nyuuzyou/wikis huggingface dataset:
Trefoil Knot in Paper forms Pentagon/Historical Note

== Historical Note on Trefoil Knot in Paper forms Pentagon ==
This construction first occurred in 's of $1682$.
It appears to have been added as an afterthought, as the rest of the work does not depend upon the construction of polygons.
This construction regularly crops up in puzzle anthologies.
== Sources ==
* : Henry van Etten: $126$
Category:Historical Notes
Trefoil Knot in Paper forms Pentagon/Mistake

== Source Work ==
: 
:The Puzzles:
::Henry van Etten
:::$126$
== Mistake ==
:The next problem occurs first in Urbino d'Aviso's treatise on the sphere (1682):
== Correction ==
The mathematician in question is .
== Sources ==
* : Henry van Etten: $126$
Category:Mistakes/Curious and Interesting Puzzles
Trefoil Knot is Homeomorphic to Circle

== Theorem ==
The trefoil knot is homeomorphic to the circle.
== Proof ==
Despite the fact that you cannot actually rearrange a trefoil knot actually into a circle in the usual $\R^3$ space, you can set up a mappping from one to the other.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.6$: Homeomorphisms: Examples $3.6.2 \ \text{(f)}$
Category:Examples of Homeomorphisms
Triangle Angle-Side-Angle Congruence

== Theorem ==
If two triangles have:
:two angles equal to two angles, respectively
:the sides between the two angles equal
then the remaining angles are equal, and the remaining sides equal the respective sides.
That is to say, if two pairs of angles and the included sides are equal, then the triangles are congruent.
== Proof ==
:
Let $\angle ABC = \angle DEF$, $\angle BCA = \angle EFD$, and $BC = EF$.
 that $AB \ne DE$.
If this is the case, one of the two must be greater.
, we let $AB > DE$.
We construct a point $G$ on $AB$ such that $BG = ED$.
Using Euclid's first postulate, we construct the segment $CG$.
Now, since we have:
:$BG = ED$
:$\angle GBC = \angle DEF$
:$BC = EF$
it follows from Triangle Side-Angle-Side Congruence that:
:$\angle GCB = \angle DFE$
But from Euclid's fifth common notion:
:$\angle DFE = \angle ACB > \angle GCB$
which is a contradiction.
Therefore, $AB = DE$.
So from Triangle Side-Angle-Side Congruence:
:$\triangle ABC = \triangle DEF$
== Also known as ==
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $\text {A}.1$: Thales (ca. B.C.)
* $(3)$
* $(3)$
Category:Triangle Angle-Side-Angle Congruence
Category:Triangles
Category:Congruence (Geometry)
Triangle Angle-Side-Angle and Side-Angle-Angle Congruence

== Theorem ==
=== Triangle Angle-Side-Angle Congruence ===
=== Triangle Side-Angle-Angle Congruence ===
== Proof ==
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.11$
Category:Triangle Angle-Side-Angle and Side-Angle-Angle Congruence
Category:Triangles
Category:Congruence (Geometry)
Triangle Angle-Side-Angle and Side-Angle-Angle Congruence/Proof

== Theorem ==
== Proof ==
Both parts of this proposition follow trivially from the other part and Sum of Angles of Triangle equals Two Right Angles.
However, it is important to note that both of these are provable without the parallel postulate, which the proof of that theorem requires.
Category:Triangle Angle-Side-Angle and Side-Angle-Angle Congruence
Category:Triangles
Triangle Congruence

== Theorem ==
=== Triangle Side-Angle-Side Congruence ===
=== Triangle Side-Side-Side Congruence ===
=== Triangle Angle-Side-Angle Congruence ===
=== Triangle Side-Angle-Angle Congruence ===
=== Triangle Right-Angle-Hypotenuse-Side Congruence ===
=== Ambiguous Case for Triangle Side-Side-Angle Congruence ===
Category:Triangles
Category:Congruence (Geometry)
Triangle Conjugacy is Mutual

== Theorem ==
Let $\CC$ be a circle.
Let $\triangle PQR$ be a triangle.
Let $\triangle P'Q'R'$ be such that:
:$P'$ is the pole of $QR$
:$Q'$ is the pole of $PR$
:$R'$ is the pole of $PQ$
with respect to $\CC$.
Then:
:$P$ is the pole of $Q'R'$
:$Q$ is the pole of $P'R'$
:$R$ is the pole of $P'Q'$
with respect to $\CC$.
That is, $\triangle PQR$ and $\triangle P'Q'R'$ are conjugate triangles with respect to $\CC$.
</onlyinclude>
== Proof ==
We have that:
:the polar of $P'$ is $QR$
:the polar of $Q'$ is $PR$ 
and so both polars pass through $R$.
Therefore:
:the polar of $R$ is $P'Q'$.
Similarly:
:the polar of $P$ is $Q'R'$
and:
:the polar of $Q$ is $P'R'$.
== Sources ==
* : Chapter $\text {III}$. The Circle: $9$. Conjugate triangles
Category:Conjugate Triangles
Triangle Inequality/Complex Numbers

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ denote the modulus of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$
=== General Result ===
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Examples ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Inequalities: $3.7.29$
* : $1$: Complex Numbers: Absolute Value: $3$
* 
* 
* 
Category:Triangle Inequality
Triangle Inequality/Complex Numbers/Examples/3 Arguments

== Example of Use of Triangle Inequality for Complex Numbers ==
For all $z_1, z_2, z_3 \in \C$:
:$\cmod {z_1 + z_2 + z_3} \le \cmod {z_1} + \cmod {z_2} + \cmod {z_3}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Triangle Inequality
Triangle Inequality/Complex Numbers/Examples/3 Arguments/Proof 1

== Example of Use of Triangle Inequality for Complex Numbers ==
== Proof ==
This is an instance of the General Triangle Inequality for Complex Numbers:
:$\cmod {z_1 + z_2 + \dotsb + z_n} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_n}$
setting $n = 3$.
Category:Triangle Inequality
Triangle Inequality/Complex Numbers/Examples/3 Arguments/Proof 3

== Example of Use of Triangle Inequality for Complex Numbers ==
== Proof ==
Let $z_1$, $z_2$ and $z_3$ be represented by the points $A$, $B$ and $C$ respectively in the complex plane.
From Geometrical Interpretation of Complex Addition, we can construct the parallelogram $OADB$ where:
:$OA$ and $OB$ represent $z_1$ and $z_2$ respectively
:$OD$ represents $z_1 + z_2$.
Also from Geometrical Interpretation of Complex Addition, we can construct the parallelogram $OCED$ where:
:$OC$ and $OD$ represent $z_3$ and $z_1 + z_2$ respectively
:$OD$ represents $z_1 + z_2 + z_3$.
:
As $OADB$ is a parallelogram, we have that $OB = AD$.
The lengths of $OA$, $AD$ and $OD$ are:
But $OA$, $OB$ and $OD$ form the sides of a triangle.
Thus from Sum of Two Sides of Triangle Greater than Third Side:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$
Similarly, as $OCED$ is a parallelogram, we have that $OD = CE$.
The lengths of $OC$, $CE$ and $OE$ are:
But $OC$, $CE$ and $OE$ form the sides of a triangle.
Thus from Sum of Two Sides of Triangle Greater than Third Side:
:$\cmod {z_1 + z_2 + z_3} \le \cmod {z_1 + z_2} + \cmod {z_3}$
The result follows.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $7 \ \text{(b)}$
Category:Triangle Inequality
Triangle Inequality/Complex Numbers/General Result

== Theorem ==
Let $z_1, z_2, \dotsc, z_n \in \C$ be complex numbers.
Let $\cmod z$ be the modulus of $z$.
Then:
:$\cmod {z_1 + z_2 + \dotsb + z_n} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_n}$
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\cmod {z_1 + z_2 + \dotsb + z_n} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_n}$
$\map P 1$ is true by definition of the usual ordering on real numbers:
:$\cmod {z_1} \le \cmod {z_1}$
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$
which has been proved in Triangle Inequality for Complex Numbers.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\cmod {z_1 + z_2 + \dotsb + z_k} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_k}$
Then we need to show:
:$\cmod {z_1 + z_2 + \dotsb + z_{k + 1} } \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_{k + 1} }$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
== Also see ==
* Triangle Inequality for Indexed Summations
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 2$. Conjugate Complex Numbers
* : $\S 2$. Geometrical Representations
* : $1$: Complex Numbers: Absolute Value: $3$
Category:Triangle Inequality
Triangle Inequality/Complex Numbers/Proof 1

== Theorem ==
== Proof ==
Let $z_1 = a_1 + i a_2, z_2 = b_1 + i b_2$.
Then from the definition of the modulus, the above equation translates into:
:$\paren {\paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2}^{\frac 1 2} \le \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} + \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}$
This is a special case of Minkowski's Inequality for Sums, with $n = 2$.
Category:Triangle Inequality
Triangle Inequality/Complex Numbers/Proof 2

== Theorem ==
== Proof ==
Let $z_1 = a_1 + i a_2, z_2 = b_1 + i b_2$.
The final statement is a tautology, and all implications are reversible.
Hence the result.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $7 \ \text{(a)}$
Category:Triangle Inequality
Triangle Inequality/Complex Numbers/Proof 3

== Theorem ==
== Proof ==
Let $z_1$ and $z_2$ be represented by the points $A$ and $B$ respectively in the complex plane.
From Geometrical Interpretation of Complex Addition, we can construct the parallelogram $OACB$ where:
:$OA$ and $OB$ represent $z_1$ and $z_2$ respectively
:$OC$ represents $z_1 + z_2$.
:
As $OACB$ is a parallelogram, we have that $OB = AC$.
The lengths of $OA$, $AC$ and $OC$ are:
But $OA$, $OB$ and $OC$ form the sides of a triangle.
The result then follows directly from Sum of Two Sides of Triangle Greater than Third Side.
== Sources ==
* : $\S 2$. Geometrical Representations: $(2.1)$
* : $1$: Complex Numbers: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $7 \ \text{(a)}$
Category:Triangle Inequality
Triangle Inequality/Examples/4 Points

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $x, y, z, t \in A$.
Then:
:$\map d {x, z} + \map d {y, t} \ge \size {\map d {x, y} - \map d {z, t} }$
== Proof ==
We have that $\map d {x, z}$, $\map d {y, t}$, $\map d {x, y}$, $\map d {z, t}$ are themselves all real numbers.
Hence the Euclidean metric on the real number line can be applied:
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 3$
Category:Examples of Triangle Inequality
Triangle Inequality/Real Numbers

== Theorem ==
Let $x, y \in \R$ be real numbers.
Let $\size x$ denote the absolute value of $x$.
Then:
:$\size {x + y} \le \size x + \size y$
=== General Result ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5==
== Examples ==
== Sources ==
* : Chapter $\text I$. Continuity: $1$. Continuous Variation
* : Chapter $\text I$: Functions and Limits: $\S 4$: Limits of Functions (footnote $\dagger$)
* : $3$: Elementary Analytic Methods: $3.2$ Inequalities: Triangle Inequalities: $3.2.5$
* : $\S 36$: Inequalities: Triangle Inequality: $36.1$
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Proposition $1.1.9$
* : $1$: Review of some real analysis: Exercise $1.5: 8$
* 
* : $\text {(ii)}$
* : $\text {(ii)}$
Category:Triangle Inequality
Triangle Inequality/Real Numbers/Examples/-1 + 3

== Examples of Use of Triangle Inequality for Real Numbers ==
:$2 = \size {-1 + 3} \le \size {-1} + \size 3 = 1 + 3 = 4$
== Sources ==
* : $\S 1$: Real Numbers: Example $\S 1.19 \ \text{(i)}$
Category:Examples of Triangle Inequality
Triangle Inequality/Real Numbers/General Result

== Theorem ==
Let $x_1, x_2, \dotsc, x_n \in \R$ be real numbers.
Let $\size x$ denote the absolute value of $x$.
Then:
:$\ds \size {\sum_{i \mathop = 1}^n x_i} \le \sum_{i \mathop = 1}^n \size {x_i}$
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \size {\sum_{i \mathop = 1}^n x_i} \le \sum_{i \mathop = 1}^n \size {x_i}$
$\map P 1$ is true by definition of the usual ordering on real numbers:
:$\size {x_1} \le \size {x_1}$
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\size {x_1 + x_2} \le \size {x_1} + \size {x_2}$
which has been proved in Triangle Inequality for Real Numbers.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \size {\sum_{i \mathop = 1}^k x_i} \le \sum_{i \mathop = 1}^k \size {x_i}$
Then we need to show:
:$\ds \size {\sum_{i \mathop = 1}^{k + 1} x_i} \le \sum_{i \mathop = 1}^{k + 1} \size {x_i}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
== Also see ==
* Triangle Inequality for Indexed Summations
== Sources ==
* : $3$: Elementary Analytic Methods: $3.2$ Inequalities: Triangle Inequalities: $3.2.6$
* : $\S 36$: Inequalities: Triangle Inequality: $36.2$
Category:Triangle Inequality
Triangle Inequality/Real Numbers/Proof 4

== Theorem ==
== Proof ==
We do a case analysis.
=== $(1): \quad x \ge 0, y \ge 0$ ===
=== $(2): \quad x \le 0, y \le 0$ ===
=== $(3): \quad x \ge 0, y \le 0$ ===
We have that $\size x = x$ and $\size y = -y$.
In this case we show:
:$\size {x + y} \le \max \set {\size x, \size y}$
Let $\size x \le \size y$.
Then:
Let $\size x \ge \size y$.
Then:
We have $\max \set {a, b} \le a + b$ for positive real numbers $a$ and $b$. 
The result follows by taking $a = \size x$ and $b = \size y$.
=== $(4): \quad x \le 0, y \ge 0$ ===
Follows by symmetry from the case $(3)$.
Category:Triangle Inequality
Triangle Inequality/Real Numbers/Proof 5

== Theorem ==
== Proof ==
From Negative of Absolute Value, it is sufficient to prove that:
:$\size x + \size y \ge x + y$
and:
:$\size x + \size y \ge -\paren {x + y}$
By definition of absolute value:
:$x \le \size x$
and:
:$y \le \size y$
Then:
:$x + y \le \size x + \size y$ 
We also have that:
Hence the result.
Category:Triangle Inequality
Triangle Inequality/Vectors in Euclidean Space

== Theorem ==
Let $\mathbf x, \mathbf y$ be vectors in the real Euclidean space $\R^n$.
Let $\norm {\, \cdot \,}$ denote vector length.
Then:
:$\norm {\mathbf x + \mathbf y} \le \norm {\mathbf x} + \norm {\mathbf y}$
If the two vectors are scalar multiples where said scalar is non-negative, an equality holds:
:$\exists \lambda \in \R, \lambda \ge 0: \mathbf x = \lambda \mathbf y \iff \norm {\mathbf x + \mathbf y} = \norm {\mathbf x} + \norm {\mathbf y}$ 
== Proof ==
Let $\mathbf x, \mathbf y \in \R^n$.
We have:
From the Cauchy-Bunyakovsky-Schwarz Inequality:
To prove that the equality holds if the vectors are scalar multiples of each other, assume:
:$\exists \lambda \in \R, \lambda \ge 0: \mathbf v = \lambda \mathbf w$
=== Sufficient Condition ===
=== Necessary Condition ===
== Sources ==
* 
* 
* 
Category:Linear Algebra
Category:Triangle Inequality
Triangle Inequality for Complex Numbers/Corollary 1

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ be the modulus of $z$.
Then:
: $\cmod {z_1 + z_2} \ge \cmod {z_1} - \cmod {z_2}$
== Proof ==
Let $z_3 := z_1 + z_2$.
Then:
== Sources ==
* : $1$: Complex Numbers: Absolute Value: $4$
Category:Complex Modulus
Triangle Inequality for Complex Numbers/Corollary 2

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ be the modulus of $z$.
Then:
:$\cmod {z_1 + z_2} \ge \cmod {\cmod {z_1} - \cmod {z_2} }$
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Inequalities: $3.7.29$
* : $1$ The complex plane: Complex numbers $\S 1.4$ Inequalities: $(3)$
Category:Complex Modulus
Triangle Inequality for Complex Numbers/Corollary 3

== Corollary to Triangle Inequality for Complex Numbers ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ denote the modulus of $z$.
Then:
:$\cmod {z_1 - z_2} \le \cmod {z_1} + \cmod {z_2}$
== Proof ==
Let $w = -z_2$.
Then:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Inequalities: $3.7.29$
Category:Triangle Inequality
Triangle Inequality for Conditional Expectation

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra.
Let $X$ be an integrable random variable.
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ given $\GG$.
Let $\expect {\size X \mid \GG}$ be a version of the conditional expectation of $\size X$ given $\GG$.
Then we have: 
:$\size {\expect {X \mid \GG} } \le \expect {\size X \mid \GG}$ almost everywhere. 
== Proof ==
From Conditional Expectation is Monotone, we have: 
:$\expect {X^+ \mid \GG} \ge 0$ almost everywhere
and:
:$\expect {X^- \mid \GG} \ge 0$ almost everywhere
where $X^+$ and $X^-$ are the positive and negative parts respectively. 
Now, almost everywhere we have:
Category:Conditional Expectation
Triangle Inequality for Generalized Sums

== Theorem ==
Let $V$ be a Banach space.
Let $\norm {\,\cdot\,}$ denote the norm on $V$.
Let $\family {v_i}_{i \mathop \in I}$ be an indexed subset of $V$.
Let the generalized sum $\ds \sum \set {v_i: i \in I}$ converge absolutely.
Then:
:$(1): \quad \ds \norm {\sum \set {v_i: i \in I} } \le \sum \set {\norm {v_i}: i \in I}$
== Proof ==
First of all, note that Absolutely Convergent Generalized Sum Converges assures us that the in $(1)$ is defined.
 there exists an $\epsilon > 0$ such that:
:$\ds \norm {\sum \set {v_i: i \in I} } > \sum \set {\norm {v_i}: i \in I} + \epsilon$
This supposition is seen to be equivalent to:
:$\ds \norm {\sum \set {v_i: i \in I} } > \sum \set {\norm {v_i}: i \in I}$
Then, by definition of a generalized sum, there necessarily exists a finite subset $F$ of $I$ with:
:$\ds \norm {\sum_{i \mathop \in F} v_i} > \norm {\sum \set {v_i: i \in I} } - \epsilon > \sum \set {\norm {v_i}: i \in I}$
However, using the standard triangle equality on this finite sum (that is, , repetitively), we also have:
:$\ds \norm {\sum_{i \mathop \in F} v_i} \le \sum_{i \mathop \in F} \norm {v_i} \le \sum \set {\norm {v_i}: i \in I}$
Here the second inequality follows from Generalized Sum is Monotone.
These two estimates constitute a contradiction, and therefore such an $\epsilon$ cannot exist.
Hence:
:$\ds \norm {\sum \set {v_i: i \in I} } \le \sum \set {\norm {v_i}: i \in I}$
Category:Generalized Sums
Category:Banach Spaces
Category:Triangle Inequality
Triangle Inequality for Indexed Summations

== Theorem ==
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $a,b$ be integers.
Let $\closedint a b$ denote the integer interval between $a$ and $b$.
Let $f : \closedint a b \to \mathbb A$ be a mapping.
Let $\size {\, \cdot \,}$ denote the standard absolute value.
Let $\vert f \vert$ be the absolute value of $f$.
Then we have the inequality of indexed summations:
:$\ds \size {\sum_{i \mathop = a}^b \map f i} \le \sum_{i \mathop = a}^b \size {\map f i}$
== Proof ==
The proof goes by induction on $b$.
=== Basis for the Induction ===
Let $b < a$.
Then all indexed summations are zero.
Because $\size 0 \le \size 0$ by definition of the standard absolute value, the result follows.
This is our basis for the induction.
=== Induction Step ===
Let $b \geq a$.
We have:
By the Principle of Mathematical Induction, the proof is complete.
== Also see ==
* Triangle Inequality for Summation over Finite Set
* Triangle Inequality for Series
Category:Summations
Category:Triangle Inequality
Triangle Inequality for Integrals

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\mu$-integrable function.
Then:
:$\ds \size {\int_X f \rd \mu} \le \int_X \size f \rd \mu$
=== Corollary ===
=== Real Number Line ===
On the real number line, the Triangle Inequality for Integrals takes the following form:
=== Complex Plane ===
In the complex plane, the Triangle Inequality for Integrals takes the following form:
=== Complex Function ===
== Proof ==
We have: 
== Also see ==
* Absolute Value of Definite Integral
* Modulus of Complex Integral
== Sources ==
* : $10.4 \ \text{(v)}$
* : $1$: Integration of complex functions: Theorem $1.33$
Category:Measure Theory
Category:Triangle Inequality
Category:Triangle Inequality for Integrals
Triangle Inequality for Integrals/Complex

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \C$ be a continuous complex function.
Then:
:$\ds \size {\int_a^b \map f t \rd t} \le \int_a^b \size {\map f t} \rd t$
where the first integral is a complex Riemann integral, and the second integral is a definite real integral.
== Proof ==
Define:
:$z \in \C$ as the value of the complex Riemann integral:
::$z = \ds \int_a^b \map f t \rd t$
:$r \in \hointr 0 \to$ as the modulus of $z$
:$\theta \in \hointr 0 {2 \pi}$ as the argument of $z$.
From Modulus and Argument of Complex Exponential:
:$z = re^{i \theta}$
Then:
As $r$ is wholly real, we have:
:$\ds 0 = \map \Im r = \int_a^b \map \Im {e^{-i \theta} \map f t} \rd t$
Then:
As $\ds r = \size {\int_a^b \map f t \rd t}$ by its definition, the result follows.
== Also see ==
* Triangle Inequality for Integrals, of which this is a special case
== Sources ==
* : $\S 2.1$
Category:Triangle Inequality for Integrals
Category:Complex Analysis
Triangle Inequality for Integrals/Complex Function

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\struct {\C, \map \BB \C}$ be the complex numbers made into a measurable space with its Borel $\sigma$-algebra.
Let $f : X \to \C$ be a $\mu$-integrable function.
Then $\cmod f$ is $\mu$-integrable and:
:$\ds \cmod {\int f \rd \mu} \le \int \cmod f \rd \mu$ 
== Proof ==
Let $\struct {\R, \map \BB \R}$ be the real numbers made into a measurable space with its Borel $\sigma$-algebra.
From Complex Modulus of Measurable Function is Measurable, $\cmod f$ is $\Sigma/\map \BB \R$-measurable.
We have:
Taking square roots, we have:
:$\cmod f \le \size {\map \Re f} + \size {\map \Im f}$
Since $f$ is $\mu$-integrable, $\map \Re f$ and $\map \Im f$ are $\mu$-integrable.
From Characterization of Integrable Functions, $\cmod {\map \Re f}$ and $\cmod {\map \Im f}$ are $\mu$-integrable.
Hence from Integral of Integrable Function is Additive, $\size {\map \Re f} + \size {\map \Im f}$ is $\mu$-integrable.
From Characterization of Integrable Functions, $\cmod f$ is then $\mu$-integrable.
Now, let:
:$\ds z = \int f \rd \mu \in \C$
By Complex Multiplication as Geometrical Transformation, there is a complex number $\alpha$ with $\cmod \alpha = 1$ such that:
:$\alpha z = \cmod z \in \R$
Let $u = \map \Re {\alpha f}$.
We have:
:$\ds \alpha \int f \rd \mu = \int \alpha f \rd \mu$
from Integral of Integrable Function is Homogeneous: Complex Function.
Since:
:$\ds \int \alpha f \rd \mu = \cmod {\int f \rd \mu} \in \R$
we have that:
:$\ds \map \Im {\int \alpha f \rd \mu} = \int \map \Im {\alpha f} \rd \mu = 0$ 
Hence:
:$\ds \int \alpha f \rd \mu = \map \Re {\int \alpha f \rd \mu} = \int \map \Re {\alpha f} \rd \mu = \int u \rd \mu$
By Modulus Larger than Real Part, we have that:
:$u \le \cmod {\alpha f} = \cmod f$
Hence by Integral of Integrable Function is Monotone, we have:
:$\ds \int u \rd \mu \le \int \cmod f \rd \mu$ 
Putting everything together we have:
:$\ds \cmod {\int f \rd \mu} \le \int \cmod f \rd \mu$ 
Category:Triangle Inequality for Integrals
Triangle Inequality for Integrals/Corollary

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\mu$-integrable function be such that: 
:$\ds \int \size f \rd \mu = 0$
Then: 
:$\ds \int f \rd \mu = 0$
== Proof ==
From Triangle Inequality for Integrals, we have: 
:$\ds \size {\int f \rd \mu} \le \int \size f \rd \mu$
We have: 
:$\ds \int \size f \rd \mu = 0$
so:
:$\ds \size {\int f \rd \mu} \le 0$
That is: 
:$\ds \size {\int f \rd \mu} = 0$
so:
:$\ds \int f \rd \mu = 0$
Category:Triangle Inequality for Integrals