Sum of Closable Densely-Defined Linear Operator and Densely-Defined Bounded Linear Operator is Closable

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\map D T, T}$ be a closable densely-defined linear operator with closure $\tuple {\map D {\overline T}, \overline T}$.
Let $\tuple {\map D S, S}$ be a densely-defined bounded linear operator with closure $\tuple {\HH, \overline S}$ such that:
:$\map D T \subseteq \map D S$
Let $\tuple {\map D T, T + S}$ be the sum of $\tuple {\map D T, T}$ and $\tuple {\map D S, S}$.
Let $\tuple {\map D {\overline T}, \overline T + \overline S}$ be the sum of $\tuple {\map D {\overline T}, \overline T}$ and $\tuple {\HH, \overline S}$.
Then $\tuple {\map D T, T + S}$ is closable densely-defined linear operator with closure $\tuple {\map D {\overline T}, \overline T + \overline S}$.
=== Corollary ===
== Proof ==
From Characterization of Closable Densely-Defined Linear Operators in terms of Closure of Graph and Closable Densely-Defined Linear Operator has Smallest Closed Extension, it is enough to show that:
:$\map \cl {\map \GG {T + S} } = \map \GG {\overline T + \overline S}$
where $\GG$ denotes the graph.
Let $\tuple {x, y} \in \map \cl {\map \GG {T + S} }$. 
By the definition of the closure, there exists a sequence $\sequence {\tuple {x_n, \paren {T + S} x_n} }_{n \mathop \in \N}$ valued in $\map \GG {T + S}$ such that:
:$\tuple {x_n, \paren {T + S} x_n} \to \tuple {x, y}$
From Convergence in Direct Product Norm, we have:
:$x_n \to x$
and:
:$\paren {T + S} x_n \to y$ 
From the definition of the closure, we have $T x_n = \overline T x_n$ and $S x_n = \overline S x_n$ for each $n \in \N$.
Since $\overline S$ is bounded from Closure of Densely-Defined Bounded Linear Operator we have $S x_n = \overline S x_n \to \overline S x$.
We therefore have:
:$\overline T x_n = T x_n \to y - \overline S x$
Since $\overline T$ is Definition:Closed Linear Transformation, we therefore obtain that:
:$x \in \map D {\overline T} \cap \HH = \map D {\overline T + \overline S}$ 
with:
:$\overline T x = y - \overline S x$
Hence we have:
:$y = \paren {\overline T + \overline S} x$
Hence we have $\tuple {x, y} \in \map \GG {\overline T + \overline S}$. 
Hence $\map \cl {\map \GG {T + S} } \subseteq \map \GG {\overline T + \overline S}$
Conversely let $\tuple {x, \paren {\overline T + \overline S} x} \in \map \GG {\overline T + \overline S}$.
From Closable Densely-Defined Linear Operator has Smallest Closed Extension, we have $\map \cl {\map \GG T} = \map \GG {\overline T}$.
So there exists $\sequence {x_n}_{n \mathop \in \N}$ in $\map D T$ such that $\tuple {x_n, T x_n} \to \tuple {x, \overline T x}$.
From Convergence in Direct Product Norm, we have $x_n \to x$ and $T x_n \to \overline T x_n$.
We have $\map D T \subseteq \map D S$, and so $x_n \in \map D S \subseteq \map D {\overline S}$ for each $n \in \N$.
Since $\overline S$ is bounded from Closure of Densely-Defined Bounded Linear Operator, we have $S x_n = \overline S x_n \to \overline S x$.
Hence we have $\paren {T + S} x_n \to \paren {\overline T + \overline S} x$ as $n \to \infty$.
Hence from Convergence in Direct Product Norm, we have $\tuple {x_n, \paren {T + S} x_n} \to \tuple {x, \paren {\overline T + \overline S} x}$ as $n \to \infty$.
Hence $\map \GG {\overline T + \overline S} \subseteq \map \cl {\map \GG {T + S} }$.
Category:Closable Densely-Defined Linear Operators
Category:Sum of Closable Densely-Defined Linear Operator and Densely-Defined Bounded Linear Operator is Closable