I know that the preimage of continuous function on closed set is also closed.

But, the preimage of continuous "functional" on closed set is also closed?

That is,

let $h:L^2 \rightarrow R$ be a continuous (bounded) linear functional, and $A \subset R$ is a closed set. Then $f^{-1}(A) =\{f\in L^2 : h(f) \in A\}$ is a closed subset in $L^2$ with $d(f,g) = \|f-g||_2$ ?

I think that it is true, but not sure....


1 Answer 1



This is a topology question: all you need here is that both $L^2$ and $\mathbb{R}$ are topological spaces. The preimage of a closed set under a continuous map is closed.


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