Even Functions in 2-Lebesgue Space form Closed Subspace of 2-Lebesgue Space

Theorem
Let $\map {L^2} \R$ be the Lebesgue $2$-space.

For all $f \in \map {L^2} \R$ denote $\map {\check f} x = \map f {-x}$.

Let $Y = \set {f \in \map {L^2} \R : f = \check f}$ be the set of all even functions in $\map {L^2} \R$.

Then $Y$ is a closed subspace of $\map {L^2} \R$.

Proof
Let $R : \map {L^2} \R \to \map {L^2} \R$ be the reflection map such that:


 * $\map R f = \check f$

Then:

Furthermore:

Thus, $R$ is a linear transformation:


 * $R \in \map L {\map {L^2} \R}$.

Moreover, because the reflection $Y$-axis preserves the total area under the curve $\map f x$:


 * $\forall f \in \map {L^2} \R : \norm {f}_2 = \norm {\check f}_2$

By contnuity of linear transformations between normed vector spaces, $R$ is continuous.

Let $I : \map {L^2} \R \to \map {L^2} \R$ be the identity map:


 * $\map I f = f$

By Identity Mapping on Normed Vector Space is Bounded Linear Operator, $I$ is a linear transformation:


 * $I \in \map L {\map {L^2} \R}$

Moreover:


 * $\forall f \in \map {L^2} \R : \norm {f}_2 = \norm {f}_2$

By contnuity of linear transformations between normed vector spaces, $I$ is continuous.

By Linear Mappings between Vector Spaces form Vector Space, $I - R$ is a linear transformation:


 * $I - R \in \map L {\map {L^2} \R}$

By definition of pointwise addition of linear operators:


 * $\map {\paren {I - R} } f = f - \check f$

Then:

By contnuity of linear transformations between normed vector spaces, $I - R$ is continuous.

Suppose $f$ is even.

Then:


 * $\check f = f$.

It follows that:


 * $\map {\paren {I - R} } f = 0$.

So the set of even functions is the kernel of $I - R$:


 * $Y = \map \ker {I - R}$

We have that $I - R$ is continuous.

By Singleton in Normed Vector Space is Closed, $\set 0$ is closed in $\map {L^2} \R$.

$Y$ is the inverse image of $\set 0$ under $I - R$.

By Mapping is Continuous iff Inverse Images of Closed Sets are Closed, $Y$ is closed.