TITLE: Prove that linear operator A has a nonzero kernel.
QUESTION [8 upvotes]: 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

REPLY [5 votes]: 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.