avewright/proofwiki-math huggingface dataset:
Theorem: Intersection Measures preserve Mutual Singularity

Let $\struct {X, \Sigma}$ be a measurable space.
Let $A \in \Sigma$.
Let $\mu$ and $\nu$ be measures such that:
:$\mu$ is mutually singular {{WRT}} $\nu$.
Let $\mu_A$ be the intersection measure of $\mu$ {{WRT}} $A$.
Then $\mu_A$ is mutually singular {{WRT}} $\nu$.

Proof:
Since $\mu$ is mutually singular with respect to $\nu$, there exists a $\nu$-null set $N$ such that $\map \mu {N^c} = 0$.
From Null Sets Closed under Subset, we have $\map \mu {A \cap N^c} = 0$.
Hence $\map {\mu_A} {N^c} = 0$. 
Hence $N$ is a $\nu$-null set such that $\map {\mu_A} {N^c} = 0$.
Hence $\mu_A$ is mutually singular {{WRT}} $\nu$.
{{qed}}
Category:Mutually Singular Measures
hc9w9ymd97ptlcy09bm1r9ksd2ppgy7
Theorem: Lebesgue Decomposition of Finite Borel Measure

Let $\map \BB \R$ be the Borel $\sigma$-algebra of the real number line $\R$.
Let $\lambda$ be the Lebesgue measure on $\struct {\R, \map \BB \R}$. 
Let $\mu$ be a finite Borel measure on $\struct {\R, \map \BB \R}$.
Then there exists:
:a discrete measure $\mu_{pp}$
:a measure $\mu_{sc}$ that is continuous and mutually singular with respect to $\lambda$
:a measure $\mu_{ac}$ that is absolutely continuous with respect to $\lambda$
such that $\mu = \mu_{pp} + \mu_{sc} + \mu_{ac}$.

Proof:
Let:
:$C = \set {x \in \R : \map \mu {\set x} > 0}$ 
We argue that $C$ is countable.
Note that:
:$\ds C = \bigcup_{n \mathop = 1}^\infty \set {x \in \R : \map \mu {\set x} > \frac 1 n}$
We show that each set:
:$\ds C_n = \set {x \in \R : \map \mu {\set x} > \frac 1 n}$
is finite.
{{AimForCont}} that $C_n$ is infinite for some $n$.
Let $D_n = \sequence {x_j}_{j \mathop \in \N}$ be a countably infinite subset of $C_n$.
Then we have:
:$\ds \map \mu {D_n} = \sum_{j \mathop = 1}^\infty \map \mu {\set {x_j} } \ge \sum_{j \mathop = 1}^\infty \frac 1 n = \infty$
This contradicts that $\mu$ is finite.
Hence $C_n$ is finite for each $n \in \N$. 
Hence from Countable Union of Finite Sets is Countable, $C$ is countable.
Define:
:$\map {\mu_{pp} } A = \map \mu {A \cap C}$
for each $A \in \Sigma$.
We have that:
:$\map {\mu_{pp} } {C^c} = \map \mu \O = 0$
By Measure is Discrete iff Concentrated on Countable Set, it follows that $\mu_{pp}$ is discrete. 
Now consider the intersection measure $\mu_{C^c}$.
From Lebesgue Decomposition Theorem, there exists finite measures $\mu_{ac}$ and $\mu_{sc}$ such that:
:$(1) \quad$ $\mu_{ac}$ is absolutely continuous with respect to $\lambda$
:$(2) \quad$ $\mu_{sc}$ and $\lambda$ are mutually singular
:$(3) \quad$ $\mu_{C^c} = \mu_{sc} + \mu_{ac}$.
We show that $\mu_{sc}$ is continuous.
Let $x \in \R$.
If $x \in C$ then we have $\set x \cap C^c = \O$ and $\map {\mu_{ac} } {\set x} = 0$ and hence $\map {\mu_{sc} } {\set x} = \map \mu \O = 0$. 
If $x \not \in C$, then we have that $\set x \cap C^c = \set x$ and $\map \mu {\set x} = \map {\mu_{ac} } {\set x} = 0$. 
Hence $\map {\mu_{sc} } {\set x} = \map \mu \O = 0$. 
Hence $\map {\mu_{sc} } {\set x} = 0$ for each $x \in \R$. 
So $\mu_{sc}$ is continuous.
We have therefore obtained the desired decomposition.
{{qed}}
Theorem: Graph of Linear Transformation is Vector Subspace

Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $X \times Y$ be the direct product of $X$ and $Y$.
Let $T : X \to Y$ be a linear transformation.
Define:
:$\map \GG T = \set {\tuple {x, T x} : x \in X} \subseteq X \times Y$ 
Then $\map \GG T$ is a vector subspace of $X \times Y$.

Proof:
Since $\map T { {\mathbf 0}_X} = {\mathbf 0}_Y$, we have:
:$\tuple { {\mathbf 0}_X, {\mathbf 0}_Y} \in \map \GG T$ 
So $\map \GG T \ne \O$.
Hence from One-Step Vector Subspace Test, it is enough to show that $u + \lambda v \in \map \GG T$ for $u, v \in \map \GG T$ and $\lambda \in K$. 
Let $\alpha \in K$ and $\tuple {x, T x}, \tuple {y, T x} \in \map \GG T$.
Then we have:
:$\tuple {x, T x} + \alpha \tuple {y, T y} = \tuple {x + \alpha y, T x + \alpha T y}$ 
Since $T$ is linear, we have that $T x + \alpha T y = \map T {x + \alpha y}$. 
Hence:
:$\tuple {x, T x} + \alpha \tuple {y, T y} = \tuple {x + \alpha y, \map T {x + \alpha y} } \in \map \GG T$ 
Hence from One-Step Vector Subspace Test, $\map \GG T$ is a vector subspace of $X \times Y$.
{{qed}}
Category:Direct Product of Vector Spaces
q7zfgmwk3ror3li36f3cdbw8qgd54it
Theorem: Characterization of Closable Densely-Defined Linear Operators in terms of Closure of Graph

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Consider $\HH \times \HH$ with the direct product norm $\norm {\, \cdot \,}_{\HH \times \HH}$.
Let $\tuple {\map D T, T}$ be a densely-defined linear operator.
Let $\map \GG T$ be the graph of $T$.
Then $T$ is closable {{iff}}:
:whenever $\tuple { {\mathbf 0}_X, y} \in \map \cl {\map \GG T}$, we have $y = {\mathbf 0}_Y$
where $\map \cl {\map \GG T}$ is the closure of $\map \GG T$ taken in $\struct {\HH \times \HH, \norm {\, \cdot \,}_{\HH \times \HH} }$.

Proof:
=== Necessary Condition ===
Suppose that $T$ is closed.
Then $\map \cl {\map \GG T} = \map \GG T$ from Set is Closed iff Equals Topological Closure.
Then if $\tuple { {\mathbf 0}_X, y} \in \map \GG T$, we have $y = \map T { {\mathbf 0}_X}$ from the definition of the graph. 
So $y = {\mathbf 0}_Y$ since $T$ is linear.
{{qed|lemma}}
Theorem: Linear Transformation defined from Graph

Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$. 
Let $X \times Y$ be the direct product of $X$ and $Y$.
Let $U$ be a vector subspace of $X \times Y$ such that:
:whenever $\tuple { {\mathbf 0}_X, y} \in U$, we have $y = {\mathbf 0}_Y$
Define:
:$\map D T = \set {x \in X : \tuple {x, y} \in U \text { for some } y \in Y}$ 
Then there exists a linear transformation $T : \map D T \to Y$ with graph $\map \GG T = U$.

Proof:
We first show that $\map D T$ is a vector subspace of $X$.
We have $\tuple { {\mathbf 0}_X, {\mathbf 0}_Y} \in U$, so ${\mathbf 0}_X \in \map D T$. 
Hence $\map D T \ne \O$. 
Let $x, y \in \map D T$ and $\alpha \in \C$.
Then there exists $u, v \in Y$ such that $\tuple {x, u} \in U$ and $\tuple {y, v} \in U$. 
Since $U$ is a vector subspace we have:
:$\tuple {x, u} + \alpha \tuple {y, v} = \tuple {x + \alpha y, u + \alpha v} \in U$
So $x + \alpha y \in \map D T$.
Hence from the One-Step Vector Subspace Test, we have that $\map D T$ is a vector subspace of $X$.
Suppose that $\tuple {x, y_1}, \tuple {x, y_2} \in U$.
Then $\tuple { {\mathbf 0}_X, y_1 - y_2} = \tuple {x, y_1} - \tuple {x, y_2} \in U$ since $U$ is a vector subspace of $X \times Y$. 
So $y_1 - y_2 = {\mathbf 0}_Y$ and hence $y_1 = y_2$. 
Hence we can define $T x$ as the unique $y$ such that $\tuple {x, y} \in U$ for $x \in \map D T$.
Clearly we will have $\map G T = U$. 
We just need to show that $T$ is linear.
Let $x, y \in \map D T$ and $\alpha \in \C$.
Then we have $\tuple {x, T x} + \alpha \tuple {y, T y} = \tuple {x + \alpha y, T x + \alpha T y} \in U$.
Since $\map T {x + \alpha y}$ is unique with $\tuple {x + \alpha y, \map T {x + \alpha y} } \in U$, we have:
:$\map T {x + \alpha y} = T x + \alpha T y$
Hence $T$ is linear.
{{qed}}
Category:Linear Transformations
3wwywxt0am32xs8cfyb7nm224pz4o57
Theorem: Closable Densely-Defined Linear Operator has Smallest Closed Extension

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.
Then there exists a closed densely-defined linear extension $\struct {\map D S, S}$ of $T$ such that:
:the graph $\map \GG S$ is $\subseteq$-minimal among the graphs of closed densely-defined linear extensions of $T$. 
Further:
:$\map \GG S = \map \cl {\map \GG T}$

Proof:
From Characterization of Closable Densely-Defined Linear Operators in terms of Closure of Graph:
:whenever $\tuple { {\mathbf 0}_\HH, y} \in \map \cl {\map \GG T}$, we have $y = {\mathbf 0}_\HH$.
Define:
:$\map D S = \set {x \in \HH : \tuple {x, y} \in \map \cl {\map \GG T} \text { for some } y \in \HH}$
From Linear Transformation defined from Graph, there exists a linear transformation $S : \map D S \to \HH$ such that $\map \GG S = \map \cl {\map \GG T}$.
Now let $S'$ be another closed densely-defined linear extensions of $T$.
We have $\map \GG T \subseteq \map \GG {S'}$.
Since $\map \GG {S'}$ is closed, we obtain:
:$\map \GG S = \map \cl {\map \GG T} \subseteq \map \GG {S'}$
from Set Closure Preserves Set Inclusion.
{{qed}}
Theorem: Closable Densely-Defined Operator is Closed iff Equal to Closure

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.
Let $\tuple {\map D {\overline T}, \overline T}$ be the closure of $\tuple {\map D T, T}$.
Then $\tuple {\map D T, T}$ is closed {{iff}} $T = \overline T$.

Proof:
Let $\struct {\HH \times \HH, \norm {\, \cdot \,}_{\HH \times \HH} }$ be the direct product of $\HH$ with itself equipped with the direct product norm. 
Suppose that $\tuple {\map D T, T}$ is closed.
From Set is Closed iff Equals Topological Closure, we have that $\map \cl {\map \GG T} = \map \GG T$.
From Closable Densely-Defined Linear Operator has Smallest Closed Extension, we have $\map \GG {\overline T} = \map \cl {\map \GG T}$ and hence:
:$\map \GG {\overline T} = \map \GG T$
So $T = \overline T$.
Now suppose that $\overline T = \overline T$. 
Then $\map \GG T = \map \GG {\overline T}$. 
Since $\map \GG {\overline T}$ is closed in $\struct {\HH \times \HH, \norm {\, \cdot \,}_{\HH \times \HH} }$, so is $\map \GG T$.
Hence $T$ is closed.
{{qed}}
Category:Closed Linear Transformations
Category:Closable Densely-Defined Linear Operators
jrcmg30u8ia7q1fav0at5iwj8a7q7t5
Theorem: Square of V Operator on Hilbert Space

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$ be the Hilbert space direct sum of $\HH$ with itself.
Define $V : \HH \times \HH \to \HH \times \HH$ by:
:$\map V {x, y} = \tuple {-y, x}$
for each $\tuple {x, y} \in \HH \times \HH$.
Then $V^2 = -I$.

Proof:
For each $x, y \in \HH$ we have:
{{begin-eqn}}
{{eqn	| l = \map {V^2} {x, y}
		| r = \map V {-y, x} 
}}
{{eqn	| r = \tuple {-x, -y}
}}
{{eqn	| r = -\tuple {x, y}
}}
{{end-eqn}}
{{qed}}
Theorem: Adjoint of Densely-Defined Linear Operator is Closed

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\map D T, T}$ be a densely-defined linear operator.
Let $\tuple {\map D {T^\ast}, T^\ast}$ be the adjoint of $\tuple {\map D T, T}$.
Then $\tuple {\map D {T^\ast}, T^\ast}$ is closed.

Proof:
Let $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$ be the Hilbert space direct sum of $\HH$ with itself.
From Equivalent Norms on Direct Product of Normed Vector Spaces, the inner product norm on $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$ is equivalent to the direct product norm on $\HH \times \HH$.
Hence from Open Sets in Vector Spaces with Equivalent Norms Coincide it is enough to show that $\map \GG {T^\ast}$ is closed in $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$.
Define $V : \HH \times \HH \to \HH \times \HH$ by:
:$\map V {x, y} = \tuple {-y, x}$
for each $\tuple {x, y} \in \HH \times \HH$.
From Graph of Adjoint of Densely-Defined Linear Operator, we have:
:$\map \GG {T^\ast} = \paren {V \sqbrk {\map \GG T} }^\bot$
where orthocomplementation is considered in the Hilbert space direct sum $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$.
From Orthocomplement is Closed Linear Subspace, $\paren {V \sqbrk {\map \GG T} }^\bot$ is closed in $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$.
Hence $\map \GG {T^\ast}$ is closed in $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$.
So $\tuple {\map D {T^\ast}, T^\ast}$ is closed.
{{qed}}
Theorem: Hilbert Space Isomorphism preserves Orthocomplements

Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces.
Let $T : \HH \to \KK$ be a Hilbert space isomorphism.
Let $U \subseteq \HH$.
Then:
:$T \sqbrk {U^\bot} = \paren {T \sqbrk U}^\bot$

Proof:
Let $x \in \KK$.
Since $T$ is a Hilbert space isomorphism, there exists $v \in \HH$ such that $x = T v$.
We then have $x \in \paren {T \sqbrk U}^\bot$ {{iff}}:
:$\innerprod x {T u}_\KK = 0$ for each $u \in U$.
That is:
:$\innerprod {T v} {T u}_\KK = 0$
Since $T$ is a Hilbert space isomorphism, this is equivalent to $\innerprod v u_\HH = 0$ for each $u \in U$.
So this is equivalent to $v \in U^\bot$.
Since $x = T v$, this is equivalent to $x \in T \sqbrk {U^\bot}$. 
So we obtain:
:$T \sqbrk {U^\bot} = \paren {T \sqbrk U}^\bot$
{{qed}}
Category:Orthocomplements
Category:Hilbert Spaces
nnzgakwgsv1m5hyvtcogna8qqtypt0q
Theorem: Solution to Quadratic Congruence

The quadratic congruence:
:$a x^2 + b x + c \equiv 0 \pmod n$
can be solved by solving the congruence:
:$y^2 \equiv \paren {b^2 - 4 a c} \pmod n$
and the linear congruence:
:$2 a x + b \equiv y \pmod n$

Proof:
{{ProofWanted|use the method of completing the square}}
Theorem: Densely-Defined Linear Operator is Closable iff Adjoint is Densely-Defined

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\map D T, T}$ be a densely-defined linear operator.
Let $\tuple {\map D {T^\ast}, T^\ast}$ be the adjoint of $\tuple {\map D T, T}$.
Then $\tuple {\map D T, T}$ is closable {{iff}} $\tuple {\map D {T^\ast}, T^\ast}$ is densely-defined. 
If $T$ is closable then the closure $\overline T$ satisfies $\overline T = T^{\ast \ast} = \paren {T^\ast}^\ast$.

Proof:
Let $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$ be the Hilbert space direct sum of $\HH$ with itself.
From Equivalent Norms on Direct Product of Normed Vector Spaces, the inner product norm on $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$ is equivalent to the direct product norm on $\HH \times \HH$.
Hence from Open Sets in Vector Spaces with Equivalent Norms Coincide, we work in $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$ without concern.
Define $V : \HH \times \HH \to \HH \times \HH$ by:
:$\map V {x, y} = \tuple {-y, x}$
for each $\tuple {x, y} \in \HH \times \HH$.
From Square of V Operator on Hilbert Space, we have $V^2 = -I$.
Hence $V^2 M = -M = M$ for any vector subspace $M$ in $\HH$.
Theorem: Adjoint of Closure of Closable Densely-Defined Linear Operator

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 adjoint $\tuple {\map D {T^\ast}, T^\ast}$. 
Let $\tuple {\map D {\overline T}, \overline T}$ be the closure of $T$ with adjoint $\tuple {\map D {\paren {\overline T}^\ast}, \paren {\overline T}^\ast}$. 
Then:
:$\tuple {\map D {T^\ast}, T^\ast} = \tuple {\map D {\paren {\overline T}^\ast}, \paren {\overline T}^\ast}$

Proof:
From Adjoint of Densely-Defined Linear Operator is Closed:
:$\tuple {\map D {T^\ast}, T^\ast}$ is closed.
From Closable Densely-Defined Operator is Closed iff Equal to Closure, we have:
:$\tuple {\map D {T^\ast}, T^\ast} = \tuple {\map D {\overline {T^\ast} }, \overline {T^\ast} }$
From Densely-Defined Linear Operator is Closable iff Adjoint is Densely-Defined, we have:
:$\tuple {\map D {\overline {T^\ast} }, T^\ast} = \tuple {\map D {T^{\ast \ast \ast} }, T^{\ast \ast \ast} }$
Writing $T^{\ast \ast \ast} = \paren {T^{\ast \ast} }^\ast$, we have:
:$\tuple {\map D {T^{\ast \ast \ast} }, T^{\ast \ast \ast} } = \tuple {\map D {\paren {\overline T}^\ast}, \paren {\overline T}^\ast}$
from Densely-Defined Linear Operator is Closable iff Adjoint is Densely-Defined.
{{qed}}
Theorem: Symmetric Densely-Defined Linear Operator is Closable

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\map D T, T}$ be a symmetric densely-defined linear operator.
Then $\tuple {\map D T, T}$ is closable.

Proof:
Let $\tuple {\map D {T^\ast}, T^\ast}$ be the adjoint of $\tuple {\map D T, T}$.
From Adjoint of Densely-Defined Linear Operator is Closed, $\tuple {\map D {T^\ast}, T^\ast}$ is closed.
From Adjoint of Symmetric Densely-Defined Linear Operator Extends Operator, $\tuple {\map D {T^\ast}, T^\ast}$ extends $\tuple {\map D T, T}$.
Hence $\tuple {\map D T, T}$ is closable.
{{qed}}
Category:Symmetric Densely-Defined Linear Operators
Category:Closable Densely-Defined Linear Operators
j0owor09inlisbdbsipohp3egkp1sv2
Theorem: Globally-Defined Symmetric Linear Operator is Bounded and Self-Adjoint

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\HH, T}$ be a symmetric linear operator.
Then $\tuple {\HH, T}$ is bounded and self-adjoint.

Proof:
Let $\tuple {\map D {T^\ast}, T^\ast}$ be the adjoint of $\tuple {\HH, T}$. 
From Adjoint of Symmetric Densely-Defined Linear Operator Extends Operator, $\tuple {\map D {T^\ast}, T^\ast}$ extends $\tuple {\HH, T}$. 
However, $\map D {T^\ast} \subseteq \HH$. 
So we have $\map D {T^\ast} = \HH = \map D T$.
Hence $T^\ast x = T x$ for each $x \in \map D T$.
So $\tuple {\HH, T} = \tuple {\HH, T^\ast}$.
So $\tuple {\HH, T}$ is self-adjoint.
From Adjoint of Densely-Defined Linear Operator is Closed, $\tuple {\HH, T}$ is also closed.
Since $\HH$ is Banach, we have that:
:$T$ is bounded
from the Closed Graph Theorem.
{{qed}}
Theorem: Injective Self-Adjoint Densely-Defined Linear Operator has Everywhere Dense Image

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\map D T, T}$ be a self-adjoint densely defined linear operator that is injective.
Then $\Img T$ is everywhere dense in $\HH$.

Proof:
From Kernel of Adjoint of Densely-Defined Linear Operator is Orthocomplement of Image, we have:
:$\map \ker {T^\ast} = \paren {\Img T}^\bot$ 
where $\tuple {\map D {T^\ast}, T^\ast}$ is the adjoint of $\tuple {\map D T, T}$.
Since $T$ is self-adjoint, we have $T^\ast = T$. 
Hence $\map \ker T = \paren {\Img T}^\bot$.
Since $T$ is injective, we have $\map \ker T = {\mathbf 0}_\HH$.
Hence:
:$\paren {\Img T}^\bot = \set { {\mathbf 0}_\HH}$
From Linear Subspace Dense iff Zero Orthocomplement we have that $\Img T$ is everywhere dense in $\HH$.
{{qed}}
Theorem: Kernel of Closed Linear Transformation is Closed

Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be Banach spaces.
Let $T : \map D T \to Y$ be a closed linear transformation.
Let $\ker T$ be the kernel of $T$.
Then $\ker T$ is closed.

Proof:
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\ker T$ converging to $x \in X$. 
We have $T x_n = {\mathbf 0}_Y$ for each $n \in \N$.
Hence from Constant Sequence in Normed Vector Space Converges, we have $T x_n \to {\mathbf 0}_Y$ as $n \to \infty$.
From Sequential Characterization of Closed Linear Transformation, we therefore have that $x \in \map D T$ and:
:$\ds T x = \lim_{n \mathop \to \infty} T x_n = {\mathbf 0}_Y$
Hence $x \in \ker T$. 
So every convergent sequence in $\ker T$ has its limit in $\ker T$.
Hence $\ker T$ is closed.
{{qed}}
Category:Closed Linear Transformations
tf36kns1yp0vq41xd32qm5ucbapfi9p
Theorem: Self-Adjoint Densely-Defined Linear Operator is Maximally Symmetric

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\map D T, T}$ be a self-adjoint densely-defined linear operator.
Then $\tuple {\map D T, T}$ is maximally symmetric.

Proof:
Let $\tuple {\map D S, S}$ be a symmetric densely-defined linear operator extending $\tuple {\map D T, T}$.
Let $\tuple {\map D {S^\ast}, S^\ast}$ and $\tuple {\map D {T^\ast}, T^\ast}$ be the adjoints of $\tuple {\map D S, S}$ and $\tuple {\map D T, T}$ respectively.
From Adjoint of Densely-Defined Linear Operator reverses Extension of Mapping:
:$\tuple {\map D {T^\ast}, T^\ast}$ extends $\tuple {\map D {S^\ast}, S^\ast}$.
Since $\tuple {\map D T, T}$ is self-adjoint, we have.
:$\tuple {\map D T, T}$ extends $\tuple {\map D {S^\ast}, S^\ast}$.
From Adjoint of Symmetric Densely-Defined Linear Operator Extends Operator, we have:
:$\tuple {\map D {S^\ast}, S^\ast}$ extends $\tuple {\map D S, S}$.
Hence $\tuple {\map D T, T}$ extends $\tuple {\map D S, S}$.
By assumption $\tuple {\map D S, S}$ extends $\tuple {\map D T, T}$.
Hence $\tuple {\map D T, T} = \tuple {\map D S, S}$.
So $\tuple {\map D T, T}$ is maximally symmetric.
{{qed}}
Theorem: Descartes's Solution to Quartic Equation

Let $P$ be the quartic equation:
:$a x^4 + b x^3 + c x^2 + d x + e = 0$
such that $a \ne 0$.
Then $P$ has solutions:
{{WIP|Work to be done to complete this}}

Proof:
First we render the quartic into monic form:
:$x^4 + \dfrac b a x^3 + \dfrac c a x^2 + \dfrac d a x + \dfrac e a = 0$
Using a '''Tschirnhaus transformation''', $x = y - \dfrac b {4 a}$, we convert $P$ into the reduced quartic:
:$y^4 + p y^2 + q y + r = 0$
This is set identically equal to:
:$\paren {y^2 + \lambda y + m} \paren {y^2 - \lambda y + n}$
Equating coefficients and eliminating $m$ and $n$ we obtain the following cubic in $\lambda^2$:
:$\lambda^6 + 2 p \lambda^4 + \paren {p^2 - 4 r} \lambda^2 - q^2 = 0$
which can be solved using Cardano's Method, for example.
Hence by solving the two quadratic equations:
{{begin-eqn}}
{{eqn | l = y^2 + \lambda y + m
      | r = 0
}}
{{eqn | l = y^2 - \lambda y + n
      | r = 0
}}
{{end-eqn}}
the solution to the reduced quartic is found.
{{qed}}
Theorem: Characterization of Completely Prime Filter in Complete Lattice

Let $\struct{L, \vee, \wedge, \preceq}$ be a complete lattice.
Let $F \subseteq L$.
Then:
:$F$ is a completely prime filter
{{iff}}
:$(1)\quad\forall A \subseteq L : \bigvee A \in F \iff \paren{\exists a \in A : a \in F}$
:$(2)\quad\forall $ finite $A \subseteq L : \bigwedge A \in F \iff \paren{\forall a \in A : a \in F}$
where:
:* $\bigvee A$ denotes the supremum of $A$ in $L$
:* $\bigwedge A$ denotes the infimum of $A$ in $L$

Proof:
=== Necessary Condition ===
Let $F$ be a completely prime filter.
{{:Characterization of Completely Prime Filter in Complete Lattice/Necessary Condition}}