Nov 2, 2022 at 17:46
• The argument is that any closed subset of Hilbert space is isomorphic to its dual. So if $H$ is HIlbert space, then take $H'$ - a closed separable subset of $H$, now $\mathcal L(H')$ is a closed subset of $\mathcal L(H)$ and not isomorphic to its dual by argument in my answer. Nov 2, 2022 at 20:19

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.