avewright/proofwiki-math huggingface dataset:
Theorem: Scalar Multiplication is Continuous in Weak Topology

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$ with weak topology $w$. 
Define $m : K \times \struct {X, w} \to \struct {X, w}$ by: 
:$\map m {\lambda, x} = \lambda x$
for each $\lambda \in K$, $x \in X$. 
Then $m$ is continuous. 
That is, scalar multiplication remains continuous when restricting to the weak topology.

Proof:
Let $X^\ast$ be the topological dual space of $X$.
From Continuity in Initial Topology, it suffices to show that for each $f \in X^\ast$ we have: 
:$f \circ m : K \times \struct {X, w} \to K$ is continuous.
Define the projections $\pr_1 : K \times {X, w} \to \struct {X, w}$ and $\pr_2 : K \times {X, w} \to \struct {X, w}$ as the projection onto the first and second factors.
Then for each $\lambda \in K$ and $x \in X$ we have: 
{{begin-eqn}}
{{eqn	| l = \map {\paren {f \circ m} } {\lambda, x}
		| r = \map f { \map m {\lambda, x} } 
		| c = {{Defof|Composition of Mappings}}
}}
{{eqn	| r = \map f {\lambda x}
		| c = Definition of $m$
}}
{{eqn	| r = \lambda \map f x
		| c = {{Defof|Linear Functional}}
}}
{{eqn	| r = \map {\pr_1} {\tuple {\lambda, x} } \map f {\map {\pr_2} {\tuple {\lambda, x} } }
}}
{{end-eqn}}
That is: 
:$f \circ m = \pr_1 \cdot \paren {f \circ \pr_2}$
From the definition of the product topology:
:$\pr_1 : K \times \struct {X, w} \to K$
and:
:$\pr_2 : K \times \struct {X, w} \to \struct {X, w}$ 
are continuous.
From the definition of the weak topology:
:$f : \struct {X, w} \to K$ is continuous.
From Composite of Continuous Mappings is Continuous:
:$f \circ \pr_2 : K \times \struct {X, w} \to K$ is continuous.
From Product of Continuous Functions on Topological Ring is Continuous:
:$\pr_1 \cdot \paren {f \circ \pr_2} : K \times \struct {X, w} \to K$ is continuous.
So we have that:
:$f \circ m : K \times \struct {X, w} \to K$ is continuous for each $f \in X^\ast$.
So from Continuity in Initial Topology:
:$m : K \times \struct {X, w} \to \struct {X, w}$ is continuous in the weak topology. 
{{qed}}
Category:Weak Topologies on Topological Vector Spaces
qa47q39y8mj71m7i33tz6kud8g5eb28
Theorem: Topological Vector Space over Topological Field remains Topological Vector Space with Weak Topology

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$ admitting a weak topology $w$. 
Then $\struct {X, w}$ is a topological vector space.

Proof:
Follows from combining: 
:Vector Addition is Continuous in Weak Topology
:Scalar Multiplication is Continuous in Weak Topology
{{qed}}
Category:Topological Vector Space over Topological Field remains Topological Vector Space with Weak Topology
Category:Weak Topologies on Topological Vector Spaces
624hwlo3fjhh236wdxruedsz3gcizhj
Theorem: Spectrum of Compact Linear Operator on Infinite-Dimensional Banach Space contains Zero

Let $X$ be an infinite-dimensional Banach space over $\C$.
Let $T : X \to X$ be a compact linear operator. 
Let $\map \sigma T$ be the spectrum of $T$.
Then $0 \in \map \sigma T$. 
That is, $T$ is not invertible as a bounded linear operator.

Proof:
Suppose $0 \not \in \map \sigma T$. 
Then $T$ is invertible as a bounded linear operator with bounded inverse $T^{-1}$ so that: 
:$T T^{-1} = I$
From Left Composition of Compact Linear Transformation with Bounded Linear Transformation is Compact, this implies that $I$ is compact.
From Identity Operator is Compact iff Finite-Dimensional Normed Vector Space, we have that $X$ is finite-dimensional.
This is a contradiction, so we have $0 \in \map \sigma T$. 
{{qed}}
Theorem: Identity Operator is Compact iff Finite-Dimensional Normed Vector Space

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space. 
Let $I : X \to X$ be the identity operator. 
Then $I$ is compact {{iff}} $X$ is finite-dimensional.

Proof:
Let $\operatorname {ball} X$ be the closed unit ball in $\struct {X, \norm {\, \cdot \,} }$.
From the definition of a compact operator, we have that $I$ is compact {{iff}}:
:$\overline {\operatorname {ball} X}$ is compact in $\struct {X, \norm \cdot_X}$.
From Closed Ball is Closed and Set is Closed iff Equals Topological Closure, this is equivalent to: 
:$\operatorname {ball} X$ is compact in $\struct {X, \norm {\, \cdot \,} }$.
From Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact, this is equivalent to:
:$X$ is finite-dimensional.
{{qed}}
Category:Finite Dimensional Vector Spaces
Category:Compact Linear Transformations
9oafp0ey08lz3vju45lzpylb7swe0dq
Theorem: Continuous Linear Operator over Infinite Dimensional Vector Space is not necessarily Invertible

Let $\struct {X, \norm {\, \cdot\,}_X}$ be a normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.
Let $I \in \map {CL} X$ be the identity element.
Let $S, T \in \map {CL} X$.
Suppose the dimension of $X$ is finite:
:$d = \dim X = \infty$
Suppose $T \circ S = I$ where $\circ$ denotes the composition of mappings.
Then $T$ and $S$ are not necessarily invertible.

Proof:
Let $X, Y$ be $2$-sequence spaces.
Let $T = L : X \to Y$ be the left shift operator.
Let $S = R : X \to Y$ be the right shift operator.
Let $x := \tuple {a_1, a_2, \ldots} \in X$.
We have that:
{{begin-eqn}}
{{eqn | l = L \circ R \circ x
      | r = \map L {\map R {\tuple{a_1, a_2, \ldots} } }
}}
{{eqn | r = \map L {\tuple {0, a_1, a_2, \dots} }
      | c = {{Defof|Right Shift Operator}}
}}
{{eqn | r = \tuple {a_1, a_2, \ldots}
      | c = {{Defof|Left Shift Operator}}
}}
{{eqn | r = I \circ x
      | c = {{Defof|Identity Element}}
}}
{{end-eqn}}
Hence:
:$L \circ R = I$
However:
{{begin-eqn}}
{{eqn | l = R \circ L \circ x
      | r = \map R {\map L {\tuple{a_1, a_2, \ldots} } }
}}
{{eqn | r = \map L {\tuple {a_2, a_3, \dots} }
      | c = {{Defof|Left Shift Operator}}
}}
{{eqn | r = \tuple {0, a_2, a_3, \ldots}
      | c = {{Defof|Right Shift Operator}}
}}
{{eqn | o = \ne
      |r = I \circ x
}}
{{end-eqn}}
Therefore, $T \circ S = I$ but $S \circ T \ne I$.
{{qed}}
Theorem: Invertible Continuous Linear Operator has Unique Inverse

Let $\struct {X, \norm {\, \cdot \,} }$ be the normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.
Let $I \in \map {CL} X$ be the identity element.
Suppose $A \in \map {CL} X$ is invertible.
Then there is a unique $B \in \map {CL} X$ such that $A \circ B = B \circ A = I$.

Proof:
Let $B_1, B_2 \in \map {CL} X$.
Suppose:
:$A \circ B_1 = I = B_1 \circ A$
:$A \circ B_2 = I = B_2 \circ A$
Then:
{{begin-eqn}}
{{eqn | l = B_1
      | r = I \circ B_1
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = B_2 \circ A \circ B_1
}}
{{eqn | r = B_2 \circ I
}}
{{eqn | r = B_2
      | c = {{Defof|Identity Element}}
}}
{{end-eqn}}
{{qed}}
Theorem: Positive Part of Darboux Integrable Function is Integrable

Let $f$ be a real function that is Darboux integrable over $\closedint a b$.
Let $f^+$ be the positive part of $f$.
Then $f^+$ is Darboux integrable over $\closedint a b$.

Proof:
Let $\epsilon > 0$ be a strictly positive real number.
By Condition for Darboux Integrability, there is a finite subdivision $P = \sequence {x_i}_{0 \mathop \le i \mathop \le n}$ such that:
:$\map {U_f} P - \map {L_f} P < \epsilon$
where $\map {U_f} P$ and $\map {L_f} P$ are the upper Darboux sum and lower Darboux sum, respectively.
Consider the terms of $\map {U_f} P$ and $\map {L_f} P$ for an arbitrary index $i$.
By Supremum does not Precede Infimum:
:$m_i^{\paren f} \le M_i^{\paren f}$
where $m_i^{\paren f}$ and $M_i^{\paren f}$ are respectively the infimum and supremum of $f$ on $\closedint {x_{i - 1} } {x_i}$.
Therefore, there are three cases to consider:
If $0 \le m_i^{\paren f} \le M_i^{\paren f}$:
:In this case, it follows from the definition of infimum that:
::$\forall x \in \closedint {x_{i - 1} } {x_i}: \map f x \ge 0$
:Therefore, on that interval, $\map {f^+} x = \map f x$.
:So:
::$M_i^{\paren {f^+} } - m_i^{\paren f} = M_i^{\paren f} - m_i^{\paren f}$
If $m_i^{\paren f} < 0 \le M_i^{\paren f}$:
:Because $\map {f^+} x \ge 0$, it follows that $m_i^{\paren {f^+} } \ge 0 > m_i^{\paren f}$.
:As $0 \le M_i^{\paren f}$, it follows that $\sup \paren {f \closedint a b \cup \set 0} = M_i^{\paren f}$.
:But $f^+ \closedint {x_{i - 1} } {x_i} \subseteq f \closedint {x_{i - 1} } {x_i} \cup \set 0$, so by Supremum of Subset:
::$M_i^{\paren {f^+} } \leq M_i^{\paren f}$
:Therefore:
::$M_i^{\paren {f^+} } - m_i^{\paren {f^+} } < M_i^{\paren f} - m_i^{\paren f}$
If $m_i^{\paren f} \le M_i^{\paren f} < 0$:
:Because $M_i^{\paren f} < 0$, it follows that $f$ is strictly negative on $\closedint {x_{i - 1} } {x_i}$.
:Thus, $\forall x \in \closedint {x_{i - 1} } {x_i}: \map {f^+} x = 0$.
:Therefore:
::$m_i^{\paren {f^+} } = M_i^{\paren {f^+} } = 0$.
:And it follows that:
::$M_i^{\paren {f^+} } - m_i^{\paren {f^+} } = 0 \le M_i^{\paren f} - m_i^{\paren f}$
Then, in every case:
:$M_i^{\paren {f^+} } - m_i^{\paren {f^+} } \le M_i^{\paren f} - m_i^{\paren f}$
But:
{{begin-eqn}}
{{eqn | l = \map {U_{f^+} } P - \map {L_{f^+} } P
      | r = \sum_i M_i^{\paren {f^+} } \paren {x_i - x_{i - 1} } - \sum_i m_i^{\paren {f^+} } \paren {x_i - x_{i - 1} }
      | c = {{Defof|Upper Darboux Sum}} and {{Defof|Lower Darboux Sum}}
}}
{{eqn | r = \sum_i \paren {M_i^{\paren {f^+} } - m_i^{\paren {f^+} } } \paren {x_i - x_{i - 1} }
}}
{{eqn | o = \le
      | r = \sum_i \paren {M_i^{\paren f } - m_i^{\paren f } } \paren {x_i - x_{i - 1} }
      | c = Inequality holds termwise
}}
{{eqn | r = \map {U_f} P - \map {L_f} P
      | c = {{Defof|Upper Darboux Sum}} and {{Defof|Lower Darboux Sum}}
}}
{{eqn | o = <
      | r = \epsilon
}}
{{end-eqn}}
Because $\epsilon$ was arbitrary, it follows from Condition for Darboux Integrability that $f^+$ is integrable on $\closedint a b$
{{qed}}
Category:Definite Integrals
60tanpk8s9crpbumoimzf42zxpgdhxh
Theorem: Positive Part of Darboux Integrable Function is Integrable/Negative Part

Let $f$ be a real function that is Darboux integrable over $\closedint a b$.
Let $f^-$ be the negative part of $f$.
Then $f^-$ is Darboux integrable over $\closedint a b$.

Proof:
$f^-$ is the positive part of $\map g x = -\map f x$.
From Linear Combination of Definite Integrals, it follows that:
:$\ds \int_a^b \map g x \rd x = -\int_a^b \map f x \rd x$
Therefore, by Positive Part of Darboux Integrable Function is Integrable, $f^-$ is Darboux integrable over $\closedint a b$.
{{qed}}
Category:Definite Integrals
poutfupt4u0zf8j6hpdus3d9k88rgrd
Theorem: Topological Evaluation Mapping is Continuous

Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
Let $f : X \to Y$ be the evaluation mapping induced by $\family{f_i}_{i \mathop \in I}$.
Then:
:$f$ is continuous

Proof:
For each $i \in I$, let:
:$\pr_i : Y \to Y_i$ denote the $i$th projection on $Y$
From Composite of Evaluation Mapping and Projection:
:$\forall i \in I : \pr_i \circ f = f_i$
By assumption:
:$\forall i \in I : \pr_i \circ f$ is continuous
From Continuous Mapping to Product Space:
:$f$ is continuous
{{qed}}
Category:Evaluation Mappings (Topological Spaces)
fblg7nkbfb09padg5cv6zz9bufupaau
Theorem: Characterization for Topological Evaluation Mapping to be Embedding

Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
Let $f : X \to Y$ be the evaluation mapping induced by $\family{f_i}_{i \mathop \in I}$.
Then:
:$f$ is an embedding
{{iff}}
:$(1)\quad$ the topology on $X$ is the initial topology with respect to $\family {f_i}_{i \mathop \in I}$
:$(2)\quad$ the family $\family {f_i}$ separates points

Proof:
=== Necessary Condition ===
Let $f$ be an embedding.
{{:Characterization for Topological Evaluation Mapping to be Embedding/Necessary Condition}}{{qed|lemma}}
Theorem: Eigenvectors Corresponding to Distinct Eigenvalues of Linear Operator are Linearly Independent

Let $K$ be a field. 
Let $X$ be a vector space over $K$. 
Let $T : X \to X$ be a linear operator.
Let $\lambda_1, \lambda_2, \ldots, \lambda_n \in K$ be distinct eigenvalues of $T$.
Let $x_1, x_2, \ldots, x_n$ be eigenvectors corresponding to $\lambda_1, \lambda_2, \ldots, \lambda_n$. 
Then $\set {x_1, \ldots, x_n}$ is a linearly independent set.

Proof:
Proof by induction:
Let $\map P n$ be the proposition:
:for any $n$ eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n \in K$ and eigenvectors $x_1, x_2, \ldots, x_n$ corresponding to $\lambda_1, \lambda_2, \ldots, \lambda_n$, we have that: 
::$\set {x_1, \ldots, x_n}$ is a linearly independent set.
Theorem: Spectrum of Bounded Linear Operator contains Point Spectrum

Let $X$ be a Banach space over $\C$. 
Let $T : X \to X$ be a bounded linear operator. 
Let $\map {\sigma_p} T$ be the point spectrum of $T$.
Let $\map \sigma T$ be the spectrum of $T$. 
Then $\map {\sigma_p} T \subseteq \map \sigma T$.

Proof:
Let $\lambda \in \map {\sigma_p} T$. 
Then there exists $x \ne \mathbf 0_X$ such that $T x = \lambda x$.
So $\paren {T - \lambda I} x = 0$ for some $x \ne \mathbf 0_X$. 
So $\ker T \ne \set {\mathbf 0_X}$.
So from Linear Transformation is Injective iff Kernel Contains Only Zero, $T$ is not injective. 
So $T$ cannot be invertible as a bounded linear operator.
So $\lambda \in \map \sigma T$.
{{qed}}
Category:Point Spectra (Linear Operators)
Category:Spectra (Bounded Linear Operators)
i0w3yzstv8menfcd4d8eowsmidwo8br
Theorem: Spectrum of Bounded Linear Operator on Finite-Dimensional Banach Space is equal to Point Spectrum

Let $X$ be a finite-dimensional Banach space over $\C$. 
Let $T : X \to X$ be a bounded linear operator.
Let $\map {\sigma_p} T$ be the point spectrum of $T$.
Let $\map \sigma T$ be the spectrum of $T$.
Then $\map \sigma T = \map {\sigma_p} T$.

Proof:
We have that $\lambda \in \map \sigma T$ {{iff}} $T - \lambda I$ is not invertible as a bounded linear transformation.
So $T - \lambda I$ is not bijective or its inverse $\paren {T - \lambda I}^{-1}$ is not bounded.
From Linear Transformations between Finite-Dimensional Normed Vector Spaces are Continuous, every linear operator $S : X \to X$ is bounded.
So $T - \lambda I$ is not invertible as a bounded linear transformation {{iff}} $T - \lambda I$ is not bijective.
From Linear Transformation from Finite-Dimensional Vector Space is Injective iff Surjective, we have that $T - \lambda I$ is not bijective {{iff}} $T - \lambda I$ is not injective. 
From Linear Transformation is Injective iff Kernel Contains Only Zero, we have $T - \lambda I$ is not injective {{iff}} $\ker T \ne \set {\mathbf 0_X}$.
That is, {{iff}} there exists $x \ne \mathbf 0_X$ such that: 
:$\paren {T - \lambda I} x = 0$
and hence $T x = \lambda x$. 
So $\lambda \in \map {\sigma_p} T$. 
{{qed}}
Theorem: Evaluation Mapping on T1 Space is Embedding if Mappings Separate Points from Closed Sets

Let $X$ be a $T_1$ topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\family {f_i}_{i \mathop \in I}$ separate points from closed sets.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
Let $f : X \to Y$ be the evaluation mapping induced by $\family{f_i}_{i \mathop \in I}$.
Then:
:$f$ is an embedding

Proof:
Let $\BB = \set{f_i^{-1} \sqbrk V : i \in I, V \text{ is open in } Y_i}$.
From Preimage of Open Sets forms Basis if Continuous Mappings Separate Points from Closed Sets:
:$\BB$ is a basis for $X$
From Analytic Basis is Analytic Sub-Basis:
:$\BB$ is a sub-basis for $X$
By definition of a $T_1$ space:
:all points of $X$ are closed
Since $\family {f_i}_{i \mathop \in I}$ separate points from closed sets then:
:$\family {f_i}_{i \mathop \in I}$ separates points
From Characterization for Topological Evaluation Mapping to be Embedding:
:the evaluation mapping $f$ is an embedding
{{qed}}
Theorem: Subspace of Product Space has Initial Topology with respect to Restricted Projections

Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\XX$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:
:$\ds \XX := \prod_{i \mathop \in I} X_i$
For each $i \in I$, let $\pr_i: \XX \to X_i$ denote the projection of $\XX$ onto $X_i$.
Let $Y \subseteq \XX$ be a non-empty subset of $\XX$.
For each $i \in I$, let $\pr_i \restriction_Y$ denote the restriction of $\pr_i$ to $Y$.
Let $\tau_\XX$ be the product topology on $\XX$.
Let $\tau_Y$ be the subspace topology on $Y$.
Then:
:$\tau_Y$ is the initial topology on $Y$ with respect to the mappings $\family {\pr_i \restriction_Y : Y \to X_i}_{i \mathop \in I}$

Proof:
By definition of product topology:
:$\tau_\XX$ is the initial topology with respect to the mappings $\family {\pr_i : \XX \to X_i}_{i \mathop \in I}$
From Subspace Topology on Initial Topology is Initial Topology on Restrictions:
:$\tau_Y$ is the initial topology on $Y$ with respect to the mappings $\family {\pr_i \restriction_Y : Y \to X_i}_{i \mathop \in I}$
{{qed}}
Category:Product Topology
Category:Topological Subspaces
Category:Initial Topology
p57f3v4facmtw3vwacrom7pviefxwnx
Theorem: Invertible Continuous Linear Operator is Bijective

Let $\struct {X, \norm {\, \cdot \,} }$ be the normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.
Let $I \in \map {CL} X$ be the identity element.
Suppose $A \in \map {CL} X$ is invertible.
Then $A$ is bijective.

Proof:
=== $A$ is injective ===
Let $x, y \in X$ be such that $\map A x = \map A y$.
Then:
:$A^{-1} \circ \map A x = A^{-1} \circ \map A y$
where $A^{-1}$ is the inverse of $A$.
By definition:
:$A^{-1} \circ A = I$
Hence:
:$x = y$
By definition, $A$ is injective.
{{qed|lemma}}
Theorem: Power Series Expansion for Reciprocal of Square Root of 1 - x

Let $x \in \R$ such that $-1 < x \le 1$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {\sqrt {1 - x} }
      | r = \sum_{k \mathop = 0}^\infty \frac {\paren {2 k}!} {\paren {2^k k!}^2} x^k
      | c = 
}}
{{eqn | r = 1 + \frac 1 2 x + \frac {1 \times 3} {2 \times 4} x^2 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^3 + \cdots
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {\sqrt {1 - x} }
      | r = \paren {1 - x}^{-\frac 1 2}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {-\frac 1 2}^{\underline k} } {k!} \paren {-x}^k
      | c = General Binomial Theorem
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-\frac 1 2} - j} } {k!} \paren {-x}^k
      | c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\ds \prod_{j \mathop = 1}^k \paren {2 j - 1} } {2^k k!} \paren {-x}^k
      | c = Translation of Index Variable of Product and simplifying
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\ds \prod_{j \mathop = 1}^k \paren {2 j - 1} \prod_{j \mathop = 1}^k \paren {2 j} } {2^k k! \ds \prod_{j \mathop = 1}^k \paren {2 j} } \paren {-x}^k
      | c = multiplying top and bottom by $\ds \prod_{j \mathop = 1}^k \paren {2 j}$
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 1}^{2 k} j} {2^k k! \paren {2^k \ds \prod_{j \mathop = 1}^k j} } x^k
      | c = simplifying
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {2 k}!} {2^k k! \paren {2^k k!} } x^k
      | c = {{Defof|Factorial}}
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {2 k}!} {\paren {2^k k!}^2} x^k
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Bijective Continuous Linear Operator is not necessarily Invertible

Let $\struct {X, \norm {\, \cdot \,} }$ be the normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.
Let $I \in \map {CL} X$ be the identity element.
Suppose $A \in \map {CL} X$ is bijective.
Then $A$ is not necessarily invertible.

Proof:
Let $\mathbb F \in \set {\R, \C}$. 
Let $\map {c_{00} } {\mathbb F}$ be the space of almost-zero sequences on $\mathbb F$.
Let $\mathbf x = \tuple {x_1, x_2, \ldots, x_N, 0, \ldots} \in c_{00}$.
Let $A : c_{00} \to c_{00}$ be a mapping such that:
:$\map A {\tuple {x_1, x_2, x_3, \ldots} } = \tuple {x_1, \dfrac {x_2} 2, \dfrac{x_3} 3, \ldots}$
Let $\struct {\map {\ell^\infty} {\mathbb F}, \norm {\, \cdot \,}_\infty}$ be the normed vector space of bounded sequences on $\mathbb F$.
We have that Space of Almost-Zero Sequences is Subspace of Space of Bounded Sequences:
:$c_{00} \subseteq \ell^\infty$
Moreover, Space of Almost-Zero Sequences with Supremum Norm is Normed Vector Space.
Theorem: Linear Transformation has Finite Rank iff Domain Quotiented by Kernel is Finite Dimensional

Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$. 
Let $T : X \to Y$ be a linear transformation. 
Let $\ker T$ be the kernel of $T$.
Let $X/\ker T$ be the quotient vector space of $X$ modulo $\ker T$.
Then $T$ has finite rank {{iff}} $X/\ker T$ is finite dimensional.

Proof:
Let $q : X \to X/\ker T$ be the quotient mapping.
Theorem: Power Series Expansion for Square Root of 1 - x

Let $x \in \R$ such that $-1 < x \le 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sqrt {1 - x}
      | r = 1 - \sum_{k \mathop = 1}^\infty \frac {\paren {2 \paren {k - 1} }!} {2^{2 k - 1} k! \paren {k - 1}!} x^k
      | c = 
}}
{{eqn | r = 1 - \frac 1 2 x - \frac 1 {2 \times 4} x^2 - \frac {1 \times 3} {2 \times 4 \times 6} x^3 - \cdots
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \sqrt {1 + x}
      | r = 1 + \frac 1 2 x - \frac 1 {2 \times 4} x^2 + \frac {1 \times 3} {2 \times 4 \times 6} x^3 - \cdots
      | c = Power Series Expansion for $\sqrt {1 + x}$
}}
{{eqn | ll= \leadsto
      | l = \sqrt {1 - x}
      | r = 1 + \frac 1 2 \paren {-x} - \frac 1 {2 \times 4} \paren {-x}^2 + \frac {1 \times 3} {2 \times 4 \times 6} \paren {-x}^3 - \cdots
      | c = substituting $x \gets -x$
}}
{{eqn | r = 1 - \frac 1 2 x - \frac 1 {2 \times 4} x^2 - \frac {1 \times 3} {2 \times 4 \times 6} x^3 - \cdots
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Non-Zero Subspace of Topological Vector Space is not von Neumann-Bounded

Let $\GF \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\GF$. 
Let $Y \ne \set {\mathbf 0_X}$ be a non-trivial subspace of $X$. 
Then $Y$ is not von Neumann-bounded.

Proof:
Let $x \in Y \setminus \set {\mathbf 0_X}$.
Then $t x \in Y$ for $t > 0$. 
Since $X$ is Hausdorff, there exists an open neighborhood $V$ of $x$ that does not contain $\mathbf 0_X$.
Then $t x \not \in t V$ for each $t > 0$. 
So we do not have $Y \subseteq t V$ for any $t > 0$. 
So $Y$ is not von Neumann-bounded.
{{qed}}