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

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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?
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