Weird subspace of a banach space, is this also banach?

Question

Trying to solve an exercise I was wondering if it would be possible to use the implicit function theorem on a tricky space in order to immediately obtain some extra properties on the solutions. However, it is not clear for me how to prove that this defines a Banach space. Let me be more specific, I would like to define a functional on the following subspace of $L^2\times L^2$: $$ \mathfrak{X}:=\{(f,g)\in L^2(\mathbb{R})\times L^2(\mathbb{R}):\ f(x)=-g(-x)\, \hbox{ a.e. on }\,\mathbb{R}\}. $$ Thus, in order to apply the implicit function theorem at least I need to prove that this defines a Banach space. Would anyone has any hint? or any argument to disprove it?

Answer 1

Hint: since $L^2 \times L^2$ is a Banach space, it is enough to show that $\mathfrak{X}$ is closed. That is, if $f_n \to f, g_n \to g$ in $L^2$ and we have $f_n(x) = -g_n(-x)$ a.e., then $f(x)=-g(-x)$ a.e. Now use the fact that a sequence converging in $L^2$ has a subsequence converging a.e. to the same limit.

Answer 2

A finite product of Banach spaces is Banach, and closed subsets of Banach spaces are Banach. So it's enough to show that $\mathfrak{X}$ is closed. Since $\mathfrak{X}$ is defined by a simple equation, it's (essentially) automatically closed. Specifically: consider the map $T: L^2(\mathbb{R}) \times L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R})$ given by $(f,g) \rightarrow h$ where $h(x) = f(x) + g(-x)$. Then $\mathfrak{X} = T^{-1}({0})$ is the inverse image of a closed set under a continuous map, and so is closed.