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

## Theorem

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

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

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

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

## Proof

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

$\map R f = \check f$

Then:

 $\ds \forall f, g \in \map {\LL^2} \R : \forall x \in \R : \ \$ $\ds \map R {\map f x + \map g x}$ $=$ $\ds \map R {\map {\paren {f + g} } x}$ Definition of Pointwise Addition of Real-Valued Functions $\ds$ $=$ $\ds \map {\paren {f + g} } {- x}$ $\ds$ $=$ $\ds \map f {-x} + \map g {-x}$ Definition of Pointwise Addition of Real-Valued Functions $\ds$ $=$ $\ds \map R {\map f x} + \map R {\map g x}$

Furthermore:

 $\ds \forall f \in \map {\LL^2} \R : \forall x, \lambda \in \R : \ \$ $\ds \map R {\lambda \map f x}$ $=$ $\ds \map R {\map {\paren {\lambda \cdot f} } x}$ Definition of Pointwise Scalar Multiplication of Real-Valued Functions $\ds$ $=$ $\ds \map {\paren {\lambda \cdot f} } {-x}$ $\ds$ $=$ $\ds \lambda \map f {-x}$ Definition of Pointwise Scalar Multiplication of Real-Valued Functions $\ds$ $=$ $\ds \lambda \map R {\map f x}$

Thus, $R$ is a linear transformation:

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

Moreover, because the reflection with respect to $y$-axis preserves the total area under the curve $\map f x$:

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

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

$\map I f = f$
$I \in \map L {\map {\LL^2} \R}$

Moreover:

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

By definition of pointwise addition of linear operators:

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

Then:

 $\ds \norm {\map {\paren {I - R} } f}_2$ $\le$ $\ds \norm {\map I f}_2 + \norm {\map R f}_2$ $\ds$ $=$ $\ds \norm {f}_2 + \norm {\check f}_2$ $\ds$ $=$ $\ds 2 \norm {f}_2$

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 $\map {\LL^2} \R$.

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

$\blacksquare$