avewright/proofwiki-math huggingface dataset:
Theorem: Self-Adjoint Densely-Defined Linear Operator has Empty Residual Spectrum

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\struct {\map D T, T}$ be a self-adjoint densely defined linear operator.
Then the residual spectrum $\map {\sigma_r} T$ is empty.

Proof:
Let $\struct {\map D {T^\ast}, T^\ast}$ be the adjoint of $\struct {\map D T, T}$. 
Since $\struct {\map D T, T}$ is self-adjoint, we have: 
:$\struct {\map D {T^\ast}, T^\ast} = \struct {\map D T, T}$
Suppose that $\map {\sigma_r} T$ is non-empty.
Then there exists $\lambda \in \map {\sigma_r} T$.
That is, there exists $\lambda \in \C$ such that $T - \lambda I$ is injective, but such that: 
:$\map {\paren {T - \lambda I} } {\map D T}$ is not everywhere dense in $\HH$.
Let:
:$R = \overline {\map {\paren {T - \lambda I} } {\map D T} }$
Then from Linear Subspace Dense iff Zero Orthocomplement, we have: 
:$R^\bot \ne \set 0$
where $R^\bot$ is the orthocomplement of $R$.
Then there exists some $y \in R^\bot$ such that $y \ne 0$. 
Then: 
:$\innerprod {\paren {T - \lambda I} x} y = 0$ for all $x \in \map D T$
from the definition of the orthocomplement.
From Inner Product is Sesquilinear, we have: 
:$\innerprod {T x} y = \lambda \innerprod x y$ for each $x \in \map D T$.
As shown in Riesz Representation Theorem (Hilbert Spaces):
:$x \mapsto \innerprod x y$ is a bounded linear functional.
So:
:$x \mapsto \innerprod {T x} y$ is a bounded linear functional.
So from the definition of the adjoint, we have: 
:$y \in \map D {T^\ast}$
We therefore have: 
{{begin-eqn}}
{{eqn	| l = 0
		| r = \innerprod {T x} y - \lambda \innerprod x y
}}
{{eqn	| r = \innerprod x {T^\ast y} - \innerprod x {\overline \lambda y}
		| c = Inner Product is Sesquilinear, {{Defof|Adjoint of Densely-Defined Linear Operator}}
}}
{{eqn	| r = \innerprod x {T y} - \innerprod x {\overline \lambda y}
		| c = {{Defof|Self-Adjoint Densely-Defined Linear Operator}}
}}
{{eqn	| r = \innerprod x {T y - \overline \lambda y}
		| c = Inner Product is Sesquilinear
}}
{{end-eqn}}
Since $\map D T$ is everywhere dense in $H$, there exists a sequence in $\map D T$ with: 
:$x_n \to T y - \overline \lambda y$ 
Then: 
:$\innerprod {x_n} {T y - \overline \lambda y}$ for each $n \in \N$.
So, taking $n \to \infty$ and using Inner Product is Continuous:
:$\innerprod {T y - \overline \lambda y} {T y - \overline \lambda y} = 0$
So from the definition of the inner product, we have: 
:$T y = \overline \lambda y$ 
So $\lambda$ is in the point spectrum $\map {\sigma_p} T$ of $T$. 
From Point Spectrum of Symmetric Densely-Defined Linear Operator is Real, we have: 
:$\overline \lambda \in \R$
so from Complex Number equals Conjugate iff Wholly Real:
:$\overline \lambda = \lambda$ 
So $\lambda \in \map {\sigma_p} T$.
But then: 
:$\paren {T - \lambda I} y = \map {\paren {T - \lambda I} } 0$
with $y \ne 0$.
So $T - \lambda I$ is not injective.
So we have a contradiction, and so $\map {\sigma_r} T = \O$.
{{qed}}
Theorem: Equivalence of Formulations of Axiom of Empty Set for Classes

In the context of class theory, the following formulations of the '''{{axiom-link|the Empty Set}}''' are equivalent:
=== Formulation 1 ===
{{:Axiom:Axiom of the Empty Set/Set Theory/Formulation 2}}
=== Formulation 2 ===
{{:Axiom:Axiom of the Empty Set (Class Theory)}}

Proof:
It is assumed throughout that the {{axiom-link|Extension|Classes}} and the {{axiom-link|Specification|Classes}} both hold.
Theorem: Successor Mapping is Progressing

Let $V$ be a basic universe.
Let $s: V \to V$ denote the '''successor mapping''' on $V$:
:$\forall x \in V: \map s x := x \cup \set x$
Then $s$ is a progressing mapping.

Proof:
Recall
By Set is Subset of Union:
:$x \subseteq x \cup \set x$
That is:
:$x \subseteq \map s x$
Thus $s$ is by definition a progressing mapping.
{{qed}}
Theorem: Closed Sets of Right Order Space on Real Numbers

Let $T = \struct {\R, \tau}$ be the right order space on $\R$.
Then $H \subseteq S$ is closed in $T$ {{iff}}:
:$H = \O$ or $\R$
or
:$H = \hointl {-\infty} a$ for some $a \in \R$.

Proof:
By definition of the right order space on $\R$, $U \subseteq S$ is open in $T$ {{iff}}:
:$U = \O$ or $\R$
or
:$U = \openint a \infty$ for some $a \in \R$.
Note that:
:$\R \setminus \O = \R$
:$\R \setminus \R = \O$
:$\R \setminus \openint a \infty = \hointl {-\infty} a$
The result follows from the definition of closed set.
{{qed}}
Category:Right Order Topologies
Category:Examples of Closed Sets
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Theorem: Right Order Topology on Real Numbers is Topology

Let $\tau$ be the '''right order topology on $\R$'''.
Then $\tau$ forms a topology on $\R$.
That is:
:$T = \struct {\R, \tau}$ is a topological space.

Proof:
Write $\O = \openint \infty \infty$ and $\R = \openint {-\infty} \infty$.
Then $\tau$ can be written as $\set {\openint j \infty: j \in \overline \R}$.
First we note that:
:$m \le n \implies \openint n \infty \subseteq \openint m \infty$
By definition we have that:
:$\O \in \tau$
Then each of the open set axioms is examined in turn:
Theorem: Image of Weakly Convergent Sequence under Compact Linear Transformation is Convergent

Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be normed vector spaces.
Let $T : X \to Y$ be a compact linear transformation.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a weakly convergent sequence with:
:$x_n \weakconv x$
Then: 
:$T x_n \to T x$
in the strong sense.

Proof:
Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively.
Let $f \in Y^\ast$.
Then $f \circ T \in X^\ast$.
Since $x_n \weakconv x$, we have: 
:$\map f {T x_n} \to \map f {T x}$ as $n \to \infty$. 
So we have: 
:$T x_n \weakconv T x$ as $n \to \infty$. 
{{AimForCont}} that $\sequence {T x_n}_{n \in \N}$ does not converge to $T x$. 
Then there exists $\epsilon > 0$ such that: 
:$\norm {T x_{n_k} - T x}_Y \ge \epsilon$
for all $k \in \N$, for a subsequence $\sequence {x_{n_k} }_{k \in \N}$.
From Weakly Convergent Sequence in Normed Vector Space is Bounded, $\sequence {x_n}_{n \in \N}$ and hence $\sequence {x_{n_k} }_{k \in \N}$ is bounded.
Since $T$ is compact, there exists a subsequence $\sequence {T x_{n_{k_j} } }_{j \in \N}$ such that: 
:$T x_{n_{k_j} } \to z \in Y$
From Convergent Sequence in Normed Vector Space is Weakly Convergent, we have: 
:$T x_{n_{k_j} } \weakconv z$
Since: 
:$T x_n \weakconv T x$ 
we have that $z = T x$ from Weak Limit in Normed Vector Space is Unique.
But then there exists $N \in \N$ such that: 
:$\norm {T x_{n_{k_j} } - T x}_Y < \epsilon$
for $j \ge N$. 
This contradicts that: 
:$\norm {T x_{n_k} - T x}_Y \ge \epsilon$
for each $k \in \N$. 
So, we have $T x_n \to T x$. 
{{qed}}
Category:Weak Convergence (Normed Vector Spaces)
Category:Compact Linear Transformations
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Theorem: Point Spectrum of Symmetric Densely-Defined Linear Operator is Real

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\struct {\map D T, T}$ be a symmetric densely-defined linear operator.
Let $\map {\sigma_p} T$ be the point spectrum of $\struct {\map D T, T}$.
Then: 
:$\map {\sigma_p} T \subseteq \R$

Proof:
If $\map {\sigma_p} T = \O$, the result is immediate.
Let $\lambda \in \map {\sigma_p} T$. 
Then, from Point Spectrum of Densely-Defined Linear Operator consists of its Eigenvalues, there exists $x \in \HH \setminus \set 0$ such that: 
:$T x = \lambda x$
Then, we have: 
:$\innerprod {T x} x = \innerprod {\lambda x} x = \lambda \innerprod x x$
and: 
{{begin-eqn}}
{{eqn	| l = \innerprod {T x} x
		| r = \innerprod x {T x}
		| c = {{Defof|Symmetric Densely-Defined Linear Operator}}
}}
{{eqn	| r = \innerprod x {\lambda x}
}}
{{eqn	| r = \overline \lambda \innerprod x x
		| c = Inner Product is Sesquilinear
}}
{{end-eqn}}
So, we have: 
:$\paren {\lambda - \overline \lambda} \innerprod x x = 0$
Since $x \ne 0$, we have $\innerprod x x \ne 0$, so:
:$\lambda = \overline \lambda$ 
From Complex Number equals Conjugate iff Wholly Real, we then have: 
:$\lambda \in \R$
So:
:$\map {\sigma_p} T \subseteq \R$
from the definition of set inclusion.
{{qed}}
Theorem: Banach Space is Reflexive iff Normed Dual is Reflexive

Let $\Bbb F \in \set {\R, \C}$. 
Let $X$ be a Banach space over $\Bbb F$.
Let $X^\ast$ be the normed dual space of $X$. 
Then: 
:$X$ is reflexive {{iff}} $X^\ast$ is reflexive.

Proof:
Let $X^{\ast \ast}$ be the second normed dual of $X$.
Let $X^{\ast \ast \ast}$ be the normed dual of $X^{\ast \ast}$.
Theorem: Partition of Spectrum of Densely-Defined Linear Operator

Let $\HH$ be a Hilbert space over $\C$. 
Let $\struct {\map D T, T}$ be a densely-defined linear operator.
Let $\map \sigma T$ be the spectrum of $T$. 
Then: 
:$\map \sigma T = \map {\sigma_p} T \cup \map {\sigma_s} T \cup \map {\sigma_r} T$
where:
:$\map {\sigma_p} T$ is the point spectrum of $T$
:$\map {\sigma_s} T$ is the continuous spectrum of $T$.
:$\map {\sigma_r} T$ is the residual spectrum of $T$

Proof:
Let $\lambda \in \map \sigma T$.
Then, from the definition of the resolvent set of $T$, at least one of the following is false: 
:$(1) \quad$ $T - \lambda I$ is injective
:$(2) \quad$ $\map {\paren {T - \lambda I} } {\map D T}$ is everywhere dense in $\HH$
:$(3) \quad$ $\paren {T - \lambda I}^{-1}$ is bounded.
Note that either $(1)$ is true and one of $(2)$ or $(3)$ is false, or $(1)$ is false.
So we have three cases:
:$(\text A) \quad$ $(1)$ is false
:$(\text B) \quad$ $(1)$ is true, $(2)$ is true, $(3)$ is false
:$(\text C) \quad$ $(1)$ is true, $(2)$ is false
Note that the last case includes both the case that $(3)$ is true and $(3)$ is false.
Clearly only one of $(\text A)$, $(\text B)$ or $(\text C)$ can be true.
$(\text A)$ is precisely the definition of the point spectrum, $(\text B)$ is precisely the definition of the continuous spectrum and $(\text C)$ is precisely the definition of the residual spectrum.
So we have the result.
{{qed}}
Theorem: Spectrum of Bounded Linear Operator equal to Spectrum as Densely-Defined Linear Operator

Let $\HH$ be a Hilbert space over $\C$. 
Let $T : \HH \to \HH$ be a bounded linear operator.
Let $\map {\sigma_1} T$ be the spectrum of $T$ as a bounded linear operator.
Let $\map {\sigma_2} T$ be the spectrum of $T$ as a densely-defined linear operator $\struct {\HH, T}$.
Then: 
:$\map {\sigma_1} T = \map {\sigma_2} T$

Proof:
From the definition of the spectrum of $T$ as a bounded linear operator, we have: 
:$\map {\sigma_1} T = \C \setminus \map {\rho_1} T$
where $\map {\rho_1} T$ is the resolvent set of $T$ as a bounded linear operator.
From the definition of the spectrum of $T$ as a densely-defined linear operator $\struct {\HH, T}$, we have: 
:$\map {\sigma_2} T = \C \setminus \map {\rho_2} T$
where $\map {\rho_2} T$ is the resolvent set of $T$ as a densely-defined linear operator $\struct {\HH, T}$.
From Resolvent Set of Bounded Linear Operator equal to Resolvent Set as Densely-Defined Linear Operator, we have: 
:$\map {\rho_1} T = \map {\rho_2} T$
So:
:$\map {\sigma_1} T = \map {\sigma_2} T$
{{qed}}
Category:Spectra (Bounded Linear Operators)
Category:Spectra (Densely-Defined Linear Operators)
aid4yvrghm33e73xqg9xdtyl7mugnk2
Theorem: Characterization of Integer Polynomial has Root in P-adic Integers

Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\map F X \in \Z \sqbrk X$ be a polynomial with integer coefficients.
Let $a \in \Z_p$.
Then:
:$\map F a  = 0$
{{iff}}
:there exists a sequence $\sequence{a_n}$ of integers: 
::$(1)\quad\ds\lim_{n \mathop \to \infty} {a_n} = a$
::$(2)\quad\map F {a_n} \equiv 0 \mod {p^{n+1}}$

Proof:
From Characterization of Polynomial has Root in P-adic Integers:
:$\map F a  = 0$
{{iff}}
:there exists a sequence $\sequence{a_n}$ of integers: 
::$(1)\quad\ds\lim_{n \mathop \to \infty} {a_n} = a$
::$(2)\quad\map F {a_n} \equiv 0 \mod {p^{n+1}\Z_p}$
By definition of a polynomial with integer coefficients:
:$\forall n \in \N_{>0} : \map F {a_n} \in \Z$
From Congruence Modulo Equivalence for Integers in P-adic Integers:
:$\map F a  = 0$
{{iff}}
:there exists a sequence $\sequence{a_n}$ of integers: 
::$(1)\quad\ds\lim_{n \mathop \to \infty} {a_n} = a$
::$(2)\quad\map F {a_n} \equiv 0 \mod {p^{n+1}}$
{{qed}}
Theorem: Characterization of Integer has Square Root in P-adic Integers

Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $a \in Z$ be an integer such that $p \nmid a$.
Then:
:$\exists x \in \Z_p : x^2 = a$
{{iff}}
:$a$ is a quadratic residue of $p$.

Proof:
Let $F \in \Z \sqbrk X$ be the polynomial:
:$\map F X = X^2 - a$
By definition of formal derivative of $F$ is:
:$\map {F'} X = 2X$
Theorem: Characterization of P-adic Unit has Square Root in P-adic Units

Let $\Z_p$ be the $p$-adic integers for some odd prime $p$.
Let $Z_p^\times$ be the set of $p$-adic units. 
Let $u \in Z_p^\times$ be a $p$-adic unit.
Let $u = c_0 + c_1 p + c_2 p^2 + \ldots$ be the $p$-adic expansion of $u$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\exists x \in \Z_p^\times : x^2 {{=}} u$}}
{{item|(2):|$c_0$ is a quadratic residue of $p$}}
{{item|(3):|$\exists y \in \Z_p : y^2 \equiv u \pmod {p \Z_p}$}}
{{end-itemize}}

Proof:
From Partial Sum Congruent to $p$-adic Integer Modulo Power of $p$:
:$u \equiv c_0 \pmod {p \Z_p}$
Theorem: Quotient Group of Quadratic Residues Modulo p of P-adic Units

Let $\Q_p$ be the $p$-adic numbers for some prime $p \ne 2$.
Let $\Q_p^\times$ denote the set of invertible elements of $\Q_p$.
Let $\paren{\Q_p^\times}^2 = \set{a^2 : a \in \Q_p^\times}$
Then the multiplicative quotient group $\Q_p^\times \mathop/ \paren{\Q_p^\times}^2$ has order $4$:
:$\exists c \in \Q_p^\times \setminus \paren{\Q_p^\times}^2 : \set{1, p, c, cp}$ is a transversal

Proof:
By definition of field:
:$\Q_p^\times = \Q_p \setminus \set{0}$ is an abelian group
From Group of Units is Group:
:$\struct{\Q_p^\times, \times}$ is a subgroup of $\struct{\Q_p^*, \times}$
From Power of Elements is Subgroup:
:$\struct{\paren{\Q_p^\times}^2, \times}$ is a subgroup of $\struct{\Q_p^\times, \times}$
By definition of quotient group, the quotient group $\Q_p^\times \mathop/ \paren{\Q_p^\times}^2$ can be formed.
{{ProofWanted}}
Theorem: Point Spectrum of Densely-Defined Linear Operator consists of its Eigenvalues

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\struct {\map D T, T}$ be a densely-defined linear operator.
Let $\map {\sigma_p} T$ be the point spectrum of $T$.
Then $\lambda \in \map {\sigma_p} T$ {{iff}} $\lambda$ is an eigenvalue of $T$.

Proof:
We have that $\lambda \in \map {\sigma_p} T$ {{iff}}:
:$T - \lambda I$ is not injective.
That is, {{iff}} there exists $x \in \map D T \setminus \set 0$ such that: 
:$\paren {T - \lambda I} x = \map {\paren {T - \lambda I} } 0 = 0$
So $\lambda \in \map {\sigma_p} T$ {{iff}} there exists $x \in \map D T \setminus \set 0$ such that: 
:$T x = \lambda I x = \lambda x$
That is, $\lambda \in \map {\sigma_p} T$ {{iff}} $\lambda$ is an eigenvalue of $T$.
{{qed}}
Category:Point Spectra (Densely-Defined Linear Operators)
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Theorem: Eigenvalue of Densely-Defined Linear Operator is Approximate Eigenvalue

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\struct {\map D T, T}$ be a densely-defined linear operator.
Let $\lambda$ be an eigenvalue of $T$.
Then $\lambda$ is an approximate eigenvalue of $T$.

Proof:
Since $\lambda$ is an eigenvalue of $T$, there exists $x \in \map D T \setminus \set 0$ such that: 
:$\paren {T - \lambda I} x = 0$
Then setting:
:$\ds x_n = \frac x {\norm x}$
we have: 
:$\paren {T - \lambda I} x_n = 0$
for each $n \in \N$, while $\norm {x_n} = 1$.
Then, we have: 
:$\paren {T - \lambda I} x_n \to 0$
So $\lambda$ is an approximate eigenvalue of $T$.
{{qed}}
Category:Approximate Eigenvalues (Densely-Defined Linear Operators)
pmp6taxsix5yes6djvmxnjaew0gr0az
Theorem: Element of Spectrum of Densely-Defined Linear Operator not in Residual Spectrum is Approximate Eigenvalue

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\struct {\map D T, T}$ be a self-adjoint densely-defined linear operator.
Let $\map \sigma T$ be the spectrum of $\struct {\map D T, T}$.
Let $\map {\sigma_r} T$ be the residual spectrum of $\struct {\map D T, T}$.
Let $\lambda \in \map \sigma T \setminus \map {\sigma_r} T$. 
Then $\lambda$ is an approximate eigenvalue of $\struct {\map D T, T}$.

Proof:
Let $\lambda \in \map \sigma T$.
From Partition of Spectrum of Densely-Defined Linear Operator, $\lambda$ is contained in either the point spectrum or continuous spectrum of $T$.
If $\lambda$ is contained in the point spectrum, we have the result from Eigenvalue of Densely-Defined Linear Operator is Approximate Eigenvalue.
Otherwise, $\lambda$ is contained in the continuous spectrum, in which case $\paren {T - \lambda I}^{-1}$ is not bounded.
That is, there exists a sequence $\sequence {y_n}_{n \mathop \in \N}$ in $\map D {\paren {T - \lambda I}^{-1} }$ with $\norm {y_n} = 1$ and: 
:$\norm {\paren {T - \lambda I}^{-1} y_n} \to \infty$
Now set: 
:$\ds x_n = \paren {T - \lambda I}^{-1} y_n$
for each $n \in \N$ and then: 
:$\ds z_n = \frac {x_n} {\norm {x_n} }$
We then have:
{{begin-eqn}}
{{eqn	| l = \paren {T - \lambda I} z_n
		| r = \frac {\paren {T - \lambda I} x_n} {\norm {x_n} }
}}
{{eqn	| r = \frac {y_n} {\norm {x_n} }
}}
{{end-eqn}}
Then: 
{{begin-eqn}}
{{eqn	| l = \norm {\paren {T - \lambda I} z_n}
		| r = \frac {\norm {y_n} } {\norm {x_n} }
}}
{{eqn	| r = \frac 1 {\norm {\paren {T - \lambda I}^{-1} y_n} }
}}
{{eqn	| o = \to
		| r = 0
		| c = from Reciprocal of Null Sequence: Corollary, since $\norm {\paren {T - \lambda I}^{-1} y_n} \to \infty$
}}
{{end-eqn}}
so we have: 
:$\paren {T - \lambda I} z_n \to 0$ 
So $\lambda$ is an approximate eigenvalue of $T$.
{{qed}}
Theorem: Separability of Normed Vector Space preserved under Isometric Isomorphism

Let $\struct {X, \norm \cdot_X}$ be a separable normed vector space.
Let $\struct {Y, \norm \cdot_Y}$ be a normed vector space that is isometrically isomorphic to $\struct {X, \norm \cdot_X}$.
Let $T : X \to Y$ be an isometric isomorphism between $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$.
Then $\struct {Y, \norm \cdot_Y}$ is separable.

Proof:
Let $\mathcal S = \set {x_n : n \in \N}$ be a countable everywhere dense subset of $X$. 
We show that $\map T {\mathcal S} = \set {T x_n : n \in \N}$ is a countable everywhere dense subset of $Y$.
Let $y \in Y$ and $\epsilon > 0$. 
Since $T$ is a bijection, there exists $x \in X$ such that $y = T x$.
Since $\mathcal S$ is everywhere dense in $X$, there exists $j \in \N$ such that: 
:$\norm {x - x_j}_X < \epsilon$
Since $T$ is a linear isometry, we have: 
:$\norm {\map T {x - x_j} }_Y = \norm {x - x_j}_X$
and so: 
:$\norm {\map T {x - x_j} }_Y < \epsilon$
That is: 
:$\norm {T x - T x_j}_Y < \epsilon$
so:
:$\norm {y - T x_j}_Y < \epsilon$
Since $\epsilon > 0$ and $y \in Y$ were arbitrary, and $T x_j \in \map T {\mathcal S}$, we have that:
:$\map T {\mathcal S}$ is everywhere dense in $Y$.
From Image of Countable Set under Mapping is Countable, we have: 
:$\map T {\mathcal S}$ is countable.
So:
:$\struct {Y, \norm \cdot_Y}$ is separable.
{{qed}}
Category:Isometric Isomorphisms (Normed Vector Spaces)
Category:Separable Spaces
3ua9jkdq84v408uefjg1w0ce0ysutfo
Theorem: Integrated Linear Differential Mapping is Continuous

Let $C^1 \closedint a b := \map {C^1} {\closedint a b, \R}$ be the space of real functions of differentiability class $C^1$.
Let $S$ be the set of differentiable functions on closed real interval vanishing at their endpoints:
:$S := \set {\mathbf h \in C^1 \closedint a b : \map {\mathbf h} a = \map {\mathbf h} b = 0}$
Let $S \subseteq C^1 \closedint a b$ be equiped with the $C^1$ norm. 
Let $\mathbf A, \mathbf B \in C \closedint a b$ be continuous real functions.
Let $L : S \to \R$ be the integrated linear differential mapping:
:$\ds \map L {\mathbf h} = \int_a^b \paren {\map {\mathbf A} t \map {\mathbf h} t + \map {\mathbf B} t \map {\mathbf h'} t} \rd t$
where $\mathbf h \in S$.
Then $L$ is continuous.

Proof:
We have that the Integrated Linear Differential Mapping is Linear. 
For $\mathbf h \in S$ we have:
{{begin-eqn}}
{{eqn | l = \size {\map L {\mathbf h} }
      | r = \size {\int_a^b \paren {\map {\mathbf A} t \map {\mathbf h} t + \map {\mathbf B} t \map {\mathbf h'} t }\rd t}
}}
{{eqn | o = \le
      | r = \int_a^b \size {\map {\mathbf A} t \map {\mathbf h} t + \map {\mathbf B} t \map {\mathbf h'} t } \rd t
}}
{{eqn | o = \le
      | r = \int_a^b \paren {\size {\map {\mathbf A} t} \size {\map {\mathbf h} t} + \size {\map {\mathbf B} t} \size {\map {\mathbf h'} t} } \rd t
}}
{{eqn | o = \le
      | r = \int_a^b \paren {\size {\map {\mathbf A} t} \norm {\mathbf h}_\infty + \size {\map {\mathbf B} t} \norm {\mathbf h'}_\infty } \rd t
      | c = {{Defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions}}
}}
{{eqn | o = \le
      | r = \int_a^b \paren {\size {\map {\mathbf A} t} \norm {\mathbf h}_{1, \infty} + \size {\map {\mathbf B} t} \norm {\mathbf h'}_\infty } \rd t
      | c = $\norm {\mathbf h}_{1, \infty} = \norm {\mathbf h}_\infty + \norm {\mathbf h'}_\infty \ge \norm {\mathbf h}_\infty$
}}
{{eqn | o = \le
      | r = \int_a^b \paren {\size {\map {\mathbf A} t} \norm {\mathbf h}_{1, \infty} + \size {\map {\mathbf B} t} \norm {\mathbf h'}_{1, \infty} } \rd t
      | c = $\norm {\mathbf h}_{1, \infty} = \norm {\mathbf h}_\infty + \norm {\mathbf h'}_\infty \ge \norm {\mathbf h'}_\infty$
}}
{{eqn | o = \le
      | r = \paren {\int_a^b \paren {\size {\map {\mathbf A} t} + \size {\map {\mathbf B} t} } \rd t} \norm {\mathbf h}_{1, \infty}
}}
{{eqn | r = M \norm {\mathbf h}_{1, \infty}
}}
{{end-eqn}}
where: 
:$\ds M : = \int_a^b \paren {\size {\map {\mathbf A} t} + \size {\map {\mathbf B} t} } \rd t$
By Continuity of Linear Transformation between Normed Vector Spaces $L$ is continuous.
{{qed}}
Theorem: Partial Sum Congruent to P-adic Integer Modulo Power of p

Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $a \in \Z_p$.
Let $\ds a = \sum_{j=0}^\infty d_j p^j$ be the $p$-adic expansion of $a$
For all $n \in \N$, let $\ds a_n = \sum_{j=0}^n d_j p^j$ be the $n$-th partial sum of the $p$-adic expansion of $a$
Then:
:$\forall n \in \N : a_n \equiv a \pmod{p^{n+1}\Z_p}$

Proof:
We have:
{{begin-eqn}}
{{eqn | q = \forall n \in \N
      | l = a - a_n
      | r = \sum_{j = 0}^\infty d_j p^j - \sum_{j = 0}^n d_j p^j
      | c = {{hypothesis}}
}}
{{eqn | r = \sum_{j = n+1}^\infty d_j p^j
      | c = Removing first $n$ terms from the series
}}
{{eqn | r = p^{n+1} \sum_{j = 0}^\infty d_{j+n+1} p^j
      | c = Extract common factor from the series
}}
{{eqn | o = \in
      | r = p^{k+1} \Z_p
      | c = {{Defof|P-adic Integer|$p$-adic Integer}}
}}
{{eqn | ll= \leadsto
      | q = \forall k \in \N
      | l = a
      | o = \equiv
      | r = a_k
      | rr= \pmod {p^{k+1} \Z_p}
      | c = {{Defof|Congruence Modulo Ideal}}
}}
{{end-eqn}}
{{qed}}
Category:P-adic Integers
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