nyuuzyou/wikis huggingface dataset:
Bound Variable/Examples/Algebraic Example

== Example of Bound Variable ==
In algebra:
:$x^2 + 2 x y + y^2 = \paren {x + y}^2$
both $x$ and $y$ are bound variables.
This is illustrated in considering the full formal statement:
:$\forall x : \forall y : \paren {x^2 + 2 x y + y^2 = \paren {x + y}^2}$
where each occurrence of $x$ and $y$ is bound by a quantifier.
Category:Bound Variables
Bound Variable/Examples/Existential Statement

== Example of Bound Variable ==
In the existential statement:
:$\exists x: \map P x$
the symbol $x$ is a bound variable.
Thus, the meaning of $\exists x: \map P x$ does not change if $x$ is replaced by another symbol.
That is, $\exists x: \map P x$ means the same thing as $\exists y: \map P y$ or $\exists \alpha: \map P \alpha$. And so on.
== Also see ==
* Bound Variable/Examples/Universal Statement
== Sources ==
* : $\S 3$: Statements and conditions; quantifiers
* : $\S 2.1$
Category:Bound Variables
Bound Variable/Examples/Family of Sets

== Example of Bound Variable ==
Let $I$ be an indexing set.
Consider the union of the indexed family of sets $\family {S_i}_{i \mathop \in I}$:
:$\ds \bigcup_{i \mathop \in I} S_i$
The variable $i$ is a bound variable such that $\ds \bigcup_{i \mathop \in I} S_i$ could as well be written $\ds \bigcup_{\alpha \mathop \in I} S_\alpha$ or $\ds \bigcup_{\gamma \mathop \in I} S_\gamma$, for example.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets
Category:Bound Variables
Bound Variable/Examples/Universal Statement

== Example of Bound Variable ==
In the universal statement:
:$\forall x: \map P x$
the symbol $x$ is a bound variable.
Thus, the meaning of $\forall x: \map P x$ does not change if $x$ is replaced by another symbol.
That is, $\forall x: \map P x$ means the same thing as $\forall y: \map P y$ or $\forall \alpha: \map P \alpha$.
And so on.
== Also see ==
* Bound Variable/Examples/Existential Statement
== Sources ==
* : $\S 3$: Statements and conditions; quantifiers
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic
* : $\S 2.1$
Category:Bound Variables
Bound for Analytic Function and Derivatives

== Lemma ==
Let $f$ be a complex function.
Let $z_0$ be a point in $\C$.
Let $r$ be a real number in $\R_{>0}$.
Let $\Gamma$ be a circle in $\C$ with center at $z_0$ and radius $r$.
Let $f$ be analytic on $\Gamma$ and its interior.
Let $t \in \C$ be such that $\cmod {t - z_0} < r$.
Then a real number $M$ exists such that, for every $n \in \N$:
:$\cmod {\map {f^{\paren n} } t} \le \dfrac {M r \, n!} {\paren {r - \cmod {t - z_0} }^\paren {n + 1} }$
== Proof ==
=== Lemma (Analytic Function Bounded on Circle) ===
We have:
:$f$ is analytic on $\Gamma$ and its interior
:$t$ is in the interior of $\Gamma$
Therefore:
:$\ds \map {f^{\paren n} } t = \frac {n!} {2 \pi i} \int_\Gamma \frac {\map f z} {\paren {z - t}^{\paren {n + 1} } } \rd z$ by Cauchy's Integral Formula for Derivatives
where $\Gamma$ is traversed counterclockwise.
We have that $f$ is bounded on $\Gamma$ by Lemma (Analytic Function Bounded on Circle).
Therefore, there is a positive real number $M$ that satisfies:
:$M \ge \cmod {\map f z}$ for every $z$ on $\Gamma$
We have $\cmod {t - z_0} < r$.
Therefore:
:$0 < r - \cmod {t - z_0}$
We observe that $r - \cmod {t - z_0}$ is the minimum distance between $t$ and $\Gamma$.
Therefore:
:$\forall z \in \Gamma: \paren {r - \cmod {t - z_0} } \le \cmod {z - t}$
We get:
Category:Analytic Functions
Bound for Analytic Function and Derivatives/Analytic Function Bounded on Circle

== Lemma ==
Let $f$ be a complex function.
Let $z_0$ be a point in $\C$.
Let $\Gamma$ be a circle in $\C$ with center at $z_0$ and radius in $\R_{>0}$.
Let $f$ be analytic on $\Gamma$.
Then $f$ is bounded on $\Gamma$.
== Proof ==
Let:
:$\map {f_{\Re} } z = \map \Re {\map f z}$
:$\map {f_{\Im} } z = \map \Im {\map f z}$
Let $\closedint a b$, $a < b$, be a closed real interval.
Let $p$ be a continuous complex-valued function defined such that:
:$\Gamma = \set {\map p u: u \in \closedint a b}$
$f$ is continuous on $\Gamma$ as $f$ is analytic on $\Gamma$ by the definition of analytic.
Also, Real and Imaginary Part Projections are Continuous.
Therefore, $f_{\Re}$ and $f_{\Im}$ are continuous by the corollary to Composite of Continuous Mappings is Continuous.
Observe that $f_{\Re}$ and $f_{\Im}$ are real-valued functions that are continuous.
Also, $p$ is a continuous function defined on a set of real numbers.
Therefore $\map {f_{\Re} } {\map p u}$ and $\map {f_{\Im} } {\map p u}$ are continuous real functions by the corollary to Composite of Continuous Mappings is Continuous.
$\map {f_{\Re} } {\map p u}$ and $\map {f_{\Im} } {\map p u}$ are bounded on $\closedint a b$ by Continuous Real Function is Bounded.
Therefore $\map f {\map p u}$ is bounded on $\closedint a b$ as $\map f {\map p u} = \map {f_{\Re} } {\map p u} + i \map {f_{\Im} } {\map p u}$ where $i = \sqrt {-1}$.
Accordingly, $f$ is bounded on $\Gamma$ as $\Gamma = \set {\map p u: u \in \closedint a b}$.
Category:Analytic Functions
Bound for Cardinality of Matroid Circuit

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $C \subseteq S$ be a circuit of $M$.
Let $\rho: \powerset S \to \Z$ denote the rank function of $M$.
Then:
:$\card C \le \map \rho S + 1$
== Proof ==
By definition of a circuit:
:$C$ is dependent
By matroid axiom $(\text I 1)$:
:$C \ne \O$
Let $x \in C$.
From Set Difference is Subset and Set Difference with Disjoint Set:
:$C \setminus \set x \subsetneq C$
From Proper Subset of Matroid Circuit is Independent and matroid axiom $(\text I 1)$:
:$C \setminus \set x \in \mathscr I$
We have:
== Sources ==
* Chapter $1.$ $\S 9.$ Circuits
Category:Matroid Circuits
Bound for Difference of Irrational Number with Convergent

== Theorem ==
Let $x$ be an irrational number.
Let $\sequence {C_n}$ be the sequence of convergents of the continued fraction expansion of $x$.
Then $\forall n \ge 1$:
:$C_n < x < C_{n + 1}$ or $C_{n + 1} < x < C_n$
:$\size {x - C_n} < \dfrac 1 {q_n q_{n + 1} }$
== Proof ==
Immediate.
Note that:
:$\size {x - C_n} < \size {C_{n + 1} - C_n} = \dfrac 1 {q_n q_{n + 1} }$
Category:Continued Fractions
Bound for Negative Part of Pointwise Sum of Functions

== Theorem ==
Let $X$ be a set. 
Let $f, g : X \to \overline \R$ be extended real-valued function.
Suppose that the pointwise sum $f + g$ is well-defined, that is: 
:there exists no $x \in X$ such that $\set {\map f x, \map g x} = \set {\infty, -\infty}$.
Then: 
:$\paren {f + g}^- \le f^- + g^-$
where $\paren {f + g}^-$, $f^-$ and $g^-$ denote the negative parts of $f + g$, $f$ and $g$ respectively.
== Proof ==
Let $x \in X$. 
From the definition of the negative part, we have: 
:$\map {f^-} x = -\min \set {\map f x, 0}$
and:
:$\map {g^-} x = -\min \set {\map g x, 0}$
Suppose first that $\map f x$ and $\map g x$ are finite.
From Minimum Function in terms of Absolute Value, we then have:
:$\ds \map {f^-} x = \frac {\size {\map f x} - \map f x} 2$
and:
:$\ds \map {g^-} x = \frac {\size {\map g x} - \map g x} 2$
We then have: 
Now suppose that $\map f x = +\infty$. 
Then $\map g x > -\infty$.
We then have:
:$\map f x + \map g x = +\infty$
So:
:$\map {\paren {f + g}^-} x = 0$
and:
:$\map {f^-} x = 0$
Since: 
:$\map {g^-} x \ge 0$
we have that: 
:$\map {\paren {f + g}^-} x \le \map {f^-} x + \map {g^-} x$
Swapping $f$ for $g$, we also get the result if $\map g x = +\infty$ and $\map f x > -\infty$.
Now suppose that $\map f x = -\infty$
Then $\map g x < \infty$.
Then: 
:$\map f x + \map g x = -\infty$
So:
:$\map {\paren {f + g}^-} x = \infty$
and:
:$\map {f^-} x = \infty$
So the inequality: 
:$\map {\paren {f + g}^-} x \le \map {f^-} x + \map {g^-} x$
holds trivially. 
Swapping $f$ for $g$, we also get the result if $\map g x = -\infty$ and $\map f x < \infty$.
Category:Negative Parts
Bound for Norm of Difference of Element with Non-Negative Multiple of Element in Unital C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra. 
Let $\le_A$ be the canonical preordering of $A$.
Let $a, b \in A$ be such that:
:${\mathbf 0}_A \le_A a \le_A b \le_A {\mathbf 1}_A$
Then we have:
:$\norm {x - b x}^2 \le \norm {x^\ast \paren { {\mathbf 1}_A - a} x}$
and:
:$\norm {x - x b}^2 \le \norm {x \paren { {\mathbf 1}_A - a} x^\ast}$
for each $x \in A$. 
== Proof ==
We have from Scalar Multiplication by Minus One reverses Preordering in Vector Space:
:$-{\mathbf 1}_A \le_A -b \le_A -a \le_A {\mathbf 0}_A$
From $(1)$ in the definition of a preordered vector space, we have:
:${\mathbf 1}_A - {\mathbf 1}_A \le_A {\mathbf 1}_A - b \le_A {\mathbf 1}_A$
so that:
:${\mathbf 0}_A \le_A {\mathbf 1}_A - b \le_A {\mathbf 1}_A$
From Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum, we have:
:$\map {\sigma_A} { {\mathbf 1}_A - b} \subseteq \closedint 0 1$
We have:
For $t \in \closedint 0 1$, we have:
:$0 \le t^2 \le t$ 
Hence we have:
:${\mathbf 0}_A \le_A \paren { {\mathbf 1}_A - b}^2 \le_A {\mathbf 1}_A - b$
from Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative: Corollary.
From Conjugation in C*-Algebra preserves Positivity: Corollary, we have:
:${\mathbf 0}_A \le_A x^\ast \paren { {\mathbf 1}_A - b}^2 x \le_A x^\ast \paren { {\mathbf 1}_A - b} x$
From Norm Preserves Ordering on Positive Elements of C*-Algebra, we have:
:$\norm {x^\ast \paren { {\mathbf 1}_A - b}^2 x} \le \norm {x^\ast \paren { {\mathbf 1}_A - b} x}$
From the inequality:
:$-b \le_A -a$
we apply $(1)$ in the definition of a preordered vector space to obtain:
:${\mathbf 1}_A - b \le_A {\mathbf 1}_A - a$
Applying Conjugation in C*-Algebra preserves Positivity: Corollary, we have:
:${\mathbf 0}_A \le_A x^\ast \paren { {\mathbf 1}_A - b} x \le_A x^\ast \paren { {\mathbf 1}_A - a} x$
From Norm Preserves Ordering on Positive Elements of C*-Algebra, we have:
:$\norm {x^\ast \paren { {\mathbf 1}_A - b} x} \le \norm {x^\ast \paren { {\mathbf 1}_A - a} x}$
Hence we have shown:
:$\norm {x - b x}^2 \le \norm {x^\ast \paren { {\mathbf 1}_A - a} x}$
To finish we now have:
== Sources ==
* : Lemma $4.2$
Category:C*-Algebras
Bound for Positive Part of Pointwise Sum of Functions

== Theorem ==
Let $X$ be a set.
Let $f, g : X \to \overline \R$ be extended real-valued function.
Suppose that the pointwise sum $f + g$ is well-defined, that is: 
:there exists no $x \in X$ such that $\set {\map f x, \map g x} = \set {\infty, -\infty}$.
Then: 
:$\paren {f + g}^+ \le f^+ + g^+$
where $\paren {f + g}^+$, $f^+$ and $g^+$ denote the positive parts of $f + g$, $f$ and $g$ respectively.
== Proof ==
Let $x \in X$. 
From the definition of the positive part, we have: 
:$\map {f^+} x = \max \set {\map f x, 0}$
and:
:$\map {g^+} x = \max \set {\map g x, 0}$
Suppose first that $\map f x$ and $\map g x$ are finite. 
From Maximum Function in terms of Absolute Value, we then have: 
:$\ds \map {f^+} x = \frac {\map f x + \size {\map f x} } 2$
and:
:$\ds \map {g^+} x = \frac {\map g x + \size {\map g x} } 2$
We then have: 
Now suppose that $\map f x = +\infty$. 
Then $\map g x > -\infty$.
We then have: 
:$\map f x + \map g x = +\infty$
So:
:$\map {\paren {f + g}^+} x = +\infty$
and: 
:$\map {f^+} x = +\infty$
So the inequality:
:$\map {\paren {f + g}^+} x \le \map {f^+} x + \map {g^+} x$
holds trivially. 
Swapping $f$ for $g$, we also get the result if $\map g x = +\infty$ and $\map f x > -\infty$. 
Now suppose that $\map f x = -\infty$
Then $\map g x < \infty$.
Then: 
:$\map f x + \map g x = -\infty$
So:
:$\map {\paren {f + g}^+} x = 0$
and: 
:$\map {f^+} x = 0$
We have that: 
:$\map {g^+} x \ge 0$
so: 
:$\map {\paren {f + g}^+} x \le \map {f^+} x + \map {g^+} x$
Swapping $f$ for $g$, we also get the result if $\map g x = -\infty$ and $\map f x < \infty$.
Category:Positive Parts
Bound for Variation of Complex Measure in terms of Jordan Decomposition

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a complex measure on $\struct {X, \Sigma}$. 
Let $\cmod \mu$ be the variation of $\mu$. 
Let $\tuple {\mu_1, \mu_2, \mu_3, \mu_4}$ be the Jordan decomposition of $\mu$.
Then: 
:$\map {\cmod \mu} A \le \map {\mu_1} A + \map {\mu_2} A + \map {\mu_3} A + \map {\mu_4} A$
for all $A \in \Sigma$.
== Proof ==
Let $A \in \Sigma$.
Let $\map P A$ be the set of finite partitions of $A$ into $\Sigma$-measurable sets.
From the definition of variation, we have: 
:$\ds \map {\cmod \mu} A = \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$
Let: 
:$\set {A_1, A_2, \ldots, A_n} \in \map P A$ 
Consider: 
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i} }$
We have, from the definition of Jordan decomposition: 
:$\mu = \mu_1 - \mu_2 + i \paren {\mu_3 - \mu_4}$
so that: 
Since: $\set {A_1, A_2, \ldots, A_n}$ is a partition of $A$, we have: 
:$\set {A_1, A_2, \ldots, A_n}$ are pairwise disjoint with: 
::$\ds A = \bigcup_{i \mathop = 1}^n A_i$
So:
Category:Complex Measures
Bound on C0 Semigroup

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a Banach space over $\GF$. 
Let $\family {\map T t}_{t \ge 0}$ be a $C_0$ semigroup. 
Let $\struct {\map B X, \norm {\, \cdot \,}_{\map B X} }$ be the space of bounded linear transformations equipped with the canonical norm.
Then there exists $M \ge 1$ and $\omega \ge 0$ such that: 
:$\norm {\map T t}_{\map B X} \le M e^{\omega t}$ 
for each $t \in \hointr 0 \infty$. 
== Proof ==
We first show that $\norm {\map T t}_{\map B X}$ is bounded on $\closedint 0 \delta$ for some $\delta > 0$.
 $\norm {\map T t}_{\map B X}$ is unbounded on $\closedint 0 \delta$ for each $\delta > 0$.
Then for each $n \in \N$ we can pick $t_n \in \closedint 0 {\frac 1 n}$ such that $\norm {\map T {t_n} }_{\map B X} \ge n$.
Then, we have: 
:$\ds \sup_{n \mathop \in \N} \norm {\map T {t_n} }_{\map B X} = \infty$ 
By definition, each $\map T {t_n}$ is a linear operator.
So we can apply Principle of Condensation of Singularities to obtain: 
:$\ds \sup_{n \mathop \in \N} \norm {\map T {t_n} x} = \infty$
for some $x \in X$.
But since: 
:$\ds \lim_{t \mathop \to 0^+} \map T {t_n} x = x$
and:
:$\ds \lim_{n \mathop \to \infty} t_n = 0$ by the Squeeze Theorem
we must have that:
:$\ds \sup_{n \mathop \in \N} \norm {\map T {t_n} x} < \infty$
by Convergent Sequence in Normed Vector Space is Bounded.
So we have a contradiction, and there exists $\delta > 0$ such that $\norm {\map T t}_{\map B X}$ is bounded on $\closedint 0 \delta$.
Suppose that: 
:$\norm {\map T t}_{\map B X} \le M$
for $t \in \closedint 0 \delta$.
Since $\map T 0 = I$, we have $M \ge 1$. 
Now for general $t \ge 0$, write: 
:$t = N \delta + r$ for $N \in \Z_{\ge 0}$ and $r \in \hointr 0 \delta$.
Then we have:
Since $t = N \delta + r$, we have: 
:$\ds \frac t \delta = N + \frac r \delta \ge N$
so we have:
Taking: 
:$\ds \omega = \frac {\log M} \delta$ 
we have:
:$\ds \norm {\map T t}_{\map B X} \le M e^{\omega t}$ 
for each $t \in \hointr 0 \infty$.
== Sources ==
* : $1.2$: Strongly Continuous Semigroups of Bounded Linear Operators
Category:C0 Semigroups
Bound on Complex Values of Gamma Function

== Theorem ==
Let $\map \Gamma z$ denote the Gamma function.
Then for any complex number $z = s + i t$, we have for $\size b \le \size t$:
:$\size {\map \Gamma {s + i t} } \le \dfrac {\size {s + i b} } {\size {s + i t} } \size {\map \Gamma {s + i b} }$
== Proof ==
From the Euler Form of the Gamma Function:
Because $\size b \le \size t$, we have that:
Using this we obtain:
Category:Gamma Function
Bound on Inverse in Canonical Preordering of Unital C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $\le_A$ be the canonical preordering on $A$.
Let $a \in A$ and $\alpha, \beta > 0$ be such that:
:$\alpha {\mathbf 1}_A \le_A a \le_A \beta {\mathbf 1}_A$
Then:
:$\beta^{-1} {\mathbf 1}_A \le_A a^{-1} \le_A \alpha^{-1} {\mathbf 1}_A$
== Proof ==
Since $\beta^{-1} {\mathbf 1}_A \le_A a$ and ${\mathbf 0}_A \le_A \beta^{-1} {\mathbf 1}_A$, we have:
:$a$ is positive
since $\le_A$ is a preordering.
In particular, $a$ is Hermitian.
From Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum, we have:
:$\map {\sigma_A} a \subseteq \closedint \alpha \beta$
Since $\alpha > 0$, we have that $0 \not \in \map {\sigma_A} a$.
Hence $a$ is invertible.
From Spectrum of Inverse of Element of Unital Algebra, we have:
:$\map {\sigma_A} {a^{-1} } = \set {t^{-1} : t \in \map {\sigma_A} a}$
Since $0 < \alpha \le t \le \beta$ for each $t \in \map {\sigma_A} a$, we have:
:$\beta^{-1} \le t \le \alpha^{-1}$
Hence we have:
:$\map {\sigma_A} {a^{-1} } \subseteq \closedint {\beta^{-1} } {\alpha^{-1} }$
Applying Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum again we have:
:$\beta^{-1} {\mathbf 1}_A \le_A a^{-1} \le_A \alpha^{-1} {\mathbf 1}_A$
Category:Canonical Preorderings of C*-Algebras
Bound on Norm of Power of Element in Normed Algebra

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$. 
Let $x \in A$ and $n \in \N$. 
Then: 
:$\norm {x^n} \le \norm x^n$ 
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\norm {x^n} \le \norm x^n$
=== Basis for the Induction ===
We have: 
:$\norm {x^1} = \norm x = \norm x^1$
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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\norm {x^k} \le \norm x^k$
from which it is to be shown that:
:$\norm {x^{k + 1} } \le \norm x^{k + 1}$ 
=== Induction Step ===
This is the induction step.
We have:
Category:Normed Algebras
Bound on Norm of Real and Imaginary Parts of Element of Banach *-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a Banach $\ast$-algebra.
Let $x \in A$.
Let $\map \Re x$ and $\map \Im x$ be the real and imaginary parts of $x$ respectively.
Then:
:$\norm {\map \Re x} \le \norm x$
and:
:$\norm {\map \Im x} \le \norm x$
for each $x \in A$. 
== Proof ==
Let $x \in A$.
Then we have:
and:
Category:Banach *-Algebras
Bound on Riemannian Distance Inside Coordinate Neighborhood

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold with or without boundary.
Let $d_g$ be the Riemannian distance.
Suppose $U \subseteq M$ is an open subset.
Let $p \in M$ be a point.
Then $p$ has a coordinate neighborhood $V \subseteq U$ such that:
:$\forall q \in V : \exists C \in \R_{> 0} : \map {d_g} {p, q} \le C \map {d_{\bar g} } {p, q}$
where $\bar g$ is the Euclidean metric in the given coordinates on $V$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Lengths and Distances
Category:Riemannian Geometry
Bound on Riemannian Distance Outside Coordinate Neighborhood

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold with or without boundary.
Let $d_g$ be the Riemannian distance.
Suppose $U \subseteq M$ is an open subset.
Let $p \in M$ be a point.
Then $p$ has a coordinate neighborhood $V \subseteq U$ such that:
:$\forall q \in M \setminus V : \exists D \in \R_{> 0} : \map {d_g} {p, q} \ge D$
where $\setminus$ denotes the set difference.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Lengths and Distances
Category:Riemannian Geometry
Bound on Survival Function of Pointwise Product

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g : X \to \overline \R$ be $\Sigma$-measurable functions.
Let $F_{f g}$ be the survival function of the pointwise product $f g$.
Let $F_f$ and $F_g$ be the survival functions of $f$ and $g$ respectively. 
Then: 
:$\ds \map {F_{f g} } {\alpha \beta} \le \map {F_f} \alpha + \map {F_g} \beta$ for all $\alpha, \beta \in \hointr 0 \infty$.
== Proof ==
Let $\alpha, \beta \in \hointr 0 \infty$.
We show that: 
:$\set {x \in X : \size {\map f x \map g x} \ge \alpha \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$
Let $x \in X$ be such that $\size {\map f x \map g x} \ge \alpha \beta$.
If $\size {\map f x \map g x} = \infty$, we have $\size {\map f x} = \infty$ or $\size {\map g x}$. 
Then we either have $\size {\map f x} \ge \alpha$ or $\size {\map g x} \ge \beta$, so that: 
:$x \in \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$.
Now suppose that $\size {\map f x \map g x} < \infty$. 
Suppose that $\size {\map f x} < \alpha$ and $\size {\map g x} < \beta$, then $\size {\map f x \map g x} = \size {\map f x} \size {\map g x} < \alpha \beta$
contradicting that: 
:$\size {\map f x \map g x} \ge \alpha \beta$ 
So we must have $\size {\map f x} \ge \alpha$ or $\size {\map g x} \ge \beta$. 
So we have:
:$x \in \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$.
again in this case.
So:
:$\set {x \in X : \size {\map f x \map g x} \ge \alpha \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$
by the definition of set inclusion. 
Then we have: 
== Sources ==
* : $1.1.1$: The Distribution Function
Category:Survival Functions
Bound on Survival Function of Pointwise Sum

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g : X \to \overline \R$ be $\Sigma$-measurable functions such that the pointwise sum $f + g$ is well-defined.
Let $F_{f + g}$ be the survival function of the pointwise scalar multiple $f + g$. 
Let $F_f$ and $F_g$ be the survival functions of $f$ and $g$ respectively. 
Then: 
:$\ds \map {F_{f + g} } {\alpha + \beta} \le \map {F_f} \alpha + \map {F_g} \beta$ for all $\alpha, \beta \in \hointr 0 \infty$.
== Proof ==
Let $\alpha, \beta \in \hointr 0 \infty$.
We show that: 
:$\set {x \in X : \size {\map f x + \map g x} \ge \alpha + \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$ 
Let $x \in X$ be such that $\size {\map f x + \map g x} \ge \alpha + \beta$.
If $\size {\map f x + \map g x} = \infty$, we have $\map f x + \map g x = \infty$ or $\map f x + \map g x = -\infty$. 
In the former case, we have $\map f x = \infty$ or $\map g x = \infty$. 
In the latter case, we have $\map f x = -\infty$ or $\map g x = -\infty$.
So in either case we have $\size {\map f x} = \infty$ or $\size {\map g x} = \infty$. 
So we either have $\size {\map f x} \ge \alpha$ or $\size {\map g x} \ge \beta$. 
So:
:$x \in \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$
in this case.
Now suppose that $\size {\map f x + \map g x} < \infty$. 
Suppose that $\size {\map f x} < \alpha$ and $\size {\map g x} < \beta$. 
Then from the Triangle Inequality, we have: 
:$\size {\map f x + \map g x} \le \size {\map f x} + \size {\map g x} < \alpha + \beta$ 
contradicting that:
:$\size {\map f x + \map g x} \ge \alpha + \beta$ 
So we must have $\size {\map f x} \ge \alpha$ or $\size {\map g x} \ge \beta$. 
So, we have: 
:$x \in \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$
again in this case.
So:
:$\set {x \in X : \size {\map f x + \map g x} \ge \alpha + \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$ for all $\alpha, \beta \in \hointr 0 \infty$. 
by the definition of set inclusion.
Then we have: 
== Sources ==
* : $1.1.1$: The Distribution Function
Category:Survival Functions