nyuuzyou/wikis huggingface dataset:
Bounded Linear Operator on Hilbert Space Direct Sum

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\sequence {\family {\HH_i, \innerprod \cdot \cdot_i} }_{i \mathop \in I}$ be a $I$-indexed family of Hilbert spaces over $\GF$.
For each $i \in I$, let $T_i : \HH_i \to \HH_i$ be a bounded linear operator.
Suppose that:
:$\ds \sup_{i \mathop \in I} \norm {T_i}_{\map B {\HH_i} } < \infty$
where $\norm {\, \cdot \,}_{\map B {\HH_i} }$ is the norm of a bounded linear operator on $\HH_i$.
Let:
:$\ds \HH = \bigoplus_{i \mathop \in I} \HH_i$
be the Hilbert space direct sum of $\sequence {\family {\HH_i, \innerprod \cdot \cdot_i} }_{i \mathop \in I}$ with norm $\norm {\, \cdot \,}$.
Define $T : \HH \to \HH$ by:
:$\map {\paren {T f} } i = \map {T_i} {\map f i}$
for each $f \in \HH$ and $i \in I$.
Then $T$ is a bounded linear operator.
Further:
:$\norm T_{\map B \HH} = \sup_{i \mathop \in I} \norm {T_i}_{\map B {\HH_i} }$
== Proof ==
We first verify that $T$ is well-defined.
Let:
:$\ds M = \sup_{i \mathop \in I} \norm {T_i}_{\map B {\HH_i} }$
For this we want to verify that for each $f \in \HH$ we have:
:$\ds \sum \set {\norm {\map {T_i} {\map f i} }_i^2 : i \in I} < \infty$
From Fundamental Property of Norm on Bounded Linear Transformation, we have:
:$\norm {\map {T_i} {\map f i} }_i \le \norm {T_i}_{\map B {\HH_i} } \norm {\map f i}_i \le M \norm {\map f i}_i$
for each $f \in \HH$.
From the definition of $\HH$, we have:
:$\ds \sum \set {\norm {\map f i}_i^2 : i \in I} < \infty$
Hence from Generalized Sum Preserves Inequality we have:
:$\ds \sum \set {\norm {\map {T_i} {\map f i} }_i^2 : i \in I} < \infty$
as desired.
We have then just shown:
Hence:
:$\norm {T f} \le M \norm f$ 
We show that $T$ is linear.
Let $f, g \in \HH$ and $\lambda \in \GF$. 
We then have for each $i \in I$:
Hence:
:$\map T {f + \lambda g} = T f + \lambda T g$ 
We finally determine the norm of $T$. 
We have already shown that:
:$\norm {T f} \le M \norm f$ for each $f \in \HH$.
Hence:
:$\norm T_{\map B \HH} \le M$ 
We now show that:
:$\norm T_{\map B \HH} \ge M$
Let $0 < \epsilon < M$. 
By the definition of supremum, there exists $i \in I$ such that:
:$\norm {T_i}_{\map \BB {H_i} } > M - \epsilon$ 
Then there exists $e_i \in \HH$ with $\norm {e_i}_i = 1$ such that:
:$\norm {T_i e_i}_{\map \BB {H_i} } > M - \epsilon$ 
Define $f \in \HH$ by $\map f i = e_i$ and $\map f j = {\mathbf 0}_\HH$ for $j \ne i$. 
We have:
So $\norm f = 1$. 
Further $\map {\paren {T f} } i = T_i e_i$, while $\map {\paren {T f} } j = {\mathbf 0}_\HH$ for $j \ne i$. 
Hence we have:
Hence we have $\norm {T f} > M - \epsilon$.
So $\norm T_{\map B \HH} > M - \epsilon$ for each $0 < \epsilon < M$.
Hence $\norm T_{\map B \HH} \ge M$. 
So we conclude $\norm T_{\map B \HH} = M$.
Category:Direct Sums of Hilbert Spaces
Bounded Linear Transformation Induces Bounded Sesquilinear Form

== Theorem ==
Let $\Bbb F$ be a subfield of $\C$. 
Let $\struct {V, \innerprod \cdot \cdot_V}$ and $\struct {U, \innerprod \cdot \cdot_U}$ be inner product spaces over $\Bbb F$. 
Let $A : V \to U$ and $B : U \to V$ be bounded linear transformations.
Let $u, v: V \times U \to \C$ be defined by:
:$\map u {h, k} := \innerprod {Ah} k_U$
:$\map v {h, k} := \innerprod h {Bk}_V$
Then $u$ and $v$ are bounded sesquilinear forms.
== Proof ==
We first show that $u$ and $v$ are sesquilinear, and then that they are bounded. 
Let $\alpha \in \mathbb F$ and $h_1, h_2 \in V$ and $k \in U$. 
We have: 
and:
Now, let $h \in V$ and $k_1, k_2 \in U$. 
We have: 
and:
So $u$ and $v$ are both sesquilinear. 
It remains to show that they are bounded.
Let $\norm \cdot_V$ be the inner product norm of $V$.
Let $\norm \cdot_U$ be the inner product norm of $U$.
Let $\norm A$ denote the norm on $A$.
We have that $A$ is a bounded linear transformation.
From Norm on Bounded Linear Transformation is Finite:
:$\norm A$ is finite.
Similarly, since $B$ is a bounded linear transformations, we have that:
:$\norm B$ is finite.
We then have, for all $h \in V$ and $k \in U$: 
so $u$ is bounded.
Similarly:
so $v$ is also bounded.
== Also see ==
* Classification of Bounded Sesquilinear Forms, which states that all sesquilinear forms are of this type.
== Sources ==
* $\S \text {II}.2$
Category:Linear Transformations on Hilbert Spaces
Bounded Linear Transformation is Into Linear Isomorphism iff Dual Operator is Surjective

== Theorem ==
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 $T^\ast$ is surjective. 
== Proof ==
Let $B_{X^\ast}^-$ and $B_{Y^\ast}^-$ be the closed unit balls of $X^\ast$ and $Y^\ast$ respectively. 
=== Necessary Condition ===
We want to show that $T$ is a linear isomorphism considered as a map $X \to T \sqbrk X$. 
Suppose that $T^\ast$ is surjective.
From Dual Operator is Bounded Linear Transformation, $T^\ast$ is a bounded linear transformation.
From Banach-Schauder Theorem, $T^\ast$ is an open mapping.
From Characterization of Open Linear Transformation between Normed Vector Spaces, there exists $\delta > 0$ such that $\delta B_{X^\ast}^- \subseteq T^\ast \sqbrk {B_{Y^\ast}^-}$.
Now, for each $x \in X$ we have: 
Hence $\map \ker T = \set { {\mathbf 0}_X}$.
So from Linear Transformation is Injective iff Kernel Contains Only Zero, we have that $T$ is injective, hence bijective as a map $X \to T \sqbrk X$.
From Characterization of Invertible Bounded Linear Transformations, $T^{-1}$ is a bounded linear transformation.
Hence $T$ is a linear isomorphism as a map $X \to T \sqbrk X$, hence an into linear isomorphism.
=== Sufficient Condition ===
Suppose that $T$ is an into linear isomorphism. 
Then $T^{-1} : T \sqbrk X \to X$ is a bounded linear transformation. 
Let $f \in X^\ast$. 
From Composite of Continuous Mappings is Continuous, $f \circ T^{-1} : T \sqbrk X \to \GF$ is continuous.
By Continuity of Linear Functionals, $f \circ T^{-1}$ is bounded.
From the Hahn-Banach Theorem, there exists $g \in Y^\ast$ such that: 
:$\map g y = \map f {T^{-1} y}$
for each $y \in T \sqbrk X$. 
Then for $x \in X$, we have:
So we have: 
:$T^\ast g = f$ 
So $T^\ast$ is surjective.
== Sources ==
* : Exercise $2.39 \text { (i)}$
Category:Linear Isomorphisms
Category:Dual Operators
Bounded Linear Transformation is Isometry iff Adjoint is Left-Inverse

== Theorem ==
Let $\HH, \KK$ be Hilbert spaces.
Let $A \in \map B{\HH, \KK}$ be a bounded linear transformation.
Then $A$ is an isometry :
:$A^*A = I_\HH$
where $A^*$ denotes the adjoint of $A$, and $I_\HH$ the identity operator on $\HH$.
== Proof ==
Let $g, h \in \HH$. Then by the definition of adjoint:
:${\innerprod {A g} {A h} }_\KK = {\innerprod {A^* A g} h}_\HH$
From the uniqueness of the adjoint, it follows that:
:${\innerprod {A g} {A h} }_\KK = {\innerprod g h}_\HH$
holds $A^*A = I_\HH$.
Hence the result by definition of isometry.
== Sources ==
* $\text {II}.2.17$
Category:Adjoints
Category:Linear Transformations on Hilbert Spaces
Bounded Linear Transformation preserves Cauchy Sequences

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces.
Let $T : X \to Y$ be a bounded linear transformation.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $X$.
Then $\sequence {T x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $Y$. 
== Proof ==
Since $T$ is a bounded linear transformation, there exists $M > 0$ such that: 
:$\norm {T x}_Y \le M \norm x_X$
for all $x \in X$. 
So, since $T$ is linear, we have:
:$\norm {T x_n - T x_m}_Y \le M \norm {x_n - x_m}_X$
Let $\epsilon > 0$.
Since $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy, there exists $N \in \N$ such that: 
:$\ds \norm {x_n - x_m}_X < \frac \epsilon M$ 
for $n, m \ge N$. 
Then: 
:$\ds \norm {T x_n - T x_m} < M \paren {\frac \epsilon M} = \epsilon$ 
for $n, m \ge N$. 
So $\sequence {T x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $Y$.
Category:Bounded Linear Transformations
Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain/Corollary 1

== Corollary to Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\Bbb F$.
Let $\map D {f_0}$ be a linear subspace of $X$.
Let $f_0 : \map D {f_0} \to \Bbb F$ be a bounded linear functional.
Then there exists a unique bounded linear functional $f : \map D f \to \Bbb F$ extending $f_0$ to $\map D f = \paren {\map D {f_0} }^-$.
== Proof ==
This is a direct consequence of Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain and: 
:Real Number Line is Banach Space if $\Bbb F = \R$
:Complex Plane is Banach Space if $\Bbb F = \C$.
Category:Bounded Linear Transformations
Category:Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain
Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain/Corollary 2

== Corollary to Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot_X}$ be a normed vector space.
Let $\struct {Y, \norm \cdot_Y}$ be a Banach space.
Let $D$ be an everywhere dense linear subspace of $X$.
Let $T_1 : X \to Y$ and $T_2 : X \to Y$ be bounded linear transformations with: 
:$T_1 x = T_2 x$ for all $x \in D$.
Then: 
:$T_1 = T_2$
== Proof ==
Define $T_0 : D \to Y$ by: 
:$T_0 x = T_1 x = T_2 x$ for all $x \in D$.
Then $T_0$ is a bounded linear transformation, and so from Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain there exists a unique bounded linear transformation extending $T_0$ to $X$.
Since $T_1$ and $T_2$ are both bounded linear transformation extending $T_0$ to $X$, so we must have $T_1 = T_2$.
Category:Bounded Linear Transformations
Category:Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain
Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain/Corollary 3

== Corollary ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\Bbb F$.
Let $\map D {T_0}$ be a linear subspace of $X$.
Let $\struct {Y, \norm \cdot_Y}$ be a Banach space over $\Bbb F$.
Let $T_0 : \map D {T_0} \to Y$ be a linear isometry.
Then there exists a unique linear isometry $T : \map D T \to Y$ extending $T_0$ to $\map D T = \paren {\map D {T_0} }^-$. 
== Proof ==
From Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain, there exists a unique bounded linear transformation $T : \map D T \to Y$ extending $T_0$ to $\map D T = \paren {\map D {T_0} }^-$. 
For $x \in \paren {\map D {T_0} }^-$ and $\sequence {x_n}_{n \mathop \in \N}$ in $\map D {T_0}$ such that $x_n \to x$ we have:
:$\ds T x = \lim_{n \mathop \to \infty} T_0 x_n$
Since $T_0$ is an isometry, we have:
:$\norm {T_0 x_n}_Y = \norm {x_n}_X$ 
Taking $n \to \infty$, we have $x_n \to x$ and $T_0 x_n \to T x$ and hence by Norm is Continuous:
:$\norm {T x}_Y = \norm x_X$
So $T$ is a linear isometry.
Category:Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain
Bounded Metric Space is not necessarily Totally Bounded

== Theorem ==
Let $M = \struct {A, d}$ be a bounded metric space.
Then it is not necessarily the case that $M$ is totally bounded.
== Proof 1 ==
== Proof 2 ==
Category:Bounded Metric Space is not necessarily Totally Bounded
Category:Bounded Metric Spaces
Category:Totally Bounded Metric Spaces
Bounded Metric Space is not necessarily Totally Bounded/Proof 1

== Theorem ==
== Proof ==
Let $M = \struct {\R, d}$ be the real number line with the Euclidean metric.
Let $M' = \struct {\R, \delta}$ be the unity-bounded metric space on $M$ where $\delta$ is defined as:
:$\delta = \dfrac d {1 + d}$
From Unity-Bounded Metric Space is Bounded, $M'$ is a bounded metric space.
From Unity-Bounded Metric Space on Real Number Line is not Totally Bounded, $M'$ is not a totally bounded metric space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Bounded Metric Space is not necessarily Totally Bounded
Bounded Metric Space is not necessarily Totally Bounded/Proof 2

== Theorem ==
== Proof ==
Let $d$ be a discrete metric on the open unit interval $\Bbb I := \openint 0 1 \subseteq \R$.
We have that for all $x \in \openint 0 1$ and for all $r \in \R_{> 1}$:
:$\map {B_r} x = \openint 0 1$
where $\map {B_r} x$ denotes the open $r$-ball of $x$.
Thus $\struct {\Bbb I, d}$ is bounded.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number such that $\epsilon < 1$.
Let $x \in \openint 0 1$ be arbitrary.
Then $\map {B_\epsilon} x$ contains $x$ alone.
So, there is no finite $\epsilon$-net for $\openint 0 1$.
By definition of totally bounded metric space, $\struct {\Bbb I, d}$ is not totally bounded.
Category:Bounded Metric Space is not necessarily Totally Bounded
Bounded Piecewise Continuous Function has Improper Integrals

== Theorem ==
Let $f$ be a real function defined on a closed interval $\closedint a b$, $a < b$. 
Let $f$ be piecewise continuous and bounded on $\closedint a b$.
Then $f$ is a piecewise continuous function with improper integrals.
== Proof ==
Let $f$ be piecewise continuous and bounded on $\closedint a b$.
It is sufficient to prove that the improper integral $\ds \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists for every $i \in \set {1, 2, \ldots, n}$.
Let $i \in \set {1, 2, \ldots, n}$.
Let $c$ be a point in $\openint {x_{i − 1} } {x_i}$.
By definition, the improper integral $\ds \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists :
:$\ds \lim_{\gamma \mathop \to {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$
and:
:$\ds \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$
both exist.
By Bounded Piecewise Continuous Function is Darboux Integrable we know that $f$ is Darboux integrable on $\closedint a b$.
Therefore, $f$ is Darboux integrable on every closed subinterval of $\closedint a b$. 
Accordingly, the following Darboux integrals:
:$\ds \int_{x_{i - 1} }^c \map f x \rd x$
:$\ds \int_{\gamma_1}^c \map f x \rd x$
:$\ds \int_c^{\gamma_2} \map f x \rd x$
:$\ds \int_c^{x_i} \map f x \rd x$
all exist where $\gamma_1 \in \openint {x_{i − 1} } c$ and $\gamma_2 \in \openint c {x_i}$.
Note that $f$ is bounded on $\closedint a b$ as $f$ is piecewise continuous and bounded.
Therefore, a bound $B$ exists for $f$ on $\closedint a b$.
We have:
which approaches $0$ as $\gamma_1$ approaches $x_{i − 1}$.
This shows that $\ds \lim_{\gamma \mathop \to x_{i - 1}+} \int_\gamma^c \map f x \rd x$ equals $\ds \int_{x_{i - 1} }^c \map f x \rd x$.
From this we gather that $\ds \lim_{\gamma \mathop \to {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$ exists.
Also:
which approaches $0$ as $\gamma_2$ approaches $x_i$.
This shows that $\ds \lim_{\gamma \mathop \to x_i-} \int_c^\gamma \map f x \rd x$ equals $\ds \int_c^{x_i} \map f x \rd x$.
From this we gather that $\ds \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$ exists.
Since $i$ is arbitrary, we have shown that $\ds \lim_{\gamma \mathop \to {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$ and $\ds \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$ exist for every $i \in \set {1, \ldots, n}$.
Therefore, $\ds \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists for every $i \in \set {1, 2, \ldots, n}$.
That is, $f$ is a piecewise continuous function with improper integrals.
== Also see ==
* Piecewise Continuous Function with Improper Integrals may not be Bounded
Category:Piecewise Continuous Functions
Bounded Piecewise Continuous Function is Darboux Integrable

== Theorem ==
Let $f$ be a real function defined on the closed interval $\closedint a b$.
Let $f$ be piecewise continuous and bounded on $\closedint a b$.
Then $f$ is Darboux integrable on $\closedint a b$.
== Proof ==
We are given that $f$ is piecewise continuous and bounded on $\closedint a b$.
Therefore, there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that for all $i \in \set {1, 2, \ldots, n}$:
:$f$ is continuous on $\openint {x_{i - 1} } {x_i}$
:$f$ is bounded on $\closedint {x_{i - 1} } {x_i}$.
Note that $n$ is the number of intervals $\openint {x_{i - 1} } {x_i}$ defined from the (finite) subdivision $\set {x_0, x_1, \ldots, x_n}$.
We shall use proof by induction on these $n$ intervals.
For all $k \in \set {1, 2, \ldots, n}$, let $\map P k$ be the proposition:
:$f$ is Darboux integrable on $\closedint {x_0} {x_k}$.
=== Basis for the Induction ===
$\map P 1$ is the case:
:$f$ is Darboux integrable on $\closedint {x_{i - 1} } {x_i}$
for an arbitrary $i \in \set {1, 2, \ldots, k}$. 
$f$ is piecewise continuous and bounded for the case $n = 1$ means that:
:$f$ is continuous on $\openint {x_{i - 1} } {x_i}$
:$f$ is bounded on $\closedint {x_{i - 1} } {x_i}$.
By Bounded Function Continuous on Open Interval is Darboux Integrable, $f$ is Darboux integrable on $\closedint {x_{i - 1} } {x_i}$.
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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$f$ is Darboux integrable on $\closedint {x_0} {x_k}$
from which it is to be shown that:
:$f$ is Darboux integrable on $\closedint {x_0} {x_{k + 1} }$.
=== Induction Step ===
This is the induction step:
By definition of a bounded piecewise continuous function, for every $i \in \set {1, 2, \ldots, k, k + 1}$:
:$f$ is continuous on $\openint {x_{i - 1} } {x_i}$
:$f$ is bounded on $\closedint {x_{i - 1} } {x_i}$.
By the induction hypothesis, $f$ is Darboux integrable on $\closedint {x_0} {x_k}$.
From the basis for the induction, $f$ is Darboux integrable on $\closedint {x_k} {x_{k + 1} }$.
We have that $f$ is Darboux integrable on $\closedint {x_0} {x_k}$ and $\closedint {x_k} {x_{k + 1} }$.
Therefore, $f$ is Darboux integrable on $\closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$ by Existence of Integral on Union of Adjacent Intervals.
We have that:
:$\closedint {x_0} {x_{k + 1} } = \closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$.
Accordingly, $f$ is Darboux integrable on $\closedint {x_0} {x_{k + 1} }$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$f$ is Darboux integrable on $\closedint a b$.
== Sources ==
* : $\S 15$: Definition of a Definite Integral: $15.1$
* : $\S 18$: Definite Integrals: Definition of a Definite Integral: $18.1$
Category:Piecewise Continuous Functions
Category:Integral Calculus
Bounded Piecewise Continuous Function may not have One-Sided Limits

== Theorem ==
Let $f$ be a real function defined on a closed interval $\closedint a b$, $a < b$. 
Let $f$ be a bounded piecewise continuous function.
Then it is not necessarily the case that $f$ is a piecewise continuous function with one-sided limits:
== Proof ==
Consider the function:
:$\map f x = \begin{cases}
0 & : x = a \\
\map \sin {\dfrac 1 {x - a} } & : x \in \hointr a b
\end{cases}$
Consider the (finite) subdivision $\set {a, b}$ of $\closedint a b$.
We observe that $\map \sin {\dfrac 1 {x - a} }$ is continuous on $\openint a b$.
Since $\map f x = \map \sin {\dfrac 1 {x - a} }$ on $\openint a b$, it follows that $f$ is continuous on $\openint a b$.
Also, $f$ is bounded by the bound $1$ on $\closedint a b$.
Therefore $f$ is a bounded piecewise continuous function on the closed interval $\closedint a b$.
We now investigate whether $f$ has one-sided limits at the endpoints of $\closedint a b$.
The function $\map \sin {\dfrac 1 {x - a} }$ varies between $-1$ and $+1$ as $x$ approaches $a$ from above.
Thus it does not converge. 
Since $\map f x = \map \sin {\dfrac 1 {x - a} }$ when $x > a$ we conclude that $\ds \lim_{x \mathop \to a^+} \map f x$ does not exist either.
$f$ has no one-sided limit at $a$.
Hence $f$ is not a piecewise continuous function with one-sided limits.
Category:Piecewise Continuous Functions
Bounded Product is Primitive Recursive

== Theorem ==
Let the function $f: \N^{k + 1} \to \N$ be primitive recursive.
Then so is the function $g: \N^{k + 1} \to \N$ defined as:
:$\map g {n_1, n_2, \ldots, n_k, z} = \begin{cases}
1 & : z = 0 \\
\ds \prod_{y \mathop = 1}^z \map f {n_1, n_2, \ldots, n_k, y} & : z > 0
\end{cases}$
== Proof ==
The function $g$ satisfies:
:$\map g {n_1, n_2, \ldots, n_k, z} = 0$
:$\map g {n_1, n_2, \ldots, n_k, z + 1} = \map g {n_1, n_2, \ldots, n_k, z} \times \map f {n_1, n_2, \ldots, n_k, z + 1}$
Hence $g$ is defined by primitive recursion from:
:the primitive recursive function $\operatorname {Add}$
:$f$, which is primitive recursive
:constants, which are primitive recursive.
Hence the result.
== Note ==
The product $\ds \prod_{y \mathop = 1}^z$ is referred to as a bounded product to distinguish it from $\ds \prod_{y \mathop = 1}^\infty$ which is not. 
Category:Primitive Recursive Functions
Bounded Rank implies Small Class

== Theorem ==
Let $S$ be a class.
Suppose the rank, denoted $\map {\operatorname{rank} } x$, of each $x \in S$ is bounded above by some ordinal $y$.
Then $S$ is a small class.
== Proof ==
Let $V$ denote the von Neumann hierarchy.
Then:
Therefore, by Axiom of Subsets Equivalents, it follows that $S$ is a small class.
== Sources ==
* : $\S 9.19$
Category:Von Neumann Hierarchy
Bounded Real-Valued Function/Examples/Sine of Reciprocal

== Example of Bounded Real-Valued Function ==
The function $g$ defined on the real numbers $\R$:
:$\forall x \in \R: g := \map \sin {\dfrac 1 x}$
is bounded.
== Sources ==
* : Chapter $\text I$: Functions and Limits: $\S 3$: Bounds of a Function
Category:Examples of Bounded Below Real-Valued Functions
Bounded Real-Valued Linear Functional is Real Part of Unique Bounded Complex-Valued Linear Functional

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\C$.
Let $\struct {X_\R, \norm {\, \cdot \,}_\R}$ be the realification of $X$ equipped with the restricted norm. 
Let $f : X_\R \to \R$ be a bounded linear functional.
Then there exists a unique bounded linear functional $g : X \to \C$ such that: 
:$\map f x = \map \Re {\map g x}$
for each $x \in X$. 
== Proof ==
Let $M > 0$ be such that: 
:$\cmod {\map f x} \le M \norm x$
for each $x \in X$. 
From Linear Functional on Complex Vector Space is Uniquely Determined by Real Part, there exists a unique linear functional $g : X \to \C$ such that: 
:$\map f x = \map \Re {\map g x}$ 
for each $x \in X$, given by: 
:$\map g x = \map f x - i \map f {i x}$ 
for each $x \in X$.
We just need to show that $g$ is bounded. 
Then we have, for each $x \in X$:
so that: 
:$\cmod {\map g x} \le M \sqrt 2 \norm x$
for each $x \in X$.
So $g$ is a bounded linear functional as required.
Category:Bounded Linear Functionals