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

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:
{{begin-eqn}}
{{eqn | ll = \forall f, g \in \map {\LL^2} \R : \forall x \in \R : 
      | l = \map R {\map f x + \map g x}
      | r = \map R {\map {\paren {f + g} } x}
      | c = {{Defof|Pointwise Addition of Real-Valued Functions}}
}}
{{eqn | r = \map {\paren {f + g} } {- x}
}}
{{eqn | r = \map f {-x} + \map g {-x}
      | c = {{Defof|Pointwise Addition of Real-Valued Functions}}
}}
{{eqn | r = \map R {\map f x} + \map R {\map g x}
}}
{{end-eqn}}
Furthermore:
{{begin-eqn}}
{{eqn | ll = \forall f \in \map {\LL^2} \R : \forall x, \lambda \in \R : 
      | l = \map R {\lambda \map f x}
      | r = \map R {\map {\paren {\lambda \cdot f} } x}
      | c = {{Defof|Pointwise Scalar Multiplication of Real-Valued Functions}}
}}
{{eqn | r = \map {\paren {\lambda \cdot f} } {-x}
}}
{{eqn | r = \lambda \map f {-x}
      | c = {{Defof|Pointwise Scalar Multiplication of Real-Valued Functions}}
}}
{{eqn | r = \lambda  \map R {\map f x}
}}
{{end-eqn}}
Thus, $R$ is a linear transformation:
:$R \in \map L {\map {\LL^2} \R}$.
Moreover, because the reflection {{WRT}} $y$-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:
{{begin-eqn}}
{{eqn | l = \norm {\map {\paren {I - R} } f}_2
      | o = \le 
      | r = \norm {\map I f}_2 + \norm {\map R f}_2
}}
{{eqn | r = \norm {f}_2 + \norm {\check f}_2
}}
{{eqn | r = 2 \norm {f}_2
}}
{{end-eqn}}
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.
{{qed}}
{{Research|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?}}