nyuuzyou/wikis huggingface dataset:
Norm of Double Centralizer Generated by Element of C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $c \in A$.
Let $\tuple {L_c, R_c}$ be the double centralizer generated by $c$.
Then:
:$\norm {L_c}_{\map B A} = \norm {R_c}_{\map B A} = \norm c$
where $\norm {L_c}_{\map B A}$ and $\norm {R_c}_{\map B A}$ denotes the norm of a bounded linear transformation.
== Proof ==
From Norm of Element of C*-Algebra as Supremum over Closed Unit Ball we have:
:$\ds \norm c = \sup_{\norm b \le 1} \norm {c b} = \sup_{\norm b \le 1} \norm {b c}$
From the definition of the double centralizer generated by $c$, we have:
:$\ds \norm c = \sup_{\norm b \le 1} \norm {\map {L_c} b} = \sup_{\norm b \le 1} \norm {\map {R_c} b}$
By the definition of the norm of a bounded linear transformation, we then have:
:$\norm c = \norm {L_c}_{\map B A} = \norm {R_c}_{\map B A}$
== Sources ==
* : $2.1$: $\text C^\ast$-Algebras
Category:Multiplier Algebras
Norm of Eisenstein Integer

== Theorem ==
Let $\alpha$ be an Eisenstein integer.
That is:
:$\alpha = a + b \omega$ for some $a, b \in \Z$
where $\omega := e^{2 \pi i /3}$ is a cube root of unity.
Then:
:$\cmod \alpha^2 = a^2 - a b + b^2$
where $\cmod {\, \cdot \,}$ denotes the modulus of a complex number.
== Proof ==
We find that:
By the definition of the polar form of a complex number:
:$\omega = \exp \paren {\dfrac {2 \pi i} 3} = \map \cos {\dfrac {2 \pi} 3} + i \, \map \sin {\dfrac {2 \pi} 3} = -\dfrac 1 2 + i \dfrac {\sqrt 3} 2$
Thus by Sum of Complex Number with Conjugate:
:$\omega + \overline \omega = 2 \cdot \paren {-\dfrac 1 2} = -1$
Also:
Therefore:
:$\cmod \alpha^2 = a^2 + \paren {\omega + \overline \omega} a b + \omega \overline \omega b^2 = a^2 - a b + b^2$
as required.
Category:Algebraic Number Theory
Norm of Element of Approximate Identity of C*-Algebra is Less Than or equal to One

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $\family {e_\lambda}_{\lambda \in \Lambda}$ be an approximate identity of $A$.
Then for each $\lambda \in \Lambda$ we have:
:$\norm {e_\lambda} \le 1$
== Proof ==
From the definition of an approximate identity, for each $\lambda \in \Lambda$ we have:
:$\map {\sigma_A} {e_\lambda} \subseteq \closedint 0 1$
So that, from the definition of spectral radius:
:$\map {r_A} {e_\lambda} \le 1$ 
By definition, each $e_\lambda$ is positive and hence Hermitian.
Hence from Spectral Radius of Normal Element of C*-Algebra Equal to Norm, we have $\norm {e_\lambda} = \map {r_A} {e_\lambda} \le 1$ for each $\lambda \in \Lambda$.
Category:Approximate Identities of C*-Algebras
Norm of Element of C*-Algebra as Supremum over Closed Unit Ball

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $c \in A$.
Then:
:$\ds \norm c = \sup_{\norm b \le 1} \norm {c b} = \sup_{\norm b \le 1} \norm {b c}$
== Proof ==
Note that if $c = {\mathbf 0}_A$, we have $\norm c = 0$ and:
:$\norm {c b} = \norm {b c} = \norm { {\mathbf 0}_A} = 0$
for each $b \in A$.
Hence we have the theorem in the case $c = {\mathbf 0}_A$.
Now take $c \ne {\mathbf 0}_A$.
From the definition of an algebra norm, we have:
:$\norm {c b} \le \norm c \norm b$
and:
:$\norm {b c} \le \norm b \norm c$
Hence we have:
:$\ds \norm c \le \sup_{\norm b \le 1} \norm {c b}$
and:
:$\ds \norm c \le \sup_{\norm b \le 1} \norm {b c}$
From the definition of a Banach $\ast$-algebra, we have:
:$\norm c = \norm {c^\ast}$
Note that:
:$\ds \norm {\frac {c^\ast} {\norm c} } \le 1$
and:
:$\norm {c b} = \frac {\norm {c c^\ast} } {\norm c}$
By the $\text C^\ast$ identity, we have:
:$\ds \frac {\norm {c c^\ast} } {\norm c} = \frac {\norm c^2} {\norm c} = \norm c$
So we have:
:$\ds \sup_{\norm b \le 1} \norm {c b} \le \norm c$ 
We conclude:
:$\ds \norm c = \sup_{\norm b \le 1} \norm {c b}$
Similarly we have:
:$\ds \frac {\norm {c^\ast c} } {\norm c} = \frac {\norm {c c^\ast} } {\norm c} = \frac {\norm c^2} {\norm c} = \norm c$
by the $\text C^\ast$ identity.
We then also obtain:
:$\ds \sup_{\norm b \le 1} \norm {b c} \le \norm c$
and hence:
:$\ds \norm c = \sup_{\norm b \le 1} \norm {b c}$
== Sources ==
* : $2.1$: $\text C^\ast$-Algebras
Category:C*-Algebras
Norm of Hermitian Operator/Corollary

== Corollary to Norm of Hermitian Operator ==
Let $\mathbb F \in \set {\R, \C}$.
Let $\HH$ be a Hilbert space over $\mathbb F$. 
Let $A : \HH \to \HH$ be a bounded Hermitian operator.
Let $\innerprod \cdot \cdot_\HH$ denote the inner product on $\HH$.
Suppose that:
:$\forall h \in \HH: \innerprod {A h} h_\HH = 0$
Then $A$ is the zero operator $\mathbf 0$.
== Proof ==
Let $\norm \cdot_\HH$ denote the inner product norm on $\HH$. 
Let $\norm A$ denote the norm of $A$.
From Norm of Hermitian Operator: 
:$\norm A = \sup \set {\size {\innerprod {A h} h_\HH}: h \in \HH, \norm h_\HH = 1}$
By definition of inner product norm:
:$\forall h \in \HH: \innerprod {A h} h_\HH = 0$
Hence, in particular: 
:$\innerprod {A h} h_\HH = 0$
for all $h \in \HH$ such that $\norm h_\HH = 1$.
So: 
:$\set {\size {\innerprod {A h} h_\HH}: h \in \HH, \norm h_\HH = 1} = \set 0$
giving: 
:$\norm A = \sup \set 0$
Hence from the definition of supremum: 
:$\norm A = 0$
So from Norm on Bounded Linear Transformation equals Zero iff Zero Operator:
:$A$ is the zero operator.
== Sources ==
* $\text {II}.2.14$
Category:Norm of Hermitian Operator
Norm of Identity Element in Normed Algebra Lower Bounded by One

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,}_R}$ be a normed division ring.
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $R$ with identity element ${\mathbf 1}_A \ne {\mathbf 0}_A$. 
Then $1 \le \norm { {\mathbf 1}_A}$. 
== Proof ==
By the definition of a algebra norm, we have:
:$\norm { {\mathbf 1}_A} = \norm { {\mathbf 1}_A^2} \le \norm { {\mathbf 1}_A}^2$
From , we have $\norm { {\mathbf 1}_A} \ne 0$, we have:
:$1 \le \norm { {\mathbf 1}_A}$
Category:Normed Algebras
Norm of Inverse of Sequence of Invertible Elements Converging to Non-Invertible Element in Unital Banach Algebra

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra.
Let $\map G A$ be the group of units of $A$.
Let $x \in A \setminus \map G A$. 
Let $\sequence {x_n}_{n \in \N}$ be a sequence in $\map G A$ such that $x_n \to x$. 
Then $\norm {x_n^{-1} } \to \infty$ as $n \to \infty$. 
== Proof ==
Note that if there existed $n \in \N$ such that: 
:$\ds \norm {x - x_n} < \frac 1 {\norm {x_n^{-1} } }$
then we would have $x \in \map G A$ from Group of Units in Unital Banach Algebra is Open. 
So, we have: 
:$\ds \frac 1 {\norm {x_n^{-1} } } \le \norm {x - x_n}$ for each $n \in \N$. 
Since we have $x_n \to x$, we have $\norm {x - x_n} \to 0$, and hence: 
:$\ds \frac 1 {\norm {x_n^{-1} } } \to 0$ as $n \to \infty$.
So:
:$\norm {x_n^{-1} } \to \infty$ as $n \to \infty$.
Category:Unital Banach Algebras
Norm of Non-Negative Increasing Continuous Function applied to Positive Element of Unital C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be positive.
Let $\map {\sigma_A} x \subseteq \hointr 0 \infty$ be the spectrum of $x$ in $A$.
Let $f : \map {\sigma_A} x \to \hointr 0 \infty$ be an increasing continuous function.
Let $\map f x$ be obtained from the continuous functional calculus.
Then we have:
:$\norm {\map f x} = \map f {\norm x}$
== Proof ==
From Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative:
:$\map f x$ is positive.
We then have:
Category:Positive Elements of C*-Algebras
Norm of Positive Linear Functional on Unital C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra with identity element ${\mathbf 1}_A$. 
Let $f : A \to \C$ be a positive linear functional.
Let $\norm {\, \cdot \,}_{A^\ast}$ be the norm of a bounded linear functional.
Then $f$ is bounded and:
:$\norm f_{A^\ast} = \map f { {\mathbf 1}_A}$ 
== Proof ==
From Positive Linear Functional on C*-Algebra is Bounded, $f$ is bounded.
We show that:
:$\cmod {\map f x} \le \map f { {\mathbf 1}_A}$
for each $x \in A$ with $\norm x \le 1$. 
Since equality is attained for $x = {\mathbf 1}_A$, we will then have $\norm f_{A^\ast} = \map f { {\mathbf 1}_A}$.
We have, for all $u \in A$:
Let $x \in A$ have $\norm x \le 1$.
Take $\theta \in \R$ such that:
:$\cmod {\map \phi x} = e^{i \theta} \map \phi x$ 
Then we have:
From Element of *-Algebra Uniquely Decomposes into Hermitian Elements, $\map \Re {e^{i \theta} x}$ is Hermitian.
From Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum, we have:
:$\map \Re {e^{i \theta} x} \le_A \norm {e^{i \theta} x} {\mathbf 1}_A$
where $\le_A$ is the canonical preordering of $A$.
From , we have:
:$\map \Re {e^{i \theta} x} \le_A \norm x {\mathbf 1}_A = {\mathbf 1}_A$
From Positive Linear Functional on C*-Algebra is Increasing on Hermitian Elements, we have:
:$\map f {\map \Re {e^{i \theta} x} } \le \map f { {\mathbf 1}_A}$
Hence:
:$\cmod {\map f x} \le \map f { {\mathbf 1}_A}$
Category:Positive Linear Functionals
Category:C*-Algebras
Norm of Refinement is no Greater than Norm of Subdivision

== Theorem ==
Let $P$ be a finite subdivision of $\closedint a b$.
Let $P'$ be a refinement of $P$.
Then:
:$\norm {P'} \le \norm P$
where $\norm P$ denotes the norm of $P$.
== Proof ==
Let $P = \set {x_0, x_1, \dotsc, x_{n - 1}, x_n}$.
Let $P' = \set {y_0, y_1, \dotsc, y_{m - 1}, y_m}$.
By definition of refinement:
:$P \subseteq P'$
Therefore, every $x_i \in P'$.
Thus, every $i \in \set {0, 1, \dotsc, n - 1, n}$, define $k_i \in \set {0, 1, \dotsc, m - 1, m}$ such that:
:$y_{k_i} = x_i$
Now, for each $j \in \set {1, 2, \dotsc, m - 1, m}$, let $i_j$ be the smallest $i$ such that $k_i \ge j$.
Since $k_{i_j} \ge j$, it follows that $y_j \le y_{k_{i_j}}$.
Therefore, by definition of $k_i$:
:$y_j \le x_{i_j}$
 $k_{i_j - 1} > j - 1$.
Then, $k_{i_j - 1} \ge j$, contradicting the fact that $i_j$ is the smallest $i$ satisfying the property.
Therefore, by Proof by Contradiction, $k_{i_j - 1} \le j - 1$.
Hence, $x_{i_j - 1} = y_{k_{i_j - 1}} \le y_{j - 1}$.
Let $Z = \norm {P'}$ be the norm of $P'$.
By definition, there is some $j \in \set {1, 2, \dotsc, m - 1, m}$ such that:
:$y_j - y_{j - 1} = Z$
But from:
:$x_{i_j - 1} \le y_{j - 1} \le y_j \le x_{i_j}$
it follows that:
:$x_{i_j} - x_{i_j - 1} \ge y_j - y_{j - 1} = Z$
Therefore, by definition of the norm of $P$:
:$\norm P \ge Z = \norm {P'}$
Category:Subdivisions (Real Analysis)
Norm of Summation

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$.
Then: 
:$\ds \norm {\sum_{k \mathop = 1}^n x_k} \le \sum_{k \mathop = 1}^n \norm {x_k}$ for all $n \in \N$.
== Proof ==
The proof proceeds by induction.
For each $n \in \N$, let $\map P n$ be the proposition:
:$\ds \norm {\sum_{k \mathop = 1}^n x_k} \le \sum_{k \mathop = 1}^n \norm {x_k}$ for all $n \in \N$.
=== Basis for the Induction ===
$\map P 1$ is the case: 
:$\ds \norm {\sum_{k \mathop = 1}^1 x_k} \le \sum_{k \mathop = 1}^1 \norm {x_k}$ 
This is immediate, since: 
:$\ds \norm {\sum_{k \mathop = 1}^1 x_k} = \norm {x_1}$
and:
:$\ds \sum_{k \mathop = 1}^1 \norm {x_k} = \norm {x_1}$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P n$ is true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true.
So this is the induction hypothesis:
:$\ds \norm {\sum_{k \mathop = 1}^n x_k} \le \sum_{k \mathop = 1}^n \norm {x_k}$
from which it is to be shown that: 
:$\ds \norm {\sum_{k \mathop = 1}^{n + 1} x_k} \le \sum_{k \mathop = 1}^{n + 1} \norm {x_k}$
=== Induction Step ===
This is the induction step:
We have: 
So $\map P n \implies \map P {n + 1}$ and the result follows by the Principle of Mathematical Induction.
Category:Normed Vector Spaces
Norm of Unit of Normed Division Algebra

== Theorem ==
Let $\struct {A_F, \oplus}$ be a normed division algebra.
Let the unit of $\struct {A_F, \oplus}$ be $1_A$.
Then:
:$\norm {1_A} = 1$
where $\norm {1_A}$ denotes the norm of $1_A$.
== Proof ==
By definition:
:$\forall a, b \in A_F: \norm {a \oplus b} = \norm a \norm b$
So:
:$\norm {1_A} = \norm {1_A \oplus 1_A} = \norm {1_A} \norm {1_A}$
So $\norm {1_A} \in \R$ is idempotent under real multiplication.
From Idempotent Elements of Ring with No Proper Zero Divisors, only $0 \in \R$ and $1 \in \R$ fit that bill.
But $\norm {1_A}$ can not be $0$ as that would make:
:$\forall a \in A_F: \norm a = \norm {1_A \oplus a} = \norm {1_A} \norm a = 0 \norm a = 0$
which contradicts the definition of the norm.
Hence the result.
== Sources ==
* 2002: : The Octonions: 1.1 Preliminaries
Category:Normed Division Algebras
Category:Norm Theory
Norm of Vector Cross Product

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vectors in the real Euclidean space $\R^3$.
Let $\times$ denote the vector cross product.
Then:
:$\norm {\mathbf a \times \mathbf b} = \norm {\mathbf a} \norm {\mathbf b} \size {\sin \theta}$
where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$, or an arbitrary number if $\mathbf a$ or $\mathbf b$ is the zero vector.
== Proof ==
Suppose either $\mathbf a$ or $\mathbf b$ is the zero vector.
Then by :
:$\norm {\mathbf a} = 0$
or:
:$\norm {\mathbf b} = 0$
By calculation, it follows that $\mathbf a \times \mathbf b$ is also the zero vector.
Hence:
:$\norm {\mathbf a \times \mathbf b} = 0$
and equality holds.
Now suppose that both $\mathbf a$ or $\mathbf b$ are non-zero vectors.
We have:
== Sources ==
* : $\S 7.5$
Category:Vector Cross Product
Norm on Bounded Linear Functional is Finite

== Theorem ==
Let $V$ be a normed vector space.
Let $L$ be a bounded linear functional on $V$.
Let $\norm L$ denote the norm on $L$ defined as:
:$\norm L = \inf \set {c > 0: \forall v \in V: \size {L v} \le c \norm v_V}$
Then:
:$\norm L < \infty$
== Proof ==
By definition of a bounded linear functional:
:$\exists c \in \R_{> 0}: \forall v \in V: \size{L v} \le c \norm v_V$
Hence:
:$\set {\lambda > 0: \forall v \in V: \size {L v} \le \lambda \norm v_V} \ne \O$
By definition:
$\set {\lambda > 0: \forall v \in V: \size {L v} \le \lambda \norm v_V}$ is bounded below by $0$.
From the Greatest Lower Bound Property:
:$\norm L = \inf \set {\lambda > 0: \forall v \in V: \size {L v} \le \lambda \norm v_V}$ exists.
We have:
The result follows.
Category:Bounded Linear Functionals
Norm on Bounded Linear Transformation is Finite

== Theorem ==
<onlyinclude>
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 $A: X \to Y$ be a bounded linear transformation.
Let $\norm A$ denote the norm of $A$ defined by:
:$\norm A = \inf \set {c > 0: \forall h \in X : \norm {A x}_Y \le c \norm x_X}$
Then:
:$\norm A < \infty$
== Proof ==
By definition of a bounded linear transformation:
:$\exists c \in \R_{> 0}: \forall x \in X : \norm{A x}_Y \le c \norm x_X$
Hence:
:$\set {\lambda > 0: \forall x \in X : \norm {A x}_Y \le \lambda \norm x_X} \ne \O$
By definition:
:$\set {\lambda > 0: \forall x \in X : \norm {A x}_Y \le \lambda \norm x_X}$ is bounded below.
From the Greatest Lower Bound Property:
:$\norm A = \inf \set {\lambda > 0: \forall x \in X : \norm {A x}_Y \le \lambda \norm x_X}$ exists.
We have:
The result follows.
Category:Norm on Space of Bounded Linear Transformations
Norm on Bounded Linear Transformation is Submultiplicative

== Theorem ==
Let $\struct {X, \norm \cdot_X}$, $\struct {Y, \norm \cdot_Y}$ and $\struct {Z, \norm \cdot_Z}$ be normed vector spaces.
Let $A : X \to Y$ and $B : Y \to Z$ be bounded linear transformations. 
Let $\norm \cdot _{\map B {X,Y} }$ be the norm for bounded linear transformations $X \to Y$.
Let $\norm \cdot _{\map B {Y,Z} }$ be the norm for bounded linear transformations $Y \to Z$.
Let $\norm \cdot _{\map B {X,Z} }$ be the norm for bounded linear transformations $X \to Z$.
Then, we have that:
:$B \circ A$ is a bounded linear transformation
with:
:$\norm {B \circ A} _{\map B {X,Z} } \le \norm B _{\map B {Y,Z} } \norm A _{\map B {X,Y} }$
That is: 
:$\norm \cdot$ is submultiplicative.
== Proof ==
From Composition of Linear Transformations is Linear Transformation, we have: 
:$B \circ A$ is a linear transformation
Let $x \in X$. 
Then, we have:
We can therefore see that: 
:$B \circ A$ is a bounded linear transformation.
So, if: 
:$\norm x_X = 1$
we have: 
:$\norm {\paren {B \circ A} x}_Z \le \norm B \norm A$
By the definition of supremum, we have: 
:$\ds \sup_{\norm x_X = 1} \norm {\paren {B \circ A} x}_Z \le \norm B \norm A$
So by the definition of the norm on the space of bounded linear transformations, we have: 
:$\norm {B \circ A} \le \norm B \norm A$
Category:Bounded Linear Transformations
Norm on C*-Algebra is Unique

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$. 
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be algebra norms on $\struct {A, \ast}$ such that:
:$\struct {A, \ast, \norm {\, \cdot \,}_1}$ and $\struct {A, \ast, \norm {\, \cdot \,}_2}$ are $\text C^\ast$-algebras.
Then:
:$\norm x_1 = \norm x_2$
for each $x \in A$. 
== Proof ==
Note that the spectrum of $x \in A$ is defined independently of the algebra norm on $A$, depending only on the invertible elements of $A$.
With that:
Category:C*-Algebras
Norm on Space of Bounded Linear Transformations is Norm

== 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 between $X$ and $Y$. 
Let $\norm {\, \cdot \,}_{\map B {X, Y} }$ be the norm on the space of bounded linear transformations.
Then $\norm {\, \cdot \,}_{\map B {X, Y} }$ is indeed a norm on $\map B {X, Y}$.
== Proof ==
From Norm on Bounded Linear Transformation is Finite, $\norm {\, \cdot \,}_{\map B {X, Y} }$ is real-valued.
=== Proof of ===
First, we have: 
:$\norm {\mathbf 0_{\map B {X, Y} } x}_Y = 0$
for each $x \in X$.
So:
:$\set {\norm {\mathbf 0_{\map B {X, Y} } x}_Y : \norm x_X = 1} = \set 0$
That is: 
:$\norm {\mathbf 0_{\map B {X, Y} } }_{\map B {X, Y} } = 0$ 
Now suppose that $T \in \map B {X, Y}$ has $\norm T_{\map B {X, Y} } = 0$.
Then from Fundamental Property of Norm on Bounded Linear Transformation, we have: 
:$\norm {T x}_Y \le \norm T_{\map B {X, Y} } \norm x_X = 0$ 
for each $x \in X$. 
Then from for $\norm {\, \cdot \,}_Y$, we have $T x = 0$ for each $x \in X$.
So $T = \mathbf 0_{\map B {X, Y} }$. 
This proves .
=== Proof of ===
Take $\lambda \in \GF$ with $\lambda \ne 0$, $T \in \map B {X, Y}$ and $x \in X$.
Take $M > 0$ such that: 
:$\norm {T x}_Y \le M \norm x_X$ 
Then:
Taking the infimum over $M > 0$ we have: 
:$\norm {\lambda T x}_Y \le \cmod \lambda \norm T_{\map B {X, Y} } \norm x_X$
So: 
:$\norm {\lambda T}_{\map B {X, Y} } \le \cmod \lambda \norm T_{\map B {X, Y} }$
Now suppose that: 
:$\norm {\lambda T}_{\map B {X, Y} } < \cmod \lambda \norm T_{\map B {X, Y} }$
Then there exists $m < \cmod \lambda \norm T_{\map B {X, Y} }$ such that: 
:$\norm {\lambda T x}_Y \le m \norm x_X$
for each $x \in X$.
Then by for ${\, \norm \,}_Y$ we have: 
:$\ds \norm {T x}_Y \le \frac m {\cmod \lambda} \norm x_X$
where:
:$\ds \frac m {\cmod \lambda} < \norm T_{\map B {X, Y} }$
This contradicts the definition of $\norm T_{\map B {X, Y} }$, so we must have: 
:$\norm {\lambda T}_{\map B {X, Y} } = \cmod \lambda \norm T_{\map B {X, Y} }$
=== Proof of ===
Let $T, S \in \map B {X, Y}$. 
Then, for each $x \in X$ we have: 
Taking the supremum over $\norm x_X = 1$, we have: 
:$\norm {T + S}_{\map B {X, Y} } \le \norm T_{\map B {X, Y} } + \norm S_{\map B {X, Y} }$
Category:Space of Bounded Linear Transformations
Category:Normed Vector Spaces
Norm on Unitization of Normed Algebra is Equivalent to Direct Product Norm

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$ that is not unital as an algebra. 
Let $\struct {A_+, \norm {\, \cdot \,}_1 }$ be the normed unitization of $\struct {A, \norm {\, \cdot \,} }$.
Let $\norm {\, \cdot \,}_2$ be the direct product norm on $A \times \GF$.
Then $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent norms. 
== Proof ==
Let $u, v \in \R$. 
We have:
:$\ds \frac 1 2 \paren {\size u + \size v} \le \frac 1 2 \paren {\max \set {\size u, \size v} + \max \set {\size u, \size v} } = \max \set {\size u, \size v}$
and:
:$\ds \max \set {\size u, \size v} \le \max \set {\size u, \size v} + \min \set {\size u, \size v} = \size u + \size v$
Let $x \in A$ and $\lambda \in \GF$. 
Taking $u = \norm x$ and $v = \cmod \lambda$ in the above we obtain:
:$\ds \frac 1 2 \paren {\norm x + \cmod \lambda} \le \max \set {\norm x, \size \lambda} \le \norm x + \cmod \lambda$
That is:
:$\ds \frac 1 2 \norm {\tuple {x, \lambda} }_1 \le \norm {\tuple {x, \lambda} }_2 \le \norm {\tuple {x, \lambda} }_1$
So $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent norms.
Category:Unitizations of Normed Algebras
Norm on Vector Space is Continuous Function

== Theorem ==
Let $V$ be a vector space with norm $\norm {\, \cdot \,}$.
The function $\norm {\, \cdot \,}: V \to \R$ is continuous.
== Proof ==
Let $x_n \to x$ in $V$.
Then we have:
:$\norm {x_n - x} \to 0$
By the Reverse Triangle Inequality:
:$\size {\norm {x_n} - \norm x} \le \norm {x_n - x}$
Hence:
:$\size {\norm {x_n} - \norm x} \to 0$
Thus:
:$\norm {x_n} \to \norm x$
Hence the result from the definition of a continuous real function.
Category:Norm Theory
Norm on Vector Space is Quasinorm

== Theorem ==
Let $\struct {R, +, \circ}$ be a division ring with norm $\norm {\,\cdot\,}_R$.
Let $V$ be a vector space over $R$.
Let $\norm \cdot$ be a norm on $V$.
Then $\norm \cdot$ is a quasinorm on $V$.
== Proof ==
=== Proof of $(\text Q 1)$ ===
This is precisely $(\text N 2)$ in the definition of a norm. 
=== Proof of $(\text Q 2)$ ===
Since $\norm \cdot$ is a norm on $V$, we have: 
:$\norm {x + y} \le \norm x + \norm y$
for all $x, y \in V$. 
In particular, there exists an $M \ge 1$ with: 
:$\norm {x + y} \le M \paren {\norm x + \norm y}$
So $(\text Q 2)$ holds. 
== Sources ==
* : Appendix G: Basic Functional Analysis
Category:Quasinorms
Norm on Vector Space is Seminorm

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$. 
Let $\norm {\, \cdot \,}$ be a norm on $X$. 
Then $\norm {\, \cdot \,}$ is a seminorm.
== Proof ==
Note that and are precisely and .
Category:Seminorms
Category:Normed Spaces
Norm satisfying Parallelogram Law induced by Inner Product/Real Case/Corollary

== Corollary ==
Let $\struct {V, \norm \cdot}$ be a normed vector space over $\R$.
Then there exists an inner product $\innerprod \cdot \cdot : V \times V \to \R$ such that: 
:$\norm x = \sqrt {\innerprod x x}$
:
:$\norm {x + y}^2 + \norm {x - y}^2 = 2 \paren {\norm x^2 + \norm y^2}$
for each $x, y \in V$.
That is, a norm is induced by an inner product it satisfies the Parallelogram Law for Inner Product Spaces.
== Proof ==
=== Necessary Condition ===
Suppose that there exists an inner product $\innerprod \cdot \cdot : V \times V \to \R$ such that: 
:$\norm x = \sqrt {\innerprod x x}$
Then $\struct {V, \innerprod \cdot \cdot}$ is an inner product space with inner product norm $\norm \cdot$. 
So from the Parallelogram Law for Inner Product Spaces, we have: 
:$\norm {x + y}^2 + \norm {x - y}^2 = 2 \paren {\norm x^2 + \norm y^2}$
for each $x, y \in V$.
=== Sufficient Condition ===
Suppose that: 
:$\norm {x + y}^2 + \norm {x - y}^2 = 2 \paren {\norm x^2 + \norm y^2}$
Then, from Norm satisfying Parallelogram Law induced by Inner Product: Real Case, there exists an inner product $\innerprod \cdot \cdot : V \times V \to \R$ such that: 
:$\norm x = \sqrt {\innerprod x x}$
== Sources ==
* $8.3$: Properties of the Induced Norms 
Category:Norm satisfying Parallelogram Law induced by Inner Product