nyuuzyou/wikis huggingface dataset:
Space of Almost-Zero Sequences is not Closed in 2-Sequence Space

== Theorem ==
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the normed 2-sequence vector space.
Let $\struct {c_{00}, \norm {\, \cdot \,}_2}$ be the normed vector space of almost-zero sequences.
Then $\struct {c_{00}, \norm {\, \cdot \,}_2}$ is not closed in $\struct {\ell^2, \norm {\, \cdot \,}_2}$.
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $c_{00}$ such that:
:$\ds x_n := \tuple {1, \frac 1 2, \ldots \frac 1 n, 0, \ldots}$
Let $\ds x := \tuple {1, \frac 1 2, \ldots, \frac 1 n, \ldots}$ with $n \in \N_{>0}$.
We have that $x \in \ell^2 \setminus c_{00}$ where $\setminus$ denotes set difference.
Then:
Pass the limit $n \to \infty$
Then:
:$\ds \lim_{n \mathop \to \infty} \norm {x_n - x} = 0$
Hence, $\struct {c_{00}, \norm {\, \cdot \,}}$ does not contain its limit points.
By definition, it is not closed.
== Sources ==
* : Chapter $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Closed Sets
Space of Bounded Linear Operators is Unital Banach Algebra

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space over $\C$.
Let $\map B X$ be the space of bounded linear operators.
Let $\norm {\, \cdot \,}_{\map B X}$ be the norm on $\map B X$.
Define multiplication $\circ : \map B X \times \map B X \to \map B X$ by:
:$\map {\paren {A \circ B} } x = \map A {B x}$
for each $x \in X$. 
Then $\struct {\map B X, \norm {\, \cdot \,}_{\map B X} }$ is a unital Banach algebra.
== Proof ==
From Space of Bounded Linear Transformations is Banach Space, $\struct {\map B X, \norm {\, \cdot \,}_{\map B X} }$ is a Banach space. 
From Norm on Bounded Linear Transformation is Submultiplicative, we assure ourselves that we have $A \circ B \in \map B X$ for each $A, B \in \map B X$. 
Further Norm on Bounded Linear Transformation is Submultiplicative gives that:
:$\norm {A \circ B}_{\map B X} \le \norm A_{\map B X} \norm B_{\map B X}$
for each $A, B \in \map B X$.
Define $I_{\map B X} : X \to X$ by:
:$I x = x$ for each $x \in X$.
We then have, for each $x \in X$ and $A \in \map B X$: 
Hence:
:$I_{\map B X} \circ A = A \circ I_{\map B X} = A$ 
Further:
Hence all that remains is to show that $\circ$ is bilinear.
Let $A, B, C \in \map B X$ and $\lambda \in \C$. 
We have, for each $x \in X$:
and:
We conclude that:
:$\struct {\map B X, \norm {\, \cdot \,}_{\map B X} }$ is a unital Banach algebra.
Category:Banach Algebras
Space of Bounded Linear Operators on Hilbert Space is Unital C*-Algebra

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B \HH$ be the space of bounded linear operators.
Let $\norm {\, \cdot \,}_{\map B \HH}$ be the operator norm on $\map B \HH$.
Let $\ast : \map B \HH \to \map B \HH$ denote the adjoint operation.
Then $\struct {\map B \HH, \ast, \norm {\, \cdot \,}_{\map B \HH} }$ is a unital $\text C^\ast$-algebra.
== Proof ==
From Space of Bounded Linear Transformations is Unital Banach Algebra, $\struct {\map B \HH, \norm {\, \cdot \,} }$ is a unital Banach algebra.
=== Proof of $(\text C^\ast 1)$ ===
From Adjoint is Involutive, we have:
:$A^{\ast \ast} = A$
for each $A \in \map B \HH$.
=== Proof of $(\text C^\ast 2)$ ===
From Adjoining is Linear, we have:
:$\paren {A + B}^\ast = A^\ast + B^\ast$
for $A, B \in \map B \HH$.
=== Proof of $(\text C^\ast 3)$ ===
From Adjoint of Composition of Linear Transformations is Composition of Adjoints, we have:
:$\paren {A B}^\ast = B^\ast A^\ast$
for each $A, B \in \map BB \HH$.
=== Proof of $(\text C^\ast 4)$ ===
From Adjoining is Linear, we have:
:$\paren {\lambda A}^\ast = \overline \lambda A^\ast$
for $A \in \map B \HH$ and $\lambda \in \C$.
=== Proof of $(\text C^\ast 5)$ ===
From Norm of Adjoint, we have:
:$\norm A^2_{\map B \HH} = \norm {A^\ast A}_{\map B \HH}$
for each $A \in \map B \HH$.
We have verified all 5 properties of a $\text C^\ast$-algebra. 
Since we have already noted that $\struct {\map B \HH, \norm {\, \cdot \,} }$ is a unital Banach algebra, we are done.
Category:C*-Algebras
Space of Bounded Linear Transformations forms Vector Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$. 
Let $\map B {X, Y}$ be the space of bounded linear transformations from $X$ to $Y$.
Let $+_B$ and $\circ_B$ be pointwise addition and pointwise scalar multiplication on $Y^X$. 
Then $\struct {\map B {X, Y}, +_B, \circ_B}_\GF$ is a vector space. 
== Proof ==
Let $\map L {X, Y}$ be the space of linear transformations between $X$ and $Y$.
From Linear Mappings between Vector Spaces form Vector Space, $\map L {X, Y}$ is a vector space over $\GF$ with pointwise addition and pointwise scalar multiplication.
It therefore suffices to show that $\map B {X, Y}$ is a vector subspace of $\map L {X, Y}$. 
From One-Step Vector Subspace Test, it suffices to show that $\map B {X, Y} \ne \O$ and: 
:$T + \lambda S \in \map B {X, Y}$
for each $\lambda \in \GF$ and $T, S \in \map B {X, Y}$.
First, we have: 
:$\norm {\mathbf 0_{\map L {X, Y} } x}_Y = 0 \le \norm x_X$
So $\mathbf 0_{\map L {X, Y} } \in \map B {X, Y}$.
Now let $T, S \in \map B {X, Y}$.
Then there exists $M, M' > 0$ such that: 
:$\norm {T x}_Y \le M \norm x_X$ for each $x \in X$
and:
:$\norm {S x}_Y \le M' \norm x_X$ for each $x \in X$.
Then for each $\lambda \in \GF$ and $x \in X$ we have: 
So $T + \lambda S \in \map B {X, Y}$.
Category:Space of Bounded Linear Transformations
Space of Bounded Sequences with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space

== Theorem ==
Let $\map {\ell^\infty} \C$ be the space of bounded sequences on $\C$.
Let $\struct {\C, +_\C, \times_\C}$ be the field of complex numbers.
Let $\paren +$ be the pointwise addition on the ring of sequences.
Let $\paren {\, \cdot \,}$ be the pointwise multiplication on the ring of sequences.
Then $\struct {\map {\ell^\infty} \C, +, \, \cdot \,}_\C$ is a vector space.
== Proof ==
Let $\sequence {a_n}_{n \mathop \in \N}, \sequence {b_n}_{n \mathop \in \N}, \sequence {c_n}_{n \mathop \in \N} \in \map {\ell^\infty} \C$.
Let $\lambda, \mu \in \C$.
Let $\sequence 0 := \tuple {0, 0, 0, \dots}$ be a complex-valued function.
Let us use real number addition and multiplication.
Define pointwise addition as:
:$\sequence {a_n + b_n}_{n \mathop \in \N} := \sequence {a_n}_{n \mathop \in \N} +_\C \sequence {b_n}_{n \mathop \in \N}$.
Define pointwise scalar multiplication as:
:$\sequence {\lambda \cdot a_n}_{n \mathop \in \N} := \lambda \times_\C \sequence {a_n}_{n \mathop \in \N}$
Let the additive inverse be $\sequence {-a_n} := - \sequence {a_n}$.
=== Closure Axiom ===
By assumption, $\sequence {a_n}_{n \mathop \in \N}, \sequence {b_n}_{n \mathop \in \N} \in \map {\ell^\infty} \C$.
By definition:
:$\ds \sup_{n \mathop \in \N} \size {a_n} < \infty$
:$\ds \sup_{n \mathop \in \N} \size {b_n} < \infty$
Consider the sequence $\sequence {a_n + b_n}$.
Then: 
Hence:
:$\sequence {a_n + b_n} \in \map {\ell^\infty} \C$
=== Commutativity Axiom ===
By Pointwise Addition on Ring of Sequences is Commutative, $\sequence {a_n} + \sequence {b_n} = \sequence {b_n} + \sequence {a_n}$
=== Associativity Axiom ===
By Pointwise Addition on Ring of Sequences is Associative, $\paren {\sequence {a_n} + \sequence {b_n} } + \sequence {c_n} = \sequence {a_n} + \paren {\sequence {b_n} + \sequence {c_n} }$.
=== Identity Axiom ===
=== Inverse Axiom ===
=== Distributivity over Scalar Addition ===
=== Distributivity over Vector Addition ===
=== Associativity with Scalar Multiplication ===
=== Identity for Scalar Multiplication ===
== Sources ==
* : $\S 1.1$: Vector Space
Category:Examples of Vector Spaces
Category:Functional Analysis
Category:Space of Bounded Sequences
Space of Bounded Sequences with Supremum Norm forms Normed Vector Space

== Theorem ==
The vector space of bounded sequences with the supremum norm forms a normed vector space.
== Proof ==
We have that:
:Space of Bounded Sequences with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space
:Supremum norm on the space of bounded sequences is a norm
By definition, $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ is a normed vector space.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Examples of Normed Vector Spaces
Space of Compact Linear Transformations is Banach Space

== Theorem ==
Let $H, K$ be Hilbert spaces.
Let $\map {B_0} {H, K}$ be the space of compact linear transformations from $H$ to $K$.
Let $\Bbb F \in \set {\R, \C}$ be the ground field of $K$.
Now $\map {B_0} {H, K} \subseteq K^H$, the set of mappings from $H$ to $K$.
Therefore, $\map {B_0} {H, K}$ can be endowed with pointwise addition ($+$) and ($\Bbb F$)-scalar multiplication ($\circ$).
Let $\norm {\, \cdot \,}$ denote the norm on bounded linear operators.
Then $\norm {\, \cdot \,}$ is a norm on $\map {B_0} {H, K}$.
Furthermore, $\map {B_0} {H, K}$ is a Banach space with respect to this norm.
== Proof ==
== Sources ==
* $\text {II.4.2 (b)}$
Category:Banach Spaces
Category:Linear Transformations on Hilbert Spaces
Category:Compact Linear Transformations
Space of Continuous Functions Vanishing at Infinity is C*-Algebra

== Theorem ==
Let $X$ be a locally compact Hausdorff space. 
Let $\map {\CC_0} X$ be the set of continuous complex-valued functions vanishing at infinity.
Equip $\map {\CC_0} X$ with pointwise addition, pointwise scalar multiplication and pointwise multiplication.
For each $f \in \map {\CC_0} X$, define:
:$\map {\overline f} x = \overline {\map f x}$
for each $x \in X$.
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $\map {\CC_0} X$.
Then $\tuple {\map {\CC_0} X, \overline {\, \cdot \,}, \norm {\, \cdot \,}_\infty}$ is a $\text C^\ast$-algebra.
== Proof ==
From Bounded Complex-Valued Continuous Functions on Topological Space form Unital C*-Algebra, $\tuple {\map {\CC_b} X, \overline {\, \cdot \,}, \norm {\, \cdot \,}_\infty}$ is a $\text C^\ast$-algebra where $\map {\CC_b} X$ is the space of bounded continuous functions on $X$ valued in $\C$.
From Closed *-Subalgebra of C*-Algebra is C*-Algebra, it suffices to show that $\map {\CC_0} X$ is a closed $\ast$-subalgebra of $\map {\CC_b} X$.
In Space of Continuous Functions on Locally Compact Hausdorff Space Vanishing at Infinity is Banach Space, $\map {\CC_0} X$ is a Banach space. 
Hence from Subspace of Complete Metric Space is Closed iff Complete, $\map {\CC_0} X$ is closed in $\map {\CC_b} X$. 
From Product of Complex-Valued Functions Vanishing at Infinity Vanishes at Infinity, we have $f g \in \map {\CC_0} X$ for each $f, g \in \map {\CC_0} X$. 
Hence $\map {\CC_0} X$ is a subalgebra of $\map {\CC_b} X$. 
Finally, from Complex Conjugate of Complex-Valued Function Vanishing at Infinity Vanishes at Infinity we have $\overline f \in \map {\CC_b} X$ whenever $f \in \map {\CC_B} X$.
Hence $\map {\CC_0} X$ is a $\ast$-subalgebra of $\map {\CC_b} X$.
Hence applying Closed *-Subalgebra of C*-Algebra is C*-Algebra, 
From Linear Combination of Complex-Valued Functions Vanishing at Infinity Vanishes at Infinity, $\tuple {\map {\CC_0} X, \overline {\, \cdot \,}, \norm {\, \cdot \,}_\infty}$ is a $\text C^\ast$-algebra.
Category:C*-Algebras
Category:Complex-Valued Functions Vanishing at Infinity
Space of Continuous Functions on Locally Compact Hausdorff Space Vanishing at Infinity is Banach Space

== Theorem ==
Let $X$ be a locally compact Hausdorff space.
Let $\struct {\map {\CC_0} X, \norm {\, \cdot \,}_\infty}$ be the space of continuous functions on $X$ vanishing at infinity.
Then $\struct {\map {\CC_0} X, \norm {\, \cdot \,}_\infty}$ is a Banach space over $\C$.
== Proof ==
We first show that $\map {\CC_0} X$ is a vector space. 
From Bounded Continuous Functions on Topological Space form Banach Space, it is enough to show that $\map {\CC_0} X$ is a linear subspace of $\map {\CC_b} X$, where $\map {\CC_b} X$ is the space of bounded continuous functions on $X$ valued in $\C$. 
From Linear Combination of Complex-Valued Functions Vanishing at Infinity Vanishes at Infinity, we have:
:$f + \lambda g \in \map {\CC_0} X$
for $f, g \in \map {\CC_0} X$ and $\lambda \in \C$.
Hence from One-Step Vector Subspace Test, $\map {\CC_0} X$ is a linear subspace of $\map {\CC_b} X$.
Hence $\map {\CC_0} X$ is a vector space. 
Now let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence in $\map {\CC_0} X$ converging to $f \in \map {\CC_b} X$.
From Uniform Limit of Sequence of Continuous Complex-Valued Functions Vanishing at Infinity Vanishes at Infinity, we have $f \in \map {\CC_0} X$. 
From Closed Subspace of Banach Space forms Banach Space, $\struct {\map {\CC_0} X, \norm {\, \cdot \,}_\infty}$ is a Banach space over $\C$.
Category:Complex-Valued Functions Vanishing at Infinity
Category:Banach Spaces
Space of Continuous Functions on Locally Compact Hausdorff Space Vanishing in Neighborhood of Infinity is Normed Vector Space

== Theorem ==
Let $X$ be a compact Hausdorff space.
Let $\struct {\map {\CC_{00} } X, \norm {\, \cdot \,}_\infty}$ be the space of continuous functions on $X$ vanishing in a neighborhood of infinity.
Then $\struct {\map {\CC_{00} } X, \norm {\, \cdot \,}_\infty}$ is a normed vector space over $\C$.
== Proof ==
We first show that $\map {\CC_{00} } X$ is a vector space. 
From Continuous Functions on Compact Space form Banach Space, it is enough to show that $\map {\CC_{00} } X$ is a linear subspace of $\map \CC X$.
From Linear Combination of Complex-Valued Functions Vanishing in Neighborhood of Infinity Vanishes in Neighborhood of Infinity, we have:
:$f + \lambda g \in \map {\CC_{00} } X$
for $f, g \in \map {\CC_{00} } X$ and $\lambda \in \C$.
Hence from One-Step Vector Subspace Test, $\map {\CC_{00} } X$ is a linear subspace of $\map \CC X$.
Hence $\map {\CC_{00} } X$ is a vector space. 
From Supremum Norm is Norm, we finally have that $\struct {\map {\CC_{00} } X, \norm {\, \cdot \,}_\infty}$ is a normed vector space over $\C$.
Category:Complex-Valued Functions Vanishing in Neighborhood of Infinity
Space of Continuous on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space

== Theorem ==
Let $I := \closedint a b$ be a closed real interval.
Let $\map C I$ be the space of real-valued functions continuous on $I$.
Let $\struct {\R, +_\R, \times_\R}$ be the field of real numbers.
Let $\paren +$ be the pointwise addition of real-valued functions.
Let $\paren {\, \cdot \,}$ be the pointwise scalar multiplication of real-valued functions.
Then $\struct {\map C I, +, \, \cdot \,}_\R$ is a vector space.
== Proof ==
Let $f, g, h \in \map C I$ such that:
:$f, g, h : I \to \R$
Let $\lambda, \mu \in \R$.
Let $\map 0 x$ be a real-valued function such that:
:$\map 0 x : I \to 0$.
Let us use real number addition and multiplication.
$\forall x \in I$ define pointwise addition as:
:$\map {\paren {f + g}} x := \map f x +_\R \map g x$.
Define pointwise scalar multiplication as:
:$\map {\paren {\lambda \cdot f}} x := \lambda \times_\R \map f x$
Let $\map {\paren {-f} } x := -\map f x$.
=== Closure Axiom ===
By Sum Rule for Continuous Real Functions, $f + g \in \map C I$
=== Commutativity Axiom ===
By Pointwise Addition on Real-Valued Functions is Commutative, $f + g = g + f$
=== Associativity Axiom ===
By Pointwise Addition is Associative, $\paren {f + g} + h = f + \paren {g + h}$.
=== Identity Axiom ===
=== Inverse Axiom ===
=== Distributivity over Scalar Addition ===
=== Distributivity over Vector Addition ===
=== Associativity with Scalar Multiplication ===
=== Identity for Scalar Multiplication ===
== Sources ==
* : $\S 1.1$: Vector Space
Category:Examples of Vector Spaces
Category:Functional Analysis
Space of Continuous on Closed Interval Real-Valued Functions with Supremum Norm forms Banach Space

== Theorem ==
Let $I = \closedint a b$ be a closed real interval.
Let $\map C I$ be the space of real-valued functions, continuous on $I$.
Let $\norm {\,\cdot\,}_\infty$ be the supremum norm on real-valued functions, continuous on $I$.
Then $\struct {\map C I, \norm {\,\cdot\,}_\infty}$ is a Banach space.
== Proof ==
A Banach space is a normed vector space, where a Cauchy sequence converges the supplied norm.
To prove the theorem, we need to show that a Cauchy sequence in $\struct {\map C I, \norm {\,\cdot\,}_\infty}$ converges.
We take a Cauchy sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\struct {\map C I, \norm {\,\cdot\,}_\infty}$.
Then we fix $t \in I$ and show, that a real Cauchy sequence $\sequence {\map {x_n} t}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$ with the limit $\map x t$.
Then we prove the continuity of $\map x t$.
Finally, we show that $\sequence {x_n}_{n \mathop \in \N}$ converges in $\struct {\map C I, \norm {\,\cdot\,}_\infty}$ with the limit $\map x t$.
=== $\sequence {\map {x_n} t}_{n \mathop \in \N}$ is a Cauchy Sequence ===
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\map C I$: 
:$\forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n, m > N : \norm {x_n − x_m}_\infty < \epsilon$
Fix $t \in I$.
Then:
Hence $\sequence {\map {x_n} t}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {\R, \size {\, \cdot \,}}$.
=== $\sequence {\map {x_n} t}_{n \mathop \in \N}$ Converges in $\struct {\R,\size {\, \cdot \,}}$ ===
From Real Number Line is Complete Metric Space, $\R$ is a complete metric space.
Therefore, $\sequence {\map {x_n} t}_{n \mathop \in \N}$ is convergent.
Denote the limit by $\map x t : I \to \R$:
:$\ds \lim_{n \mathop \to \infty} \map {x_n} t = \map x t$
=== $\map x t$ is Continuous ===
Choose $N$ such that: 
:$\forall n, m > N : \norm{x_n - x_m} < \dfrac \epsilon 3$
Let $\tau \in I$.
Then $\forall n > N$ and $m = N + 1 > N$:
Take the limit $n \to \infty$:
By assumption, $\map {x_{N + 1} } t \in \map C I$.
Then:
:$\forall t, \tau \in I : \exists \delta > 0: \size {\tau - t} < \delta \implies \size {\map {x_{N + 1} } t - \map {x_{N + 1} } \tau} < \dfrac \epsilon 3$
Thus:
Hence, $\map x t$ is continuous in $I$.
=== $\sequence {x_n}_{n \mathop \in \N}$ Converges to $x$===
Let $\epsilon > 0$.
Choose $N$ such that:
:$\forall n, m > N : \norm {x_n - x_m}_\infty < \epsilon$
Fix $n > N$.
Let $t \in I$.
Then, $\forall m > N$:
Take the limit $m \to \infty$:
This holds for every $n > N$.
Repeat the argument for all $\epsilon > 0$:
:$\forall \epsilon > 0 : \exists N \in \N : \forall n \in \N : \forall n > N : \norm {x_n - x}_\infty < \epsilon$.
Therefore, $x_n$ converges to $x$ in $\struct {\map C I, \norm {\,\cdot\,}_\infty}$:
:$\ds \lim_{x \mathop \to \infty} x_n = x$
== Sources ==
* : Chapter $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Functional Analysis
Category:Banach Spaces
Space of Continuous on Closed Interval Real-Valued Functions with Supremum Norm forms Normed Vector Space

== Theorem ==
Let $I := \closedint a b$ be a closed real interval.
The space of continuous real-valued functions on $I$ with supremum norm forms a normed vector space.
== Proof ==
We have that:
:Space of Continuous on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space
:Supremum Norm is Norm/Continuous on Closed Interval Real-Valued Function
By definition, $\struct {\map C I, \norm {\, \cdot \,}_\infty}$ is a normed vector space.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Examples of Normed Vector Spaces
Space of Continuously Differentiable on Closed Interval Real-Valued Functions with C^1 Norm forms Normed Vector Space

== Theorem ==
Space of Continuously Differentiable on Closed Interval Real-Valued Functions with $C^1$ norm forms a normed vector space.
== Proof ==
Let $I := \closedint a b$ be a closed real interval.
We have that:
:Space of Continuously Differentiable on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space
:$\map {C^1} I$ norm on the space of continuously differentiable on closed interval real-valued functions is a norm
By definition, $\struct {\map {C^1} I, \norm {\, \cdot \,}_{1, \infty} }$ is a normed vector space.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Examples of Normed Vector Spaces
Space of Continuously Differentiable on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space

== Theorem ==
Let $I := \closedint a b$ be a closed real interval.
Let $\map C I$ be a space of real-valued functions continuous on $I$.
Let $\map {C^1} I$ be a space of continuously differentiable functions on $I$.
Let $\struct {\R, +_\R, \times_\R}$ be the field of real numbers.
Let $\paren +$ be the pointwise addition of real-valued functions.
Let $\paren {\, \cdot \,}$ be the pointwise scalar multiplication of real-valued functions.
Then $\struct {\map {C^1} I, +, \, \cdot \,}_\R$ is a vector space.
== Proof ==
From Space of Continuous on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space:
:$\struct {\map C I, +, \, \cdot \,}_\R$ is a vector space.
By Differentiable Function is Continuous:
:$\map {C^1} I \subset \map C I$
Let $f, g \in \map {C^1} I$.
Let $\alpha \in \R$.
Let $\map 0 x$ be a real-valued function such that:
:$\map 0 x : I \to 0$
Restrict $\paren +$ to $\map {C^1} I \times \map {C^1} I$.
Restrict $\paren {\, \cdot \,}$ to $\R \times \map {C^1} I$.
=== Closure under vector addition ===
By Sum Rule for Derivatives:
:$f + g \in \map {C^1} I$
=== Closure under scalar multiplication ===
By Derivative of Constant Multiple:
:$\alpha \cdot f \in \map {C^1} I$
=== Nonemptiness ===
By Derivative of Constant, a constant mapping is differentiable.
Hence, $\map 0 x \in \map {C^1} I$.
We have that $\map {C^1} I$ is closed under restrictions of $\paren +$ and $\paren {\, \cdot \,}$ to $\map {C^1} I$.
Also, $\map {C^1} I$ is non-empty.
By definition, $\struct {\map {C^1} I, +, \, \cdot \,}_\R$ is a vector subspace of $\struct {\map C I, +, \, \cdot \,}_\R$.
Since $\struct {\map {C^1} I, +, \, \cdot \,}_\R$ satisfies vector space axioms under given restrictions, it is a vector space.
== Sources ==
* : $\S 1.1$: Vector Space
Category:Examples of Vector Spaces
Category:Functional Analysis
Space of Integrable Functions Under Pointwise Addition forms Abelian Group

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\LL^1} \mu$ be the space of real-valued $\mu$-integrable functions.
Then $\map {\LL^1} \mu$, endowed with pointwise addition, forms an abelian group over $\map {\LL^1} \mu$.
== Proof ==
Suppose $f, g \in \map {\LL^1} \mu$.
By Pointwise Sum of Integrable Functions is Integrable Function:
:$f + g \in \map {\LL^1} \mu$
So $\map {\LL^1} \mu$ is closed under pointwise addition.
Hence the result.
== Also see ==
* Space of Integrable Functions is Vector Space
Category:Integrable Functions (Measure Space)‎
Category:Examples of Abelian Groups
Space of Integrable Functions is Vector Space

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\LL^1} \mu$ be the space of real-valued $\mu$-integrable functions.
Then $\map {\LL^1} \mu$, endowed with pointwise $\R$-scalar multiplication and pointwise addition, forms a vector space over $\R$.
== Proof ==
== Sources ==
* : $\S 10$
Category:Integrable Functions (Measure Space)‎
Category:Examples of Vector Spaces
Space of Lipschitz Functions is Banach Space/Shift of Finite Type

== Theorem ==
Let $\struct {X _\mathbf A, \sigma_\mathbf A}$ be a shift of finite type.
Let $\theta \in \openint 0 1$.
Let $F_\theta$ be the Lipschitz space on $X _\mathbf A$.
Let $\norm \cdot_\theta$ be the Lipschitz norm on $F_\theta$.
Then $\struct {F_\theta, \norm \cdot_\theta}$ is a Banach space.
== Proof ==
== Sources ==
* : Chapter $1$: Subshifts of Finite Type and Function Spaces
Category:Ergodic Theory
Category:Functional Analysis