nyuuzyou/wikis huggingface dataset:
Nonconstant Periodic Function with no Period is Discontinuous Everywhere

== Theorem ==
Let $f$ be a real periodic function that does not have a period.
Then $f$ is either constant or discontinuous everywhere.
== Proof ==
Let $f$ be a real periodic function that does not have a period.
Let $x \in \R$.
If there is no $y \in \R$ such that $\map f x \ne \map f y$, then the result is verified by Constant Function has no Period.
If $f$ is non-constant, then let $y$ be such a value.
Let $\LL_{f > 0}$ be the set of all positive periodic elements of $f$.
This set is non-empty by Absolute Value of Periodic Element is Peroidic.
It is seen that $<$ forms a right-total relation on $\LL_{f > 0}$, for if not then $f$ would have a period.
By the Axiom of Dependent Choice there must exist a strictly-decreasing sequence $\sequence {L_n}$ in $\LL_{f > 0}$.
Since this sequence is bounded below by zero, it follows via Monotone Convergence Theorem that this sequence converges.
Also, by Convergent Sequence is Cauchy Sequence the sequence is Cauchy.
Since $\sequence {L_n}$ is Cauchy, the sequence $\sequence {d_n}_{n \mathop \ge 1}$ formed by taking $d_n = L_n - L_{n - 1}$ is null.
From Combination of Periodic Elements it follows that $\sequence {d_n}_{n \mathop \ge 1}$ is contained in $\LL_{f > 0}$, for every $d_n$ is a periodic element of $f$.
Consider the sequence $\sequence {\paren {x - y} \bmod d_n}_{n \mathop \ge 1}$.
It is seen by Limit of Modulo Operation that this sequence is also null.
Let $\epsilon \in \R_{> 0}$ such that $\epsilon < \size {\map f x - \map f y}$.
For any $\delta \in \R_{> 0}$, there is a $n \in \N_{> 0}$ such that:
But:
And so $f$ is not continuous at $x$.
But our choice of $x$ was completely arbitrary.
Hence the result.
Category:Periodic Functions
Nonconstructive Proof/Examples/Natural Number

== Example of Nonconstructive Proof ==
This is the shape of a proof which is nonconstructive:
:It is false that every number $n$ lacks the property $P$.
:Therefore there exists at least one number $n_0$ that has property $P$.
Unless an example of such a $n_0$ can be constructed, such an argument is not allowed in a constructive proof.
== Sources ==
* 
Category:Examples of Constructive Proofs
Category:Examples of Nonconstructive Proofs
Nondeterministic Turing Machine is Equivalent to Deterministic Turing Machine

== Theorem ==
A nondeterministic turing machines and a deterministic turing machines are equivalent.
== Proof ==
=== Every Deterministic Turing Machine can be run by a Nondeterministic Turing Machine ===
The machine definitions only differ in whether $\tuple {q', Y', d } \in \map \delta {q, Y}$ or $\tuple {q', Y', d} = \map \delta {q, Y}$.
Define singletons $\map \delta {q, Y} = \set {\tuple {q', Y', d} }$ from the deterministic mapping and run your DTM on your NTM.
=== Every Nondeterministic Turing Machine can be run by a Deterministic Turing Machine ===
Albeit slow: sequentialize the parallel search breadth-first.
Define yourself a universal turing machine that can single-step an instantaneous description of your NTM and follow a given NTM transition table like this:
::1. for each applicable transition, copy the instantaneous description to the end of the tape and perform its step.
::2. when the transition table is exhausted, mark the original instantaneous description processed and attend to the next as in 1., unless you halt for simulating an accepting state or having run out of unprocessed configurations.
Category:Turing Machines
Nonempty Grothendieck Universe contains Von Neumann Natural Numbers

== Theorem ==
Let $\mathbb U$ be a non-empty Grothendieck universe.
Let $\N$ denote the set of von Neumann natural numbers.
Then $\N$ is a subset of $\mathbb U$.
== Proof ==
We prove the claim by induction.
=== Basis for the Induction===
By Empty Set is Element of Nonempty Grothendieck Universe $\O \in \mathbb U$.
=== Induction hypothesis ===
For some fixed $n \in \N$, we have $n \in \mathbb U$.
=== Induction Step ===
We have to show, that $n + 1 \in \mathbb U$.
Category:Grothendieck Universes
Nonexistence of Complex Matrices whose Commutator equals Identity

== Theorem ==
Let $d \in \N_{> 0}$ be a positive natural number.
Let $\mathbf A, \mathbf B \in \C^{d \times d}$ be complex matrices.
Let $\mathbf I$ be the $d \times d$ identity matrix.
Then there is no $\mathbf A$, $\mathbf B$ such that $\mathbf A \mathbf B - \mathbf B \mathbf A = \mathbf I$.
== Proof ==
We have that complex numbers form a commutative ring.
 there are $\mathbf A$, $\mathbf B$ such that $\mathbf A \mathbf B - \mathbf B \mathbf A = \mathbf I$.
Then:
This is a contradiction.
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Category:Matrices
Category:Commutativity
Nonexistence of Continuous Linear Transformations over Finite Dimensional Vector Space whose Commutator equals Identity

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be the space of continuous linear transformations.
Let $A, B \in \map {CL} {X, X}$ be continuous linear transformations.
Let $I$ be the identity operator.
Then there is no $A, B$ such that $A \circ B - B \circ A = I$
== Proof ==
 there is $A, B$ such that $A \circ B - B \circ A = I$.
=== Lemma ===
$\forall n \in \N_{> 0} : A \circ B^n - B^n \circ A = n B^{n - 1}$
where
:$B^n = \underbrace{B \circ B \circ \ldots \circ B}_{n \text{ times} }$
and $B^0 = I$.
=== Proof ===
This will be a proof by mathematical induction.
==== Basis for the Induction ====
Let $n = 1$
Then:
==== Induction Hypothesis ====
This is the induction hypothesis:
:$A \circ B^n - B^n \circ A = n B^{n - 1}$
It is to be demonstrated that it follows that:
:$A \circ B^{n + 1} - B^{n + 1} \circ A = \paren {n + 1} B^n$
==== Induction Step ====
Take the supremum operator norm of $A \circ B^n - B^n \circ A = n B^{n - 1}$:
=== Lemma ===
:$\forall n \in \N_{> 0} : B^{n - 1} \ne \mathbf 0$
=== Proof ===
 $\exists n \in \N_{> 0} : B^{n - 1} = \mathbf 0$
For $n = 1$ we have $B^0 = I \ne \mathbf 0$.
Hence, $n \ne 1$.
Suppose:
:$\exists n \in \N_{> 0} : B^{n - 1} \ne \mathbf 0$
If $B^n = \mathbf 0$ then:
Altogether:
:$\forall n \in \N_{> 0} : \paren {B^{n - 1} \ne \mathbf 0} \implies \paren {B^n \ne \mathbf 0}$
This is a contradiction.
Hence: 
:$\forall n \in \N_{> 0} : \norm {B^{n - 1} } > 0$.
Thus: 
:$\forall n \in \N_{> 0} : n \le 2 \norm A \norm B$
This holds only if $\norm A \norm B = \infty$.
However, the image of $\norm {\, \cdot \,}$ is a subset of $\R$:
:$\Img {\norm {\, \cdot \,}} \subseteq \R$
and does not contain the infinity.
This is a contradiction.
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Category:Continuous Linear Transformations
Category:Commutativity
Nonlimit Ordinal Cofinal to One

== Theorem ==
Let $x$ be a nonlimit non-empty ordinal.
Let $\operatorname{cof}$ denote the cofinal relation. 
Let $1$ denote the ordinal one.
Then:
:$\operatorname{cof} \left({x, 1}\right)$
== Proof ==
Since $1 = 0^+$, $1$ is not a limit ordinal.
Let $\le$ denote the subset relation.
It follows that $0 < 1$ by Ordinal is Less than Successor.
Moreover, $1 \le x$ follows by the fact that $0 < x$ and Successor of Element of Ordinal is Subset.
Thus we have $0 < 1 \le x$ and so by Condition for Cofinal Nonlimit Ordinals:
:$\operatorname{cof} \left({x, 1}\right)$
== Sources ==
* : $\S 10.54$
Category:Ordinals
Nonparametric Method/Examples/Arbitrary Example 1

== Example of Nonparametric Method ==
Suppose we have random samples from two populations with unspecified cumulative distribution functions $\map F x$ and $\map G x$.
A test for:
:the hypothesis $\map F x = \map G x$ for all $x$
against:
:the alternative hypothesis that $\map F x < \map G x$ for all $x$
would be a test that is both nonparametric and distribution-free.
== Sources ==
* 
* 
Category:Examples of Nonparametric Methods
Nonparametric Method/Examples/Arbitrary Example 2

== Example of Nonparametric Method ==
Suppose we have random samples from two populations with unspecified cumulative distribution functions $\map F x$ and $\map G x$.
A test for:
:the hypothesis $\map G x = \map F x$ for all $x$
against:
:the alternative hypothesis that $\map G x < \map F {x - \theta}$ for all $x$, where $\theta$ is some parameter
would be a test that is distribution-free but not nonparametric, as it involves a parameter $\theta$.
== Sources ==
* 
* 
Category:Examples of Nonparametric Methods
Nonsingular Matrix corresponds to Automorphism

== Theorem ==
Let $R$ be a ring with unity.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $G$ be an $n$-dimensional $R$-module.
Let $\map {\MM_R} n$ be the $n \times n$ matrix space over $R$.
Let $\map {\LL_R} G$ be the set of all linear operators on $G$.
Then the nonsingular elements of the ring of square matrices $\struct {\map {\MM_R} n, +, \times}$ correspond directly to automorphisms of $\map {\LL_R} G$.
== Proof ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Nonsingular Matrices
Nontrivial Zeroes of Riemann Zeta Function are Symmetrical with respect to Critical Line

== Theorem ==
The nontrivial zeroes of the Riemann $\zeta$ function are distributed symmetrically the critical line.
That is, suppose $s_1 = \sigma_1 + i t$ is a nontrivial zero of $\map \zeta s$.
Then there exists another nontrivial zero $s_2$ of $\map \zeta s$ such that:
:$s_2 = 1 - s_1$
== Proof ==
From Functional Equation for Riemann Zeta Function, we have:
:$\ds \pi^{-s/2} \map \Gamma {\dfrac s 2} \map \zeta s = \pi^{\paren {s/2 - 1/2 } } \map \Gamma {\dfrac {1 - s} 2} \map \zeta {1 - s}$
We suppose $s_1 = \sigma_1 + i t$ is a nontrivial zero of $\map \zeta s$.
Then we have:
It remains to be shown that of the three terms on the , $\map \zeta {1 - s_1}$ MUST equal zero.
We can rewrite the first term in terms of the exponential function
:$\ds \pi^{\paren {s_1 / 2 - 1 / 2} } = \map \exp {\map \ln {\pi^{\paren {s_1 / 2 - 1 / 2} } } }$
From the definition of the exponential function, we know $\map \exp {\map \ln {\pi^{\paren { {s_1}/2 - 1/2 } } }}$ never equals zero.
From Zeroes of Gamma Function, we know $\map \Gamma {\dfrac {1 - s_1} 2}$ never equals zero.
Therefore:
:$\map \zeta {1 - s_1} = 0$
== Also see ==
* Riemann Hypothesis: If this is true, all nontrivial zeroes are already on the critical line.
== Sources ==
* : $0,5$
Category:Riemann Zeta Function
Nonzero Eigenvalue of Compact Operator has Finite Dimensional Eigenspace

== Theorem ==
Let $H$ be a Hilbert space.
Let $T \in \map {B_0} H$ be a compact operator.
Let $\lambda \in \map {\sigma_p} T, \lambda \ne 0$ be a nonzero eigenvalue of $T$.
Then the eigenspace for $\lambda$ has finite dimension.
== Proof ==
$\def \sequence#1{\left({#1}\right)}$
Note that in the following, the notation for a sequence of terms in the form $e_n$ has been changed for this particular page to be $\sequence {e_n}$ rather than the standard $\left\langle{e_n}\right\rangle$.
This is in order to avoid potential confusion with the notation $\innerprod {\lambda e_{n_k} } {\lambda e_{n_p} }$ for inner product.
Assume that there is an infinite orthonormal sequence $\sequence {e_j}$ in my eigenspace for the eigenvalue $\lambda$.
That is:
:$\sequence {e_j: j \in J} \subset \map \ker {T - \lambda}$
Since $\norm {e_j} = 1$, then $\sequence {T e_n}$ will be in the closure image of $T$ of the closed unit ball.
Since the closure of this image is compact by assumption, then there is a subsequence $\sequence {T e_{n_k} }$ that converges, hence $\sequence {T e_{n_k} }$ is Cauchy.
Notice that
:$\norm {T e_{n_k} - Te_{n_p} }^2 = \norm {\lambda e_{n_k} - \lambda e_{n_p} }^2 = 2 \size \lambda^2 > 0$
for $k \ne p$, we are using the fact that $\lambda \ne 0$ and $\innerprod {\lambda e_{n_k} } {\lambda e_{n_p} } = 0$ (since $\sequence {e_j}$ is an orthonormal sequence).
Since $\lambda$ is constant, then it contradicts the fact that $\sequence {e_{n_k} }$ is Cauchy.
Thus $\map \ker {T - \lambda}$ is finite-dimensional.
== Sources ==
* $\text{II}.4.13$
Category:Linear Transformations on Hilbert Spaces
Nonzero Ideal of Polynomial Ring over Field has Unique Monic Generator

== Theorem ==
Let $K$ be a field.
Let $K \sqbrk x$ be the polynomial ring in one variable over $K$.
Let $I \subseteq K \sqbrk x$ be a nonzero ideal.
Then $I$ is generated by a unique monic polynomial.
== Proof ==
Category:Polynomial Rings
Nonzero Natural Number is Successor

== Theorem ==
Let $\N$ be the $0$-based natural numbers:
:$\N = \set {0, 1, 2, \ldots}$
Let $s: \N \to \N: \map s n = n + 1$ be the successor mapping.
Then:
:$\forall n \in \N \setminus \set 0 \paren {\exists m \in \N: \map s m = n}$
== Proof ==
The proof will proceed by the Principle of Finite Induction on $\N \setminus \set 0$.
=== Basis for the Induction ===
:$\map s 0 = 1$
So $1$ is the immediate successor of $0$.
This is our basis for the induction.
=== Induction Hypothesis ===
This is our induction hypothesis:
:$\exists m \in \N: \map s m = k$
Then we need to show:
:$\exists m' \in \N: \map s {m'} = k + 1$
=== Induction Step ===
This is our induction step:
Hence the result by the Principle of Finite Induction.
Category:Proofs by Induction
Category:Natural Numbers
Norm Equivalence Preserves Completeness

== Theorem ==
Let $R$ be a normed division ring.
Let $X$ be a vector space over $R$. 
Let $\norm \cdot$ and $\norm \cdot'$ be equivalent norms on $X$.
Then $\struct {X, \norm \cdot}$ is a Banach space $\struct {X, \norm \cdot'}$ is a Banach space. 
== Proof ==
Suppose that $\struct {X, \norm \cdot}$ is a Banach space.
Let $\sequence {x_n}_{n \in \N}$ be a Cauchy sequence in $\struct {X, \norm \cdot'}$.
From Cauchy Sequences in Vector Spaces with Equivalent Norms Coincide, $\sequence {x_n}_{n \in \N}$ is a Cauchy sequence in $\struct {X, \norm \cdot}$.
Since $\struct {X, \norm \cdot}$ is a Banach space, there exists $x \in X$ such that $x_n \to x$ in $\struct {X, \norm \cdot}$.
From Convergent Sequences in Vector Spaces with Equivalent Norms Coincide, we also have $x_n \to x$ in $\struct {X, \norm \cdot'}$.
So if $\struct {X, \norm \cdot}$ is a Banach space, then so is $\struct {X, \norm \cdot'}$.
Swapping $\struct {X, \norm \cdot}$ and $\struct {X, \norm \cdot'}$ in the previous argument shows that if $\struct {X, \norm \cdot'}$ is a Banach space, so is $\struct {X, \norm \cdot}$.
Category:Banach Spaces
Norm Equivalence is Equivalence

== Theorem ==
Let $X$ be a vector space.
Let $\norm {\, \cdot \,}_a$ and $\norm {\, \cdot \,}_b$ be equivalent norms on $X$.
Denote this relation by $\sim$: 
:$\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$.
Then $\sim$ is an equivalence relation.
== Proof ==
=== Reflexivity ===
Let $\norm {\, \cdot \,}$ be a norm on $X$.
Then for all $x \in X$ we have that:
:$\norm x = 1 \cdot \norm {x}$
Therefore:
:$1 \cdot \norm x \le \norm x \le 1 \cdot \norm x$
Hence:
:$\norm {\, \cdot \,} \sim \norm {\, \cdot \,}$.
=== Symmetry ===
Suppose, $\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$.
Then:
:$\exists m, M \in \R_{> 0} : m \le M : \forall x \in X : m \norm x_b \le \norm x_a \le M \norm x_b$
Consider two inequalities, obtained by division by $M$ and $m$:
:$\dfrac m M \norm x_b \le \dfrac 1 M \norm x_a \le \norm x_b$
:$\norm x_b \le \dfrac 1 m \norm x_a \le \dfrac M m \norm x_b$
Notice, that: 
:$\dfrac 1 M \norm x_a \le \norm x_b \le \dfrac 1 m \norm x_a$
We have that $m \le M$ implies $\dfrac 1 m \ge \dfrac 1 M$.
Define $C := \dfrac 1 m$ and $c := \dfrac 1 M$.
Hence, $c \le C$ and:
:$\norm {\, \cdot \,}_b \sim \norm {\, \cdot \,}_a$
=== Transitivity ===
Suppose, $\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$ and $\norm {\, \cdot \,}_b \sim \norm {\, \cdot \,}_c$.
 
Then:
:$\exists m_{a b}, M_{a b} \in \R_{> 0} : m_{a b} \le M_{a b} : \forall x \in X : m_{a b} \norm x_b \le \norm x_a \le M_{a b} \norm x_b$ 
:$\exists m_{b c}, M_{b c} \in \R_{> 0} : m_{b c} \le M_{b c} : \forall x \in X : m_{b c} \norm x_c \le \norm x_b \le M_{b c} \norm x_c$
Generate $2$ more inequalities by multiplying the second inequality by $m_{a b}$ and $M_{a b}$:
:$m_{a b} m_{b c} \norm x_c \le m_{a b} \norm x_b \le m_{a b} M_{b c} \norm x_c$
:$M_{a b} m_{b c} \norm x_c \le M_{a b} \norm x_b \le M_{a b} M_{b c} \norm x_c$
From above it follows that:
Define $k := m_{a b} m_{b c}$ and $K := M_{a b} M_{b c}$.
Then $k \le K$ and:
:$k \norm x_c \le \norm x_a \le K \norm x_c$
Therefore:
:$\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_c$
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Examples of Equivalence Relations
Category:Norm Theory
Category:Vector Spaces
Norm Sequence of Cauchy Sequence has Limit

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$ be a Cauchy sequence in $R$.
Then $\sequence {\norm {x_n} }$ has a limit in $\R$.
That is,
:$\exists l \in \R: \ds \lim_{n \mathop \to \infty} \norm {x_n} = l$
== Proof ==
It is first shown that $\sequence {\norm {x_n} }$ is a real Cauchy sequence in $\R$.
Let $\epsilon \in \R_{>0}$ be given.
By the definition of Cauchy sequence then:
:$\exists N \in \N: \forall n, m > N, \norm {x_n - x_m} < \epsilon$
By Reverse Triangle Inequality on Normed Division Ring, then:
:$\forall n, m > N: \cmod {\norm {x_n} - \norm {x_m} } \le \norm {x_n - x_m} < \epsilon$
From the definition of a real Cauchy sequence it follows that $\sequence {\norm {x_n} }$ is a real Cauchy sequence in $\R$.
By Cauchy's Convergence Criterion on Real Numbers $\sequence {\norm {x_n} }$ has a limit in $\R$.
== Sources ==
Category:Cauchy Sequences
Category:Normed Division Rings
Norm Topology Induced by Metric Induced by Norm

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$. 
Let $d$ be the metric induced on $X$ by $\norm {\, \cdot \,}$.
Then the open sets of $\struct {X, d}$ as a metric space are precisely the open sets of $\struct {X, \norm {\, \cdot \,} }$ as a normed vector space. 
== Proof ==
For $\epsilon > 0$ and $x \in X$, let $\map {B_\epsilon^d} x$ be the open ball in $\struct {X, d}$ with radius $\epsilon$ and center $x$. 
For $\epsilon > 0$ and $x \in X$, let $\map {B_\epsilon^{\norm {\, \cdot \,} } } x$ be the open ball in $\struct {X, \norm {\, \cdot \,} }$ with radius $\epsilon$ and center $x$. 
It is therefore sufficient to show that $\map {B_\epsilon^d} x = \map {B_\epsilon^{\norm {\, \cdot \,} } } x$.
Let $y \in X$.
We have $y \in \map {B_\epsilon^d} x$ :
:$\map d {x, y} < \epsilon$
This is equivalent to:
:$\norm {x - y} < \epsilon$
This is then equivalent to:
:$y \in \map {B_\epsilon^{\norm {\, \cdot \,} } } x$
Hence we have: 
:$\map {B_\epsilon^d} x = \map {B_\epsilon^{\norm {\, \cdot \,} } } x$
Category:Normed Vector Spaces
Category:Metric Spaces
Norm in terms of Normed Dual Space

== Theorem ==
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: 
:$\ds \norm x = \sup_{f \in B_{X^\ast}^-} \cmod {\map f x}$
Category:Normed Dual Spaces
Norm is Complete Iff Equivalent Norm is Complete

== Theorem ==
Let $R$ be a division ring.
Let $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ be equivalent norms on $R$.
Then:
:$\struct {R,\norm {\,\cdot\,}_1}$ is complete $\struct {R,\norm {\,\cdot\,}_2}$ is complete.
== Proof ==
By Cauchy Sequence Equivalence, for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_1$ $\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_2$.
By Convergent Equivalence, for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ converges in $\norm{\,\cdot\,}_1$ $\sequence {x_n}$ converges in $\norm{\,\cdot\,}_2$.
Hence:
:every Cauchy sequence in $\norm{\,\cdot\,}_1$ converges in $\norm{\,\cdot\,}_1$ every Cauchy sequence in $\norm{\,\cdot\,}_2$ converges in $\norm{\,\cdot\,}_2$.
The result follows.
Category:Normed Division Rings
Category:Complete Metric Spaces
Norm is Continuous

== Theorem ==
Let $\struct {V, \norm {\,\cdot\,} }$ be a normed vector space.
Then the mapping $x \mapsto \norm x$ is continuous.
Here, the metric used is the metric $d$ induced by $\norm {\,\cdot\,}$.
== Proof ==
Since $\norm x = \map d {x, \mathbf 0}$, the result follows directly from Distance Function of Metric Space is Continuous.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 1.$ Elementary Properties and Examples
Category:Norm Theory
Category:Continuous Mappings
Norm of Adjoint

== Theorem ==
Let $H, K$ be Hilbert spaces.
Let $A \in \map B {H, K}$ be a bounded linear transformation.
Let $A^* \in \map B {K, H}$ be the adjoint of $A$.
Then $A$ and $A^*$ satisfy:
:$\norm A_{\map B {H, K} }^2 = \norm {A^*}_{\map B {K, H} }^2 = \norm {A^* A}_{\map B {H, H} }$
where:
:$\norm \cdot_{\map B {H, K} }$ denotes the operator norm on $\map B {H, K}$
:$\norm \cdot_{\map B {K, H} }$ denotes the operator norm on $\map B {K, H}$
:$\norm \cdot_{\map B {H, H} }$ denotes the operator norm on $\map B {H, H}$
== Proof ==
Let $h \in H$ such that $\norm h_H \le 1$.
Then:
By definition $(1)$ for $\norm A$, it follows that:
:$\norm A^2 \le \norm {A^* A} \le \norm {A^*} \norm A$
That is:
:$\norm A \le \norm {A^*}$.
By substituting $A^*$ for $A$, and using $A^{**} = A$ from Adjoint is Involutive, the reverse inequality is obtained.
Hence $\norm A^2 = \norm {A^* A} = \norm {A^*}^2$.
== Sources ==
* $\text {II}.2.7$
* $13.1$: Existence of Hilbert Adjoint
Category:Adjoints
Norm of Bounded Linear Transformation in terms of Closed Unit Ball

== 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 $T : X \to Y$ 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 $\norm T_{\map B {X, Y} } \le M$ :
:$T \sqbrk {B_X^-} \subseteq M B_Y^-$
== 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: 
:$T \sqbrk {B_X^-} \subseteq M B_Y^-$
=== Sufficient Condition ===
Suppose that:
:$T \sqbrk {B_X^-} \subseteq M B_Y^-$
Let $x \in X \setminus \set { {\mathbf 0}_X}$
Then, we have: 
:$\ds \frac x {\norm x_X} \in B_X^-$
and hence: 
:$\ds \norm {\map T {\frac x {\norm x_X} } }_Y \le M$
From , we have:
:$\ds \norm {T x}_Y \le M \norm x_X$
Clearly this also holds for $x = {\mathbf 0}_X$.
So $T$ is bounded with $\norm T_{\map B {X, Y} } \le M$.
Category:Bounded Linear Transformations
Norm of Continuous Function is Continuous

== 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 $f : X \to Y$ be a continuous mapping.
Define $\norm f_Y : X \to \hointr 0 \infty$ by: 
:$\map {\paren {\norm f_Y} } x = \norm {\map f x}_Y$
for each $x \in X$. 
Then $\norm f_Y$ is continuous.
== Proof ==
Follows immediately from combining Norm on Vector Space is Continuous Function and Composite of Continuous Mappings is Continuous.
Category:Normed Vector Spaces