Exercise 12.1: Closedness of Nullspace
Since $L$ is bounded, it is coninuous at zero, hence continuous. Therefore, $\operatorname{null}(L)$ is the preimage of a closed set under a continuous function, which is closed.
Since $L$ is bounded, it is coninuous at zero, hence continuous. Therefore, $\operatorname{null}(L)$ is the preimage of a closed set under a continuous function, which is closed.