avewright/proofwiki-math huggingface dataset:
Theorem: Weak-* Dense Subset of Normed Dual Space Separates Points

Let $\GF \in \set {\R, \C}$.
Let $X$ be a normed vector space over $\GF$. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $D$ be everywhere dense in $\struct {X^\ast, w^\ast}$. 
Then $D$ separates points.

Proof:
Suppose that $D$ is everywhere dense in $\struct {X^\ast, w^\ast}$.
Let $x, y \in X$ be such that: 
:$\map f x = \map g x$ for each $f, g \in D$.
Then: 
:$\map {x^\wedge} f = \map {x^\wedge} g$ for each $f, g \in D$. 
From Characterization of Continuity of Linear Functional in Weak-* Topology, $x^\wedge : \struct {X^\ast, w^\ast} \to \GF$ is continuous. 
From Metric Space is Hausdorff, $\GF$ is Hausdorff.
By hypothesis, $D$ is everywhere dense in $\struct {X^\ast, w^\ast}$.
Hence from Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide, we obtain: 
:$\map {x^\wedge} f = \map {x^\wedge} g$ for each $f, g \in X^\ast$.
Then: 
:$\map f x = \map g x$ for each $f, g \in X^\ast$.
From Normed Dual Space Separates Points, we obtain $x = y$.
{{qed}}
Category:Weak-* Topologies
l11e3m1uzs6k96h3pc9jjqfii6ct1hy
Theorem: Characterization of Paracompactness in T3 Space/Lemma 19

:$\UU$ is a open cover of $X$ in $T$.

Proof:
Let $s \in X$.
By definition of discrete:
:$\exists U \in \tau : x \in U : \size {\set{B \in \BB : U \cap B} } \le 1$
Hence:
:$U \in \UU$
Since $x$ was arbitrary:
:$\forall x \in X : \exists U \in \UU : x \in U$
It follows that $\UU$ is an open cover of $X$ in $T$ by definition.
{{qed}}
Category:Characterization of Paracompactness in T3 Space
suy2ds6qmfz0oflhdki0jh1jm20rdbs
Theorem: Vector Subspace of Normed Dual Space is Weak-* Dense iff Separates Points

Let $\GF \in \set {\R, \C}$.
Let $X$ be a normed vector space over $\GF$. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $F \subseteq X^\ast$ be a vector subspace of $X^\ast$.
Then $F$ is everywhere dense in $\struct {X^\ast, w^\ast}$ {{iff}} it separates points.

Proof:
=== Necessary Condition ===
Let $F$ be everywhere dense in $\struct {X^\ast, w^\ast}$.
From Weak-* Dense Subset of Normed Dual Space Separates Points, $F$ separates points.
{{qed|lemma}}
Theorem: Characterization of Paracompactness in T3 Space/Lemma 21

:$\forall B \in \BB, x \in X : \map W x \cap W \sqbrk B \ne \O \leadsto \map {W \circ W} x \cap B \ne \O$

Proof:
Let $B \in \BB$.
Let $x \in X$.
Let $y \in \map W x \cap W \sqbrk B$.
By definition of intersection:
:$y \in \map W x$
and
:$y \in W \sqbrk B$
By definition of image of element:
:$\tuple{x, y} \in W$
By definition of symmetric:
:$\tuple{y, x} \in W$
By definition of image of subset:
:$\exists z \in B : \tuple{z, y} \in W$
By definition of composite relation:
:$\exists z \in B : \tuple{z, x} \in W \circ W$
By definition of image of element:
:$z \in \map {W \circ W} x \cap B$
Hence:
:$\map {W \circ W} x \cap B \ne \O$
Since $B$ and $x$ were arbitrary, then:
:$\forall B \in \BB, x \in X : \map W x \cap W \sqbrk B \ne \O \leadsto \map {W \circ W} x \cap B \ne \O$
{{qed}}
Category:Characterization of Paracompactness in T3 Space
020zsc4ecnjdnt8mcobnwnk1nobuyf9
Theorem: Injective Linear Transformation between Normed Vector Spaces sends Closed Unit Ball to Closed Unit Ball iff Isometric Isomorphism

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 $T : X \to Y$ be a injective linear transformation.
Let $B_X^-$ be the closed unit ball of $X$.
Let $B_Y^-$ be the closed unit ball of $Y$. 
Then $T$ is an isometric isomorphism {{iff}}:
:$T \sqbrk {B_X^-} = B_Y^-$

Proof:
=== Necessary Condition ===
Suppose that $T$ is an isometric isomorphism.
Then: 
:$\norm {T x}_Y = \norm x_X$ for each $x \in X$. 
Hence if $\norm x_X \le 1$, we have $\norm {T x}_Y \le 1$.
So $T \sqbrk {B_X^-} \subseteq B_Y^-$.
Conversely, let $y \in B_Y^-$.
Since $T$ is an isometric isomorphism:
:$T$ is bijective and there exists $x \in X$ such that $y = T x$. 
Since $T$ is a linear isometry, we have: 
:$\norm x_X = \norm {T x}_Y \le 1$
So we have $x \in B_X^-$. 
So we have $T \sqbrk {B_X^-} = B_Y^-$.
{{qed|lemma}}
Theorem: Image of Point under Open Neighborhood of Diagonal is Open Neighborhood of Point

Let $T = \struct{X, \tau}$ be a topological space.
Let $\tau_{X \times X}$ denote the product topology on the cartesian product $X \times X$.
Let $U$ be an open neighborhood of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct {X \times X, \tau_{X \times X} }$.
Then:
:$\forall x \in X : \map U x$ is an open neighborhood of $x$ in $T$

Proof:
Let $x \in X$.
By definition of open neighborhood:
:$U$ is a neighborhood of $\Delta_X$
From Image of Point under Neighborhood of Diagonal is Neighborhood of Point:
:$\map U x$ is a neighborhood of $x$ in $T$
It remains to show that $\map U x$ is open in $T$, that is, $\map U x \in \tau$.
Let:
:$\WW = \set{W \in \tau : \exists V \in \tau : x \in V, V \times W \subseteq U}$.
Let $y \in \map U x$.
By definition of image:
:$\tuple {x, y} \in U$
By definition of product topology:
:$\BB = \set{V \times W : V, W \in \tau}$ is a basis for  $\tau_{X \times X}$
By definition of basis:
:$\exists V, W \in \tau : \tuple{x, y} \in V \times W \subseteq U$
By definition of $\WW$:
:$W \in \WW$
Since $y$ was arbitrary:
:$\forall y \in \map U x : \exists W \in \WW : y \in W$
Let $W \in \WW$.
By definition of $\WW$:
:$\exists V \in \tau : x \in V, V \times W \subseteq U$
From Cartesian Product of Subsets:
:$\set x \times W \subseteq V \times W$
From Subset Relation is Transitive:
:$\set x \times W \subseteq U$
By definition of image:
:$W \subseteq \map U x$
Since $W$ was arbitrary:
:$\forall W \in \WW : W \subseteq \map U x$
From Characterization of Set Equals Union of Sets:
:$\map U x = \bigcup \WW$
By {{Open-set-axiom|1}}:
:$\map U x \in \tau$
Since $x$ was arbitrary:
:$\forall x \in X : \map U x$ is an open neighborhood of $x$ in $T$
{{qed}}
Category:Open Sets
Category:Product Spaces
Category:Relations
6mdk4p5ph8taiph1n87ocb2h959k4p8
Theorem: Image of Subset under Open Neighborhood of Diagonal is Open Neighborhood of Subset

Let $T = \struct{X, \tau}$ be a topological space.
Let $\tau_{X \times X}$ denote the product topology on the cartesian product $X \times X$.
Let $U$ be an open neighborhood of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct {X \times X, \tau_{X \times X} }$.
Then:
:$\forall A \subseteq X : U \sqbrk A$ is an open neighborhood of $A$ in $T$

Proof:
Let $A \subseteq X$.
From Image of Subset under Neighborhood of Diagonal is Neighborhood of Subset:
:$U \sqbrk A$ is a neighborhood of $A$ in $T$
From Image of Subset under Relation equals Union of Images of Elements:
:$U \sqbrk A = \ds \bigcup_{x \in A} \map U x$
From Image of Point under Open Neighborhood of Diagonal is Open Neighborhood of Point:
:$\forall x \in A : \map U x$ is an open neighborhood of $x$ in $T$
By {{Open-set-axiom|1}}:
:$U \sqbrk A \in \tau$
Since $A$ was arbitrary:
:$\forall A \subseteq X : U \sqbrk A$ is an open neighborhood of $A$ in $T$
{{qed}}
Category:Open Sets
Category:Product Spaces
Category:Relations
5s9p4dwqwc5hjpafytvh5qqwvphvaq7
Theorem: Primitive of Reciprocal of a x squared + b by Root of c x squared + d/Lemma

:$\ds \int \dfrac {\d x} {\paren {a x^2 + b} \sqrt {c x^2 + d} } = \dfrac {\sqrt c} {2 a} \int \dfrac {\d u} {\paren {\sqrt {\paren {u - \frac d 2}^2 - \frac d 4} } \paren {u - \frac {a d + b c} a} }$
where $u := c x^2 + d$.

Proof:
{{begin-eqn}}
{{eqn | l = u
      | r = c x^2 + d
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \sqrt {\dfrac {u - d} c}
      | c = 
}}
{{eqn | lo= \text {and}
      | l = \frac {\d u} {\d x}
      | r = 2 c x
      | c = Primitive of Power
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \int \dfrac {\d x} {\paren {a x^2 + b} \sqrt {c x^2 + d} }
      | r = \int \dfrac {\d u} {2 c x \paren {a x^2 + b} \sqrt {c x^2 + d} }
      | c = Integration by Substitution
}}
{{eqn | r = \int \dfrac {\d u} {2 c \sqrt {\frac {u - d} c} \paren {a \paren {\frac {u - d} c} + b} \sqrt u}
      | c = substituting for $x$
}}
{{eqn | r = \int \dfrac {c \sqrt c \d u} {2 c \sqrt {u - d} \paren {a u - a d + b c} \sqrt u}
      | c = simplifying
}}
{{eqn | r = \dfrac {\sqrt c} {2 a} \int \dfrac {\d u} {\paren {\sqrt {\paren {u - \frac d 2}^2 - \frac d 4} } \paren {u - \frac {a d - b c} a} }
      | c = completing the square
}}
{{end-eqn}}
{{qed}}
Category:Primitive of Reciprocal of a x squared + b by Root of c x squared + d
gbn2xzg779xxezfkqew8i51f94v71qb
Theorem: Argument of Complex Conjugate equals Negative of Argument

Let $z \in \C$ be a complex number.
Then:
:$\arg {\overline z} = -\arg z$
where:
:$\arg$ denotes the argument of a complex number
:$\overline z$ denotes the complex conjugate of $z$.

Proof:
Let $z$ be expressed in polar form:
:$z := r \paren {\cos \theta + i \sin \theta}$
Then:
{{begin-eqn}}
{{eqn | l = \overline z
      | r = r \paren {\cos \theta - i \sin \theta}
      | c = Polar Form of Complex Conjugate
}}
{{eqn | r = r \paren {\map \cos {-\theta} + i \map \sin {-\theta} }
      | c = Cosine Function is Even, Sine Function is Odd
}}
{{end-eqn}}
The result follows by definition of the argument of a complex number
{{qed}}
Theorem: Complex Division as Product with Conjugate over Square of Modulus

Let $z_1$ and $z_2$ be complex numbers.
Then the operation of division can be expressed as:
:$\dfrac {z_1} {z_2} = \dfrac {z_1 \overline {z_2} } {\cmod {z_2}^2}$
where:
:$\overline {z_2}$ denotes the complex conjugate of $z_2$
:$\cmod {z_2}$ denotes the complex modulus of $z_2$.

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac {z_1} {z_2}
      | r = \dfrac {z_1 \overline {z_2} } {z_2 \overline {z_2} }
      | c = multiplying top and bottom by $\overline {z_2}$
}}
{{eqn | r = \dfrac {z_1 \overline {z_2} } {\cmod {z_2}^2}
      | c = Modulus in Terms of Conjugate
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Complex Number

Let $z = a + i b$ be a complex number.
Then its square is given by:
:$z^2 = a^2 - b^2 + i \paren {2 a b}$

Proof:
{{begin-eqn}}
{{eqn | l = z^2
      | r = \paren {a + i b}^2
      | c = {{hypothesis}}
}}
{{eqn | r = \paren {a + i b} \paren {a + i b}
      | c = {{Defof|Square Function}}
}}
{{eqn | r = \paren {a \times a - b \times b} + i \paren {a \times b + b \times a}
      | c = {{Defof|Complex Multiplication}}
}}
{{eqn | r = a^2 - b^2 + i \paren {2 a b}
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Cube of Complex Number

Let $z = a + i b$ be a complex number.
Then its cube is given by:
:$z^3 = a^3 - 3 a b^2 + i \paren {3 a^2 b - b^3}$

Proof:
{{begin-eqn}}
{{eqn | l = z^3
      | r = \paren {a + i b}^3
      | c = {{hypothesis}}
}}
{{eqn | r = \paren {a + i b}^2 \paren {a + i b}
      | c = {{Defof|Cube (Algebra)}}
}}
{{eqn | r = \paren {a^2 - b^2 + i \paren {2 a b} } \paren {a + i b}
      | c = Square of Complex Number
}}
{{eqn | r = \paren {\paren {a^2 - b^2} a - b \paren {2 a b} } + i \paren {\paren {a^2 - b^2} b + \paren {2 a b} a}
      | c = {{Defof|Complex Multiplication}}
}}
{{eqn | r = \paren {a^3 - a b^2 - 2 a b^2} + i \paren {a^2 b - b^3 + 2 a^2 b}
      | c = multiplying out
}}
{{eqn | r = a^3 - 3 a b^2 + i \paren {3 a^2 b - b^3}
      | c = gathering like terms
}}
{{end-eqn}}
{{qed}}
Theorem: Fourth Power of Complex Number

Let $z = a + i b$ be a complex number.
Then its fourth power is given by:
:$z^4 = a^4 - 6 a^2 b^2 + b^4 + i \paren {4 a^3 b - 4 a b^3}$

Proof:
{{begin-eqn}}
{{eqn | l = z^4
      | r = \paren {a + i b}^4
      | c = {{hypothesis}}
}}
{{eqn | r = a^4 + 4 a^3 \paren {i b} + 6 a^2 \paren {i b}^2 + 4 a \paren {i b}^3 + \paren {i b}^4
      | c = Fourth Power of Sum
}}
{{eqn | r = a^4 + 4 i a^3 b + 6 i^2 a^2 b^2 + 4 i^3 a b^3 + i^4 b^4
      | c = Complex Multiplication is Commutative
}}
{{eqn | r = a^4 + 4 i a^3 b - 6 a^2 b^2 - 4 i a b^3 + b^4
      | c = {{Defof|Imaginary Unit}}
}}
{{eqn | r = a^4 - 6 a^2 b^2 + b^4 + i \paren {4 a^3 b - 4 a b^3}
      | c = gathering like terms
}}
{{end-eqn}}
{{qed}}
Theorem: Fifth Power of Complex Number

Let $z = a + i b$ be a complex number.
Then its fifth power is given by:
:$z^5 = a^5 - 10 a^3 b^2 + 5 a b^4 + i \paren {5 a^4 b - 10 a^2 b^3 + b^5}$

Proof:
{{begin-eqn}}
{{eqn | l = z^5
      | r = \paren {a + i b}^5
      | c = {{hypothesis}}
}}
{{eqn | r = a^5 + 5 a^4 \paren {i b} + 10 a^3 \paren {i b}^2 + 10 a^2 \paren {i b}^3 + 5 a \paren {i b}^4 + \paren {i b}^5
      | c = Fifth Power of Sum
}}
{{eqn | r = a^5 + 5 i a^4 b + 10 i^2 a^3 b^2 + 10 i^3 a^2 b^3 + 5 i^4 a b^4 + i^5 b^5
      | c = Complex Multiplication is Commutative
}}
{{eqn | r = a^5 + 5 i a^4 b - 10 a^3 b^2 - 10 i a^2 b^3 + 5 a b^4 + i b^5
      | c = {{Defof|Imaginary Unit}}
}}
{{eqn | r = a^5 - 10 a^3 b^2 + 5 a b^4 + i\paren {5 a^4 b - 10 a^2 b^3 + b^5}
      | c = gathering like terms
}}
{{end-eqn}}
{{qed}}
Theorem: Riesz-Kakutani Representation Theorem

Let $X$ be a Hausdorff compact space.
Let $\map \BB X$ be the Borel $\sigma$-algebra on $X$.
Let $\map C {X, \R}$ be the space of real-valued continuous functions.
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $X$.
Let $\struct {C', \norm {\, \cdot \,}_{C'} }$ be the normed dual space of $\struct {\map C {X, \R}, \norm {\, \cdot \,}_\infty}$.
Then $\ell \in C'$ {{iff}} there is a unique signed measure $\mu$ on $\struct {X, \map \BB X}$ such that:
:the variation $\size \mu$ of $\mu$ is finite
:$\ds \forall f \in \map C {X, \R} : \map \ell f = \int_X f \rd \mu$
In addition:
:$\norm \ell_{C'} = \map {\size \mu} X$

Proof:
{{ProofWanted}}
{{Namedfor|Frigyes Riesz|cat = Riesz F|name2 = Shizuo Kakutani|cat2 = Kakutani}}
Theorem: Arbitrary Power of Complex Number

Let $z = a + i b$ be a complex number.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
{{begin-eqn}}
{{eqn | l = z^n
      | r = \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ even} } } \paren {-1}^{j / 2} \dbinom n j a^{n - j} b^j} + i \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom n j a^{n - j} b^j}
      | c = 
}}
{{eqn | r = \paren {a^n - \dbinom n 2 a^{n - 2} b^2 + \dbinom n 4 a^{n - 4} b^4 - \cdots}
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = i \paren {\dbinom n 1 a^{n - 1} b - \dbinom n 3 a^{n - 3} b^3 + \cdots} 
      | c = 
}}
{{end-eqn}}

Proof:
=== Lemma ===
{{:Arbitrary Power of Complex Number/Lemma}}{{qed|lemma}}
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds z^n = \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ even} } } \paren {-1}^{j / 2} \dbinom n j a^{n - j} b^j} + i \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom n j a^{n - j} b^j}$
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = z^1
      | r = \paren {a + i b}^1
      | c = 
}}
{{eqn | r = \dbinom 1 0 a^1 b^0 + i \dbinom 1 1 a^0 b^1
      | c = 
}}
{{eqn | r = \dbinom 1 0 a^{1 - 0} b^0 + i \dbinom 1 1 a^{1 - 1} b^1
      | c = 
}}
{{eqn | r = \paren {\sum_{\substack {0 \mathop \le j \mathop \le 1 \\ \text {$j$ even} } } \paren {-1}^{j / 2} \dbinom 1 j a^{1 - j} b^j} + i \paren {\sum_{\substack {0 \mathop \le j \mathop \le 1 \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom 1 j a^{1 - j} b^j}
      | c = 
}}
{{end-eqn}}
Thus $\map P 1$ is seen to hold.
Theorem: Neumann Series Theorem

Let $X$ be a Banach space.
Let $\map {CL} X$ be the continous linear transformation space.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $A \in \map {CL} X$ be such that $\norm A < 1$.
Let $\circ$ be the composition of mappings.
Let $I$ be the identity mapping.
Then:
:$I - A$ is invertible in $\map {CL} X$
:$\ds \paren {I - A}^{-1} = \sum_{n \mathop = 0}^\infty A^n$
:$\norm {\paren{I - A}^{-1} } \le \dfrac 1 {1 - \norm A}$

Proof:
Let $\ds S_k := \sum_{n \mathop = 0}^k A^n$.
Theorem: Image of Bounded Linear Transformation is Everywhere Dense iff Dual Operator is Injective

Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$. 
Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively. 
Let $T : X \to Y$ be a bounded linear transformation.
Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator of $T$. 
Then $T \sqbrk X$ is everywhere dense in $Y$ {{iff}} $T^\ast$ is injective.

Proof:
From Annihilator of Image of Bounded Linear Transformation is Kernel of Dual Operator, we have:
:$T \sqbrk X^\bot = \map \ker {T^\ast}$
where $T \sqbrk X^\bot$ denotes the annihilator of $T \sqbrk X$.
From Linear Transformation is Injective iff Kernel Contains Only Zero, we then have that $T^\ast$ is injective {{iff}}:
:$T \sqbrk X^\bot = \set { {\mathbf 0}_{Y^\ast} }$
From Annihilator of Subspace of Banach Space is Zero iff Subspace is Everywhere Dense, this is equivalent to:
:$T \sqbrk X$ is everywhere dense in $Y$.
This was the demand.
{{qed}}
Theorem: Characterization of Paracompactness in T3 Space/Lemma 9

Let $\BB$ be a discrete set of subsets of $X$.
Then there exists an open neighborhood $W$ of the diagonal $\Delta_X$ of $X \times X$ in $T \times T$:
:$\forall x \in X : \card {\set{B \in \BB : \map W x \cap W \sqbrk B \ne \O}} \le 1$

Proof:
Let:
:$\UU = \set{ U \in \tau : \card {\set{B \in \BB : U \cap B} } \le 1}$
Theorem: Characterization of Paracompactness in T3 Space/Lemma 10

:$\WW$ is an open $\sigma$-discrete refinement of $\UU$

Proof:
=== $\WW$ is Set of Open Sets ===
Let:
:$W \in \WW$
By definition of $\WW$:
:$\exists n \in \N, A \in \AA : W = U_A \cap V_n \sqbrk A$
We have {{hypothesis}}:
:$U_A \in \tau$
From Image of Subset under Open Neighborhood of Diagonal is Open Neighborhood of Subset:
:$V_n \sqbrk A \in \tau$
By {{Open-set-axiom|2}}:
:$W \in \tau$
Since $W$ was arbitrary, it follows that:
:$\WW$ is a set of open sets
{{qed|lemma}}
Theorem: Real and Imaginary Parts of Integer Power of Complex Number are Harmonic

Let $z \in \C$ be a complex number.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $z^n$ denote $z$ raised to the $n$th power.
Then both the real part $\map \Re {z^n}$ and the imaginary part $\map \Im {z^n}$ of $z^n$ are harmonic polynomials.

Proof:
=== Real Part of Integer Power of Complex Number is Harmonic ===
{{:Real Part of Integer Power of Complex Number is Harmonic}}{{qed|lemma}}
Theorem: Characterization of Open Linear Transformation between Normed Vector Spaces

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 $T : X \to Y$ be a linear transformation.
Let $B_X^-$ and $B_Y^-$ be the closed unit balls of $X$ and $Y$ respectively. 
{{TFAE}}
:$(1) \quad$ $T$ is an open map
:$(2) \quad$ there exists $\delta > 0$ such that $\delta B_Y^- \subseteq T \sqbrk {B_X^-}$
:$(3) \quad$ there exists $M > 0$ such that for all $y \in Y$ there exists $x \in T^{-1} \sqbrk {\set y}$ such that $\norm x_X \le M \norm y_Y$

Proof:
Let $B_X^O$ and $B_Y^O$ be the open unit balls of $X$ and $Y$ respectively.
Theorem: Linear Transformation between Normed Vector Spaces is Open iff Image of Open Unit Ball is Open

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 $T : X \to Y$ be a linear transformation.
Let $B_X^O$ be the open unit ball of $X$.
Then $T$ is open {{iff}} $T \sqbrk {B_X^O}$ is open.

Proof:
=== Necessary Condition ===
Suppose that $T$ is open.
From Open Ball is Open Set, $B_X^O$ is open in $\struct {X, \norm {\, \cdot \,}_X}$.
So $T \sqbrk {B_X^O}$ is  open in $\struct {Y, \norm {\, \cdot \,}_Y}$.
{{qed|lemma}}
Theorem: Norm of Bounded Linear Transformation in terms of Closed Unit Ball

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 $T : X \to Y$ be a linear transformation.
Let $B_X^-$ and $B_Y^-$ be the closed unit balls of $X$ and $Y$ respectively. 
Let $M > 0$.
Then $T$ is bounded with $\norm T_{\map B {X, Y} } \le M$ {{iff}}:
:$T \sqbrk {B_X^-} \subseteq M B_Y^-$

Proof:
=== Necessary Condition ===
Let $T$ be a bounded linear transformation with $\norm T_{\map B {X, Y} } \le M$.
Then for each $x \in X$ we have: 
:$\norm {T x}_Y \le M \norm x_X$
Then if $x \in B_X^-$, we have $\norm x_X \le 1$ and hence: 
:$\norm {T x}_Y \le M$
So we have: 
:$T \sqbrk {B_X^-} \subseteq M B_Y^-$
{{qed|lemma}}
Theorem: Norm in terms of Normed Dual Space

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$. 
Let $X^\ast$ be the normed dual of $X$.
Let $B_{X^\ast}^-$ be the closed unit ball of $X^\ast$. 
Then: 
:$\ds \norm x = \sup_{f \in B_{X^\ast}^-} \cmod {\map f x}$

Proof:
From Fundamental Property of Norm on Bounded Linear Functional, we have: 
:$\ds \cmod {\map f x} \le \norm x$
for each $x \in X$.
From Existence of Support Functional, there exists $f \in B_{X^\ast}^-$ such that $\map f x = \norm x$. 
Hence we conclude: 
:$\ds \norm x = \sup_{f \in B_{X^\ast}^-} \cmod {\map f x}$
{{qed}}
Category:Normed Dual Spaces
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Theorem: Reciprocal of Complex Number in terms of Conjugate and Modulus

Let $z \in \C$ be a complex number.
The reciprocal of $z$ can be expressed as:
:$\dfrac 1 z = \dfrac {\overline z} {\cmod z^2}$
where:
:$\overline z$ denotes the complex conjugate of $z$
:$\cmod z^2$ denotes the modulus of $z$.

Proof:
Let $z$ be defined as:
:$z = a = i b$
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 z
      | r = \dfrac {a - i b} {a^2 + b^2}
      | c = Reciprocal of Complex Number
}}
{{eqn | r = \dfrac {a - i b} {\paren {\sqrt {a^2 + b^2} }^2}
      | c = rearranging
}}
{{eqn | r = \dfrac {\overline z} {\paren {\sqrt {a^2 + b^2} }^2}
      | c = {{Defof|Complex Conjugate}}
}}
{{eqn | r = \dfrac {\overline z} {\cmod z^2}
      | c = {{Defof|Complex Modulus}}
}}
{{end-eqn}}
{{qed}}
Theorem: Reciprocal of Complex Number

Let $z = a + i b$ be a complex number.
The reciprocal of $z$ is:
:$\dfrac 1 z = \dfrac {a - i b} {a^2 + b^2}$

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac 1 z
      | r = \dfrac 1 {a + i b}
      | c = Definition of $z$
}}
{{eqn | r = \dfrac {a - i b} {\paren {a + i b} \paren {a - i b} }
      | c = multiplying top and bottom by $a - i b$
}}
{{eqn | r = \dfrac {a - i b} {a^2 + b^2}
      | c = Product of Complex Number with Conjugate
}}
{{end-eqn}}
{{qed}}
Theorem: Reciprocal of Power of Complex Number

Let $z \in \C$ be a complex number.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $z^n$ denote $z$ raised to the $n$th power.
The reciprocal of $z^n$ can be expressed as:
:$\dfrac 1 {z^n} = \dfrac {\overline z^n} {\cmod z^{2 n} }$
where:
:$\overline z$ denotes the complex conjugate of $z$
:$\cmod z^2$ denotes the modulus of $z$.

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {z^n}
      | r = \paren {\dfrac 1 z}^n
      | c = 
}}
{{eqn | r = \paren {\dfrac {\overline z} {\cmod z^2} }^n
      | c = 
}}
{{eqn | r = \dfrac {\paren {\overline z}^n} {\paren {\cmod z^2}^n}
      | c = 
}}
{{eqn | r = \dfrac {\overline z^n} {\cmod z^{2 n} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Bounded Linear Transformation is Into Linear Isomorphism iff Dual Operator is Surjective

Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be Banach spaces over $\GF$. 
Let $X^\ast$ and $Y^\ast$ be the normed duals of $X$ and $Y$ respectively.
Let $T : X \to Y$ be a bounded linear transformation.
Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator of $T$. 
Then $T$ is an into linear isomorphism {{iff}} $T^\ast$ is surjective.

Proof:
Let $B_{X^\ast}^-$ and $B_{Y^\ast}^-$ be the closed unit balls of $X^\ast$ and $Y^\ast$ respectively.