avewright/proofwiki-math huggingface dataset:
Theorem: Norm on Bounded Linear Transformation is Submultiplicative

Let $\struct {X, \norm \cdot_X}$, $\struct {Y, \norm \cdot_Y}$ and $\struct {Z, \norm \cdot_Z}$ be normed vector spaces.
Let $A : X \to Y$ and $B : Y \to Z$ be bounded linear transformations. 
Let $\norm \cdot _{\map B {X,Y} }$ be the norm for bounded linear transformations $X \to Y$.
Let $\norm \cdot _{\map B {Y,Z} }$ be the norm for bounded linear transformations $Y \to Z$.
Let $\norm \cdot _{\map B {X,Z} }$ be the norm for bounded linear transformations $X \to Z$.
Then, we have that:
:$B \circ A$ is a bounded linear transformation
with:
:$\norm {B \circ A} _{\map B {X,Z} } \le \norm B _{\map B {Y,Z} } \norm A _{\map B {X,Y} }$
That is: 
:$\norm \cdot$ is submultiplicative.
{{explain|Which norm is this $\norm \cdot$?}}

Proof:
From Composition of Linear Transformations is Linear Transformation, we have: 
:$B \circ A$ is a linear transformation
Let $x \in X$. 
Then, we have:
{{begin-eqn}}
{{eqn	| l = \norm {\paren {B \circ A} x}_Z
	| o = \le
	| r = \norm B \norm {A x}_Y
	| c = Fundamental Property of Norm on Bounded Linear Transformation
}}
{{eqn	| o = \le
	| r = \norm B \norm A \norm x_X
	| c = Fundamental Property of Norm on Bounded Linear Transformation
}}
{{end-eqn}}
We can therefore see that: 
:$B \circ A$ is a bounded linear transformation.
So, if: 
:$\norm x_X = 1$
we have: 
:$\norm {\paren {B \circ A} x}_Z \le \norm B \norm A$
By the definition of supremum, we have: 
:$\ds \sup_{\norm x_X = 1} \norm {\paren {B \circ A} x}_Z \le \norm B \norm A$
So by the definition of the norm on the space of bounded linear transformations, we have: 
:$\norm {B \circ A} \le \norm B \norm A$
{{qed}}
Category:Bounded Linear Transformations
lgkbdilysy23gc599d3zgirypged2in
Theorem: Rational Sequence Converges in P-adic Numbers iff Sequence Represents Limit

Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a$ be a $p$-adic number, that is left coset, in $\Q_p$.
Let $\sequence{x_n}$ be a rational sequence.
Then:
:$\sequence{x_n}$ converges to $a$ {{iff}} $\sequence{x_n}$ is a representative of $a$
=== Corollary ===
{{:Rational Sequence Converges in P-adic Numbers iff Sequence Represents Limit/P-adic Expansion}}

Proof:
Let $\norm {\,\cdot\,}^\Q_p$ be the p-adic norm on the rationals $\Q$.
By definition of the $p$-adic numbers:
:$\Q_p$ is the quotient ring $\CC \, \big / \NN$
where:
:$\CC$ is the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}^\Q_p}$.
and
:$\NN$ is the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}^\Q_p}$.
By definition of the $p$-adic numbers, $\norm {\, \cdot \,}_p:\Q_p \to \R_{\ge 0}$ is the norm on the quotient ring $\Q_p$ defined by:
:$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN }_p = \lim_{n \mathop \to \infty} \norm{x_n}_p$
From Rational Numbers are Dense Subfield of P-adic Numbers, the mapping $\phi: \Q \to \Q_p$ defined by:
:$\map \phi r = \tuple {r, r, r, \dotsc} + \NN$
::where $\tuple {r, r, r, \dotsc}$ is the constant sequence
embeds $\Q$ as a dense subfield of $\Q_p$.
The theorem follows immediately from Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit.
{{qed}}
Category:P-adic Number Theory
Category:Rational Sequence Converges in P-adic Numbers iff Sequence Represents Limit
5b6qofi5bghd970xl22cjnpq002wna6
Theorem: Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit

Let $\struct {R, \norm{\,\cdot\,}_R}$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Let $Q = \CC  / \NN$ where $\CC  / \NN$ denotes a quotient ring.
Let $\norm {\, \cdot \,}_Q: Q \to \R_{\ge 0}$ be the norm on the quotient ring $Q$ defined by:
:$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN }_Q = \lim_{n \mathop \to \infty} \norm{x_n}_R$
Let $\phi: R \to Q$ be the mapping from $R$ to the quotient ring $Q$ defined by:
:$\forall a \in R: \map \phi a = \tuple {a, a, a, \ldots} + \NN$
where $\sequence {a, a, a, \ldots} + \NN$ is the left coset in $Q$ that contains the constant sequence $\sequence {a, a, a, \ldots} $.
Let $\sequence{x_n}$ be a sequence in $R$.
Let $y \in Q$.

Proof:
=== Lemma 1 ===
{{:Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit/Lemma 1}}{{qed|lemma}}
Theorem: Vanishing Distributional Derivative of Distribution implies Distribution is Constant

Let $T \in \map {\DD'} \R$ be a Schwartz distribution.
Let $\mathbf 0$ be the zero distribution.
Suppose the distributional derivative of $T$ vanishes:
:$\ds \dfrac \d {\d x} T = \mathbf 0$
Then $T$ is a constant distribution.

Proof:
Let $\phi \in \map \DD \R$ be a test function.
Then:
{{begin-eqn}}
{{eqn | l = 0
      | r  = \map {\mathbf 0} \phi
}}
{{eqn | r = \map {T'} \phi
      | c = Assumption of the Theorem
}}
{{eqn | r = - \map T {\phi'}
      | c = {{Defof|Distributional Derivative}}
}}
{{end-eqn}}
Hence:
:$\set {\phi' : \phi \in \map \DD \R} \subseteq \ker T$
where $\ker$ denotes the kernel.
Let $\mathbf 1$ be a constant mapping such that $\mathbf 1 : \R \to 1$.
Then the associated Schwartz distribution reads:
:$\ds \map {T_{\mathbf 1}} \phi = \int_{-\infty}^\infty \map \phi x \rd x$
Furthermore:
{{begin-eqn}}
{{eqn | l = \ker T_{\mathbf 1}
      | r = \set {\psi \in \map \DD \R : \int_{-\infty}^\infty \map \psi x \rd x = 0}
}}
{{eqn | r = \set {\phi' : \phi \in \map \DD \R}
      | c = Characterization of Derivative of Test Function
}}
{{eqn | o = \subseteq
      | r = \ker T
}}
{{end-eqn}}
We have that Test Function Space with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space.
Let $V = \map \DD \R$, $L = T$ and $\ell = T_{\mathbf 1}$.
By Kernel of Linear Transformation contained in Kernel of different Linear Transformation implies Transformations are Proportional:
:$\exists c \in \C : T = c T_{\mathbf 1}$ 
By definition of multiplication of a Schwartz distribution by a smooth function:
:$c T_{\mathbf 1} = T_c$
{{qed}}
Theorem: Product Rule for Distributional Derivatives of Distributions multiplied by Smooth Functions

Let $\alpha \in \map {C^\infty} \R$ be a smooth real function.
Let $T \in \map {\DD'} \R$ be a Schwartz distribution.
Then in the distributional sense it holds that:
:$\paren {\alpha T}' = \alpha' T + \alpha T'$

Proof:
Let $\phi \in \map \DD \R$ be a test function.
By the Product Rule for Derivatives:
:$\paren {\alpha \phi}' = \alpha' \phi + \alpha \phi'$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\paren {\alpha T}'} \phi
      | r = -\map {\paren {\alpha T} } {\phi'}
      | c = {{Defof|Distributional Derivative}}
}}
{{eqn | r = -\map T {\alpha \phi'}
      | c = {{Defof|Multiplication of Schwartz Distribution by Smooth Function}}
}}
{{eqn | r = -\map T {\paren {\alpha \phi}' - \alpha' \phi}
      | c = Product Rule for Derivatives
}}
{{eqn | r = -\map T {\paren {\alpha \phi}'} + \map T {\alpha' \phi}
      | c = {{Defof|Schwartz Distribution}}: Linearity
}}
{{eqn | r = \map {T'} {\alpha \phi} + \map T {\alpha' \phi}
      | c = {{Defof|Distributional Derivative}}
}}
{{eqn | r = \alpha \map {T'} \phi + \alpha' \map T \phi
      | c = {{Defof|Multiplication of Schwartz Distribution by Smooth Function}}
}}
{{eqn | r = \map {\paren {\alpha T' + \alpha' T} } \phi
      | c = Addition of Distributions
}}
{{end-eqn}}
{{qed}}
Theorem: Distributional Solution to y' - k y = 0

Let $f \in \map {C^1} \R$ be a continuously differentiable function.
Let $T \in \map {\DD'} \R$ be a Schwartz distribution.
Let $T_f$ be a Schwartz distribution associated with $f$.
Let $\mathbf 0 \in \map {\DD'} \R$ be the zero distribution.
Let $T$ be a distributional solution to the following distributional differential equation:
:$\paren {\dfrac \d {\d x} - k} T = \mathbf 0$
{{explain|Unless a "distributional solution" actually is the same thing as a "weak solution" (I believe not), then we need the former to have its own page, where it is specifically defined. At the very least it should be a redirect.}}
Then $T = T_f$ where $f$ is a classical solution to the ordinary differential equation:
{{explain|General or particular? From its form I suspect the latter. If it's actually "technically neither" because of complications that I haven't perceived, then we need to set up a page for "classical solution", or we are in serious danger of further muddying the already rather turbid water around this area. In any case, the exact nature of $f$ really needs to be explained further. The source does not discuss this, but the classical solution $f$ is the solution with zero RHS, so I guess general.}}
:$\paren {\dfrac \d {\d x} - k} f = 0$

Proof:
In the distributional sense we have:
{{begin-eqn}}
{{eqn | l = \paren {\map \exp {- k x} T}'
      | r = -\map \exp {-k x} T + \map \exp {-k x} T'
      | c = Product Rule for Distributional Derivatives of Distributions multiplied by Smooth Functions
}}
{{eqn | r = \map \exp {-k x} \paren {-k T + T'}
}}
{{eqn | r = \map \exp {-k x} \paren {\dfrac \d {\d x} - k} T
}}
{{eqn | r = \mathbf 0
}}
{{end-eqn}}
By Vanishing Distributional Derivative of Distribution implies Distribution is Constant:
:$\exists c \in \C : \map \exp {-k x} T = T_c$
In other words:
{{begin-eqn}}
{{eqn | l = T
      | r = \map \exp {k x} T_c
}}
{{eqn | r = T_{c \map \exp {k x} }
}}
{{end-eqn}}
If $f = c \map \exp {k x}$, then:
:$\paren {\dfrac \d {\d x} - k} f = 0$
{{qed}}
Theorem: Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit/Lemma 1

:$\sequence{x_n}$ is a Cauchy sequence {{iff}} $\sequence{\map \phi {x_n}}$ is a Cauchy sequence

Proof:
From Embedding Division Ring into Quotient Ring of Cauchy Sequences:
:the mapping $\phi: R \to Q$ is a distance-preserving monomorphism
We have
{{begin-eqn}}
{{eqn | r = \sequence{x_n} \text{is a Cauchy sequence}
      | o = 
}}
{{eqn | ll = \leadstoandfrom
      | o = 
      | r = \forall \epsilon > 0 : \exists N \in \N : \forall m, n \ge N: \norm{x_n - x_m}_Q < \epsilon
      | c = {{Defof|Cauchy Sequence in Normed Division Ring|Cauchy Sequence}}
}}
{{eqn | ll = \leadstoandfrom
      | o = 
      | r = \forall \epsilon > 0 : \exists N \in \N : \forall m, n \ge N: \norm{\map \phi {x_n} - \map \phi {x_m} }_Q < \epsilon
      | c = {{Defof|Distance-Preserving Mapping}}
}}
{{eqn | ll = \leadstoandfrom
      | o = 
      | r = \sequence{x_n} \text{is a Cauchy sequence}
      | c = {{Defof|Cauchy Sequence in Normed Division Ring|Cauchy Sequence}}
}}
{{end-eqn}}
{{qed}}
Category:Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit
17himrvvzg1ljv4vi0b1e5r26god4go
Theorem: Dx - k is Hypoelliptic

Let $f \in \map {C^\infty} \R$ be a smooth real function.
Let $T \in \map {\DD'} \R$ be a Schwartz distribution.
Let $T_f$ be the Schwartz distribution associated with $f$.
Let $k \in \R$.
Suppose in the distributional sense it holds that:
:$\paren {\dfrac \d {\d x} - k} T = T_f \quad (1)$
Then there is a $c \in \R$ such that:
:$T = T_{F + c \map \exp {k x} }$
where $F \in \map {C^\infty} \R$ is a solution to:
{{explain|General or particular? From its form it looks the latter. The source does not discuss this, but $F$ looks like a particular solution.}}
:$\paren {\dfrac \d {\d x} - k} F = f$
In other words, $\dfrac \d {\d x} - k$ is a hypoelliptic operator.

Proof:
By assumption $f \in \map {C^\infty} \R$.
By Solution to Linear First Order ODE with Constant Coefficients:
:$\exists F \in \map {C^\infty} \R : \paren {\dfrac \d {\d x} - k} F = f$
By Differentiable Function as Distribution and multiplication of Schwartz distribution by a smooth function:
:$\exists F \in \map {C^\infty} \R : \paren {\dfrac \d {\d x} - k} T_F = T_f \quad (2)$
Subtract $(2)$ from $(1)$:
{{begin-eqn}}
{{eqn | l = \paren {\dfrac \d {\d x} - k} \paren {T - T_F}
      | r = T_f - T_f 
}}
{{eqn | r = \mathbf 0
}}
{{end-eqn}}
where $\mathbf 0$ is the zero distribution.
This is solved by:
:$T - T_F = T_{c \map \exp {k x} }$
or:
:$T = T_{F + c \map \exp {k x} }$
where $c \in \C$.
Finally:
:$\ds \paren {\dfrac \d {\d x} - k} \paren {F + c \map \exp {k x} } = f$
{{qed}}
Theorem: Riemann Uniformization Theorem

Every connected $2$-dimensional manifold has a complete Riemannian metric with the constant Gaussian curvature.
{{Stub|More precise statement}}

Proof:
{{ProofWanted}}
{{Namedfor|Georg Friedrich Bernhard Riemann|cat = Riemann}}
Theorem: Cartan-Hadamard Theorem

Let $M$ be a complete connected $n$-dimensional Riemannian manifold.
Suppose all sectional curvatures of $M$ are less than or equal to zero.
Then the universal covering space of $M$ is diffeomorphic to $\R^n$.
{{explain|What is the universal covering space of $M$? Does such a space exist? The book uses this theorem as an example of generalization of Gauss-Bonet theorem. Not all details are explained in this or any other chapter of the book, so I cannot say more.}}

Proof:
{{ProofWanted}}
{{Namedfor|Élie Joseph Cartan|name2 = Jacques Salomon Hadamard|cat = Cartan|cat2 = Hadamard}}
Theorem: Bonnet-Myers Theorem

Let $M$ be a complete connected Riemannian manifold.
Suppose all the sectional curvatures of $M$ are bounded below by a positive constant.
Then $M$ is compact and has a finite fundamental group.

Proof:
{{ProofWanted}}
{{Namedfor|Pierre Ossian Bonnet|name2 = Sumner Byron Myers|cat = Bonnet|cat2 = Myers}}
Theorem: Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit/Lemma 2

For each $n \in \N$:
:$\norm{\map \phi {x_n} - y}_Q = \ds \lim_{m \mathop \to \infty} \norm{x_n - x_m}_R$

Proof:
Let $n \in \N$.
We have:
:$\tuple {x_n, x_n, x_n, \ldots} \in \map \phi {x_n}$
:$\sequence{x_m} \in y$
From Element of Group is in Unique Coset of Subgroup:
:$\tuple {x_n, x_n, x_n, \ldots}  + \NN = \map \phi {x_n}$
:$\sequence{x_m} + \NN = y$
Then:
{{begin-eqn}}
{{eqn | l = \sequence{x_n - x_m}_{m \in \N} + \NN
      | r = \paren{\tuple {x_n, x_n, x_n, \ldots} - \sequence{x_m}_{m \in \N} } + \NN
      | c = Difference Rule for Cauchy Sequences in Normed Division Ring
}}
{{eqn | r = \paren{\tuple {x_n, x_n, x_n, \ldots} + \NN} - \paren{\sequence{x_m}_{m \in \N} + \NN}
      | c = {{Defof|Quotient Ring}}
}}
{{eqn | r = \map \phi {x_n} - y
}}
{{end-eqn}}
From Element of Group is in its own Coset:
:$\sequence{x_n - x_m}_{m \in \N} \in \map \phi {x_n} - y$
By definition of induced norm:
:$\norm{\map \phi {x_n} - y}_Q = \ds \lim_{m \mathop \to \infty} \norm{x_n - x_m}_R$
{{qed}}
Category:Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit
13e29w0413qaw25scfftbca6mam96u6
Theorem: Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit/Necessary Condition

{{:Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit}}
If $\sequence{\map \phi {x_n}}$ converges to $y$ then $\sequence{x_n} \in y$

Proof:
Let $\sequence{\map \phi {x_n}}$ converge to $y$.
From Convergent Sequence in Normed Division Ring is Cauchy Sequence:
:$\sequence{\map \phi {x_n}}$ is a Cauchy Sequence
From Lemma 1:
:$\sequence{x_n}$ is a Cauchy Sequence
Let $y'$ be the left coset that contains $\sequence{x_n}$.
From sufficient condition:
:$\ds \lim_{n \mathop \to \infty} \norm{\map \phi {x_n} }_Q = y'$
From Convergent Sequence in Metric Space has Unique Limit:
:$y = y'$
Theorem: Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit/Sufficient Condition

{{:Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit}}
If $\sequence{x_n} \in y$ then $\sequence{\map \phi {x_n}}$ converges to $y$

Proof:
Let $\sequence{x_n} \in y$.
Then $\sequence{x_n}$ is a Cauchy Sequence by definition of $y$.
Let $\epsilon > 0$ be arbitrary.
By definition of a Cauchy sequence:
:$\exists N \in \N: \forall n, m \ge N : \norm{x_n - x_m}_R < \dfrac \epsilon 2$
Let $n \ge N$ be arbitrary.
From Difference Rule for Cauchy Sequences in Normed Division Ring, the sequence $\sequence{x_n - x_m}_{m \in \N}$ is a Cauchy sequence.
Hence:
:$\forall m \ge N : \norm{x_n - x_m}_R < \dfrac \epsilon 2$
From Norm Sequence of Cauchy Sequence has Limit:
:$\ds \lim_{m \mathop \to \infty} \norm{x_n - x_m}_R$ exists
From Inequality Rule for Real Sequences:
:$\ds \lim_{m \mathop \to \infty} \norm{x_n - x_m}_R \le \dfrac \epsilon 2 < \epsilon$
From Lemma 2:
:$\norm{\map \phi {x_n} - y}_Q < \epsilon$
By definition of convergent sequence:
:$\ds \lim_{n \mathop \to \infty} \norm{\map \phi {x_n} }_Q = y$
Theorem: Gram-Schmidt Orthogonalization/Corollary 2

Let $\struct {V, \innerprod \cdot \cdot}$ be an $n$-dimensional inner product space over $\R$ or $\C$.
Let $\tuple {v_1, \ldots, v_n}$ be any ordered basis for $V$.
Then there is an orthonormal ordered basis $\tuple {b_1, \ldots, b_n}$ satisfying the following conditions:
:$\forall k \in \set {1, \ldots, n} : \span \set {v_1, \ldots v_k} = \span \set {b_1, \ldots, b_k} $

Proof:
By the definition of basis, it follows that $\set{ v_1, \ldots, v_n }$ is a linearly independent subset of $V$.
From {{Corollary|Gram-Schmidt Orthogonalization|1}}, it follows that there exists an orthonormal subset $\set {b_1, \ldots, b_n}$ of $V$ such that:
:$\forall k \in \set {1, \ldots, n}: \span \set {v_1, \ldots v_k} = \span \set {b_1, \ldots, b_k} $
where $\span$ denotes linear span.
From Orthogonal Set is Linearly Independent Set, it follows that $\set {b_1, \ldots, b_n}$ is a linearly independent subset of $V$.
By Sufficient Conditions for Basis of Finite Dimensional Vector Space, it follows that $\tuple {b_1, \ldots, b_n}$ is an ordered basis of $V$.
{{qed}}
{{Namedfor|Jørgen Pedersen Gram|name2 = Erhard Schmidt|cat = Gram|cat2 = Schmidt}}
Theorem: Rational Sequence Converges in P-adic Numbers iff Sequence Represents Limit/P-adic Expansion

Let $\ds \sum_{n \mathop = m}^\infty d_n p^n$ be a $p$-adic expansion.
Then:
:$\ds \sum_{n \mathop = m}^\infty d_n p^n$ converges to $a$ {{iff}} $\ds \sum_{n \mathop = m}^\infty d_n p^n$ is a representative of $a$

Proof:
By definition of a $p$-adic expansion:
:$\ds \sum_{n \mathop = m}^\infty d_n p^n$ is a rational sequence.
The theorem follows from Rational Sequence Converges in P-adic Numbers iff Sequence Represents Limit.
{{qed}}
Category:Rational Sequence Converges in P-adic Numbers iff Sequence Represents Limit
rqerqmrf1d8slcacex4h2enaarbaez0
Theorem: Solution to Distributional Ordinary Differential Equation with Constant Coefficients

Let $D$ be an ordinary differential operator with constant complex coefficients:
:$\ds D = \sum_{k \mathop = 0}^n a_k \paren {\dfrac \d {\d x}}^k$
Let $f \in \map {C^\infty} \R$ be a smooth real function.
Let $T \in \map {\DD'} \R$ be a Schwartz distribution.
Let $T_f$ be a Schwartz distribution associated with $f$.
Suppose $T$ is a distributional solution to $D T = T_f$.
Then $T = T_F$ where $F \in \map {C^\infty} \R$ is a classical solution to $D F = f$.
{{explain|Does this really mean a general solution? At first glance it looks like a particular solution. The source is not concerned whether this is a general or particular solution. Rename it according to the standard here.}}

Proof:
Let $\map P \xi$ be a polynomial over complex numbers such that:
:$\ds \map P \xi = \sum_{k \mathop = 0}^n a_k \xi^k = a_n \prod_{k \mathop = 0}^n \paren {\xi - \lambda_k}$
where $a_n \ne 0$.
Then there exists a polynomial $\map Q \xi$ such that:
:$\map P \xi = \paren {\xi - \lambda_n} \map Q \lambda$
Let:
:$\ds D = \sum_{k \mathop = 0}^n a_k \paren {\dfrac \d {\d x}}^k$
Then:
:$D = \map P {\dfrac \d {\d x} }$
Furthermore:
:$D = \paren {\dfrac \d {\d x} - \lambda_n} D_1$
where:
:$D_1 := \map Q {\dfrac \d {\d x} }$
Now we will use the principle of mathematical induction to show that:
:$\paren {DT = T_f, f \in \map {C^\infty} \R} \implies \paren {T = T_F, F \in \map {C^\infty} \R}$
Theorem: Vector Space over Division Subring is Vector Space/Special Case

Let $\struct {R, +, \circ}$ be a ring with unity whose unity is $1_R$.
Let $S$ be a division subring of $R$, such that $1_R \in S$.
The vector space $\struct {R, +, \circ_S}_S$ over $\circ_S$ is a $S$-vector space.

Proof:
A vector space over a division ring $D$ is by definition a unitary module over $D$. 
$S$ is a division ring by assumption.
$\struct {R, +, \circ_S}_S$ is a unitary module by Subring Module is Module/Special Case.
{{qed}}
Theorem: Intersection of Submodules is Submodule

Let $R$ be a ring.
Let $\struct {G, +_G}$ be an abelian group.
Let $M = \struct {G, +, \circ}_R$ be an $R$-module.
Let $H$ and $K$ be submodules of $M$.
Then $H \cap K$ is also a submodule of $M$.

Proof:
This is a special case of the General Result with $S = \set {H, K}$.
The proof follows immediately from the proof of the General Result.
{{qed}}
Theorem: Intersection of Set of Submodules containing Subset is Smallest Submodule

Let $R$ be a ring.
Let $\struct {G, +_G}$ be an abelian group.
Let $M = \struct {G, +, \circ}_R$ be an $R$-module.
Let $S \subset M$ be a subset of $M$.
Let $T$ be the set of all submodules of $M$ which contain $S$ as a subset.
Then the intersection $\bigcap T$ is the smallest submodule of $M$ containing $S$.

Proof:
By hypothesis, we have:
:$\ds T = \bigcap \set { M' \subseteq M : S \subseteq M', \textrm {$M'$ is a submodule of $M$} }$
From Intersection of Submodules is Submodule:General Result, it follows that $\bigcap T$ is a submodule of $M$.
As $S \subseteq M'$ for all $M' \in T$, it follows that $S \subseteq \bigcap T$.
Let $M' \in T$ be a submodule of $M$ such that $S \subseteq M'$.
From Intersection is Subset:General Result, it follows that $M' \subseteq \bigcap T$.
Hence the result.
{{qed}}
Theorem: Fundamental Solutions to Distributional Homogeneous ODE with Constant Coefficients differ by Classical Solution

Let $E_*, E, T \in \map {\DD'} \R$ be Schwartz distributions.
Let $D$ be an ordinary differential operator with constant coefficients.
Let $f$ be a function differentiable by $D$.
Let $T_f \in \map {\DD'} \R$ be a Schwartz distribution associated with $f$.
Let $\delta$ be the Dirac delta distribution.
Let $E_*$ be the fundamental solution to $D E_* = \delta$
Then $E$ is a fundamental solution to $DE = \delta$ {{iff}} $E = E_* + T_F$ where:
:$DE_* = \delta$
:$DF = 0$

Proof:
=== Necessary Condition ===
Suppose both $E_*$ and $E$ are fundamental solutions:
:$DE = \delta$
:$DE_* = \delta$
Taking the difference yields:
:$D \paren {E - E_*} = \mathbf 0$
where $\mathbf 0 \in \map {\DD'} \R$ is the zero distribution.
By Solution to Distributional Ordinary Differential Equation with Constant Coefficients:
:$E - E_* = T_F$
where $DF = 0$.
Hence:
:$E = E_* + T_F$
{{qed|lemma}}
Theorem: Elements of Submodule form Subgroup

Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {G, +_G}$ be an abelian group.
Let $\struct {G, +_G, \circ_G}_R$ be an $R$-module.
Let $\struct {H, +_H, \circ_H}_R$ be a submodule of $\struct {G, +_G, \circ_G}_R$.
Then $\struct {H, +_H}$ is a subgroup of $\struct {G, +_G}$.

Proof:
By definition of submodule, $\struct {H, +_H}$ is an abelian group.
The result follows by definition of subgroup.
{{qed}}
Category:Submodules
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