# 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 $\map f x$:

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

By Continuity of Linear Transformation between Normed Vector Spaces, $R$ is continuous.

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

- $\map I f = f$

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

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

Moreover:

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

By Continuity of Linear Transformation between Normed Vector Spaces, $I$ is continuous.

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

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

By definition of pointwise addition of linear operators:

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

Then:

\(\ds \norm {\map {\paren {I - R} } f}_2\) | \(\le\) | \(\ds \norm {\map I f}_2 + \norm {\map R f}_2\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \norm {f}_2 + \norm {\check f}_2\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 2 \norm {f}_2\) |

By Continuity of Linear Transformation 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 $\map {\LL^2} \R$.

$Y$ 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.

$\blacksquare$

Further research is required in order to fill out the details.In particular: I need to state that p-Lebesgue space with p-norm is normed vector space, but elsewhere norm is replaced by semi-norm, so what is true?You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by finding out more.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Research}}` from the code. |

## Sources

- 2017: Amol Sasane:
*A Friendly Approach to Functional Analysis*... (previous) ... (next): Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$