Prove that linear operator A has a nonzero kernel.

Question

H is a separable Hilbert space, E is an inseparable Hilbert space, A is a continuous linear operator from E to the space L (H) of continuous operators on H with an operator norm. $A:E\to L(H).$ Prove that A has a nonzero kernel.

Please help

Answer 1

Hint: find a countable set of linear functionals $\ell_k$ in $L(H)^*$ which separate points, i.e. if $\ell_{k}(T)=0$ for all $k$ then $T=0$. (It may help if you think of linear operators on $H$ as infinite-dimensional matrices.) Now consider the adjoint map $A^* : L(H)^* \to E$. Since $E$ is not separable, you can find some nonzero $x \in E$ which is orthogonal to all the $A^* \ell_k$. Show that $Ax=0$.

At a more abstract level, the obstruction is that $L(H)^*$ is weak-* separable and $E^*$ is not. If $A$ were injective then the image of $A^*$ would be dense, and in particular weak-* dense, and a continuous linear map from a separable TVS to a non-separable one can never have dense image.