Brezis Functional Analysis Exercise 2.10

banach-spacesfunctional-analysis

Hi I'm trying the following exercise from the book.

Let $E$ and $F$ be two Banach spaces and let $T\in\mathcal{L}(E,F)$ be surjective

  1. Let $M$ be any subset of $E$. Prove that $T(M)$ is closed in F iff $M+N(T)$ is closed in $E$.

I'm stuck with the converse.

The hint from the book says that since $T$ is surjective then
$$
T((M+N(T))^c)=(T(M))^c.
$$

But I don't understand why, since the property says that if a function $f$ is surjective then
$$
(f(A))^c \subset f(A^c).
$$
Could anyone explain me? Or give another hint?
Thanks

Best Answer

  • For any function $f \colon X \to Y$ and any subset $B$ of $Y$, $$ f^{-1}(B^c) = f^{-1}(B)^c. $$
  • For any surjective function $f \colon X \to Y$ and any subset $B$ of $Y$, $$ f(f^{-1}(B)) = B. $$
  • If $V$ and $W$ are vector spaces over the same field, then for any linear map $L \colon V \to W$ and any subspace $S$ of $V$, $$ L^{-1}(L(S)) = S+N(L). $$

These are easy exercises that has nothing to do with functional analysis.

So, if $M+N(T) = T^{-1}(T(M))$ is closed in $E$, then $$ (M+N(T))^c = T^{-1}(T(M)^c) $$ is open in $E$, and so $$ T((M+N(T))^c) = T(T^{-1}(T(M)^c)) = T(M)^c $$ is open in $F$ since $T$ is surjective.

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