nyuuzyou/wikis huggingface dataset:
Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication/Examples/7

== Example of Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication ==
== Proof ==
A direct application of Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication:
== Sources ==
* : $18$
* : $18$
Category:Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication
Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication/Examples/9

== Example of Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication ==
== Proof ==
A direct application of Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication:
== Sources ==
* : $18$
* : $18$
Category:Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication
Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication/Examples/99

== Example of Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication ==
== Proof ==
A direct application of Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication:
== Sources ==
* : $18$
Category:Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication
Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication/Historical Note

== Historical Note on Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication ==
 reports in of $1986$ that this result was discussed in Volume $10$ of by , but further corroborating evidence for this has not yet been found.
== Sources ==
* : $18$
Category:Historical Notes
Reversal of Limits of Definite Integral

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \R$ be an integrable real function.
Let:
:$\ds \int_a^b \map f x \rd x$
be the definite integral of $f$ over $\closedint a b$.
Let $a \le b$.
Then:
:$\ds \int_a^b \map f x \rd x = -\int_b^a \map f x \rd x$
== Proof ==
== Sources ==
* : $\S 15$: General Formulas involving Definite Integrals: $15.10$
* : $6$. Integral Calculus: Appendix: Rules and Techniques of Integration: $1.3$
* : $\S 13.2$
* : $\S 13.16$
* 
* 
* : $\S 18$: Definite Integrals: General Formulas involving Definite Integrals: $18.10$
Category:Definite Integrals
Reversal of Number Multiplied by 11

== Theorem ==
Let $n \in \N$ be a number for which, when written in decimal notation, no two adjacent digits total to more than $9$.
Let $n'$ denote the reversal of $n$.
Then $n \times 11$ is the reversal of $n' \times 11$.
== Proof ==
By Basis Representation Theorem, there exists one and only one sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le t}$ such that:
:$(1): \quad \ds n = \sum_{k \mathop = 0}^t r_k 10^k$
:$(2): \quad \ds \forall k \in \closedint 0 t: r_k \in \N_{10}$
:$(3): \quad r_t \ne 0$
Since no two adjacent digits of $n$ total to more than $9$, we have:
:$\paren {r_i + r_{i - 1} } \in \N_{10}$ for $i = 1, 2, \dots, t$
Now:
Since $r_0, r_t$ and each $r_k + r_{k - 1}$ is in $\N_{10}$, the above is the unique representation of $n \times 11$.
The reversal of $n$, $n'$, is given by:
:$\ds n = \sum_{k \mathop = 0}^t r_{t - k} 10^k$
Hence:
Since $r_t, r_0$ and each $r_{t - k} + r_{t - k + 1}$ is in $\N_{10}$, the above is the unique representation of $n' \times 11$.
Its reversal is given by:
:$\ds r_0 + \paren {r_0 + r_1} 10^1 + \paren {r_1 + r_2} 10^2 + \dots + \paren {r_{t - 1} + r_t} 10^t + r_t 10^{t + 1}$
which is:
:$\ds r_0 + \sum_{k \mathop = 1}^t \paren {r_{k - 1} + r_k} 10^k + r_t 10^{t + 1}$
which can be seen to be equal to $n \times 11$.
== Examples ==
== Sources ==
* : $2615$
* : $2615$
Category:Reversals
Category:11
Category:Reversal of Number Multiplied by 11
Reversal of Number Multiplied by 11/Examples

== Examples of Reversal of Number Multiplied by 11 ==
=== Reversal of $2615$ Multipled by $11$ ===
=== Reversal of $45 \, 173$ Multipled by $11$ ===
=== Reversal of $2 \, 363 \, 511 \, 509$ Multipled by $11$ ===
Category:Reversal of Number Multiplied by 11
Reversal of Number Multiplied by 11/Examples/2,363,511,509

== Example of Reversal of Number Multiplied by 11 ==
$11$ multiplied by the reversal of $2 \, 363 \, 511 \, 509$ equals the reversal of the product of $2 \, 363 \, 511 \, 509$ and $11$.
== Proof ==
 2,363,511,509 9,051,153,632
 x 11 x 11
 -------------- --------------
 2,363,511,509 9,051,153,632
 23,635,115,090 90,511,536,320
 -------------- --------------
 25,998,626,599 99,562,689,952
 -------------- --------------
== Sources ==
* : $2615$
* : $2615$
Category:Reversal of Number Multiplied by 11
Reversal of Number Multiplied by 11/Examples/2615

== Example of Reversal of Number Multiplied by 11 ==
$11$ multiplied by the reversal of $2615$ equals the reversal of the product of $2615$ and $11$.
== Proof ==
 2615 5162
 x 11 x 11
 ----- -----
 2615 5162
 26150 51620
 ----- -----
 28765 56782
 ----- -----
== Sources ==
* : $2615$
* : $2615$
Category:Reversal of Number Multiplied by 11
Reversal of Number Multiplied by 11/Examples/45,173

== Example of Reversal of Number Multiplied by 11 ==
$11$ multiplied by the reversal of $45 \, 173$ does not equal the reversal of the product of $45 \, 173$ and $11$.
== Proof ==
 45,173 37,154
 x 11 x 11
 ------- -------
 45,173 37,154
 451,730 371,540
 ------- -------
 496,903 408,694
 ------- -------
== Sources ==
* : $2615$
* : $2615$
Category:Reversal of Number Multiplied by 11
Reversal of Order of Vertices of Triangle causes Reversal of Sign of Area

== Theorem ==
Let $\triangle ABC$ be a triangle embedded in the plane.
Let $\Area \triangle ABC = \AA$.
Then:
:$\Area \triangle CBA = -\AA$.
== Proof ==
$\triangle CBA$ is the same as $\triangle ABC$ but with its vertices in the reverse order.
We have that:
:if $\triangle ABC$ is traversed anticlockwise going $AB \to BC \to CA$, then $\triangle CBA$ is traversed clockwise going $CB \to BA \to AC$
:if $\triangle ABC$ is traversed clockwise going $AB \to BC \to CA$, then $\triangle CBA$ is traversed anticlockwise going $CB \to BA \to AC$.
The result follows by definition of sign of $\Area \triangle ABC$
== Sources ==
* : Chapter $\text I$. Coordinates: $7$. Sign of an area
Category:Triangles
Reverse Fatou's Lemma/Integrable Functions

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\sequence {f_n}_{n \mathop \in \N} \in \LL^1$, $f_n: X \to \R$ be a sequence of $\mu$-integrable functions.
Let $\ds \limsup_{n \mathop \to \infty} f_n: X \to \overline \R$ be the pointwise limit superior of the $f_n$.
Suppose that there exists an $\mu$-integrable $f: X \to \R$ such that for all $n \in \N$, $f_n \le f$ pointwise.
Then:
:$\ds \limsup_{n \mathop \to \infty} \int f_n \rd \mu \le \int \limsup_{n \mathop \to \infty} f_n \rd \mu$
where:
:the integral sign denotes $\mu$-integration
:the limit inferior is taken in the extended real numbers $\overline \R$.
== Proof ==
== Sources ==
* : $\S 10$: Problem $10.8 \ \text{(ii)}$
Category:Reverse Fatou's Lemma
Reverse Fatou's Lemma/Positive Measurable Functions

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$, $f_n: X \to \overline \R$ be a sequence of positive measurable functions.
Suppose that there exists a positive measurable function $f: X \to \overline \R$ such that:
:$\ds \int f \rd \mu < +\infty$
:$\forall n \in \N: f_n \le f$
where $\le$ signifies a pointwise inequality.
Let $\ds \limsup_{n \mathop \to \infty} f_n: X \to \overline \R$ be the pointwise limit superior of the $f_n$.
Then:
:$\ds \limsup_{n \mathop \to \infty} \int f_n \rd \mu \le \int \limsup_{n \mathop \to \infty} f_n \rd \mu$
where:
:the integral sign denotes $\mu$-integration
:the limit superior is taken in the extended real numbers $\overline \R$.
== Proof ==
== Also see ==
* Fatou's Lemma for Integrals
== Sources ==
* : $\S 9$: Problem $8$
Category:Reverse Fatou's Lemma
Reverse Hölder's Inequality for Sums

== Theorem ==
Let $p, q \in \R_{>0}$ be strictly positive real numbers such that $\dfrac 1 p - \dfrac 1 q = 1$.
Suppose that the sequences $\mathbf x = \sequence {x_n}$ and $\mathbf y = \sequence {y_n}$ in $\C$ (or $\R$) are such that the series
:$\ds \paren {\sum_{n \mathop = 1}^\infty \size {x_n}^p}^{1/p}$
and
:$\ds \paren {\sum_{n \mathop = 1}^\infty \size {y_n}^{-q} }^{-1/q}$
are convergent.
Denote these values by $\norm {\mathbf x}_p$ and $\norm {\mathbf y}_{-q}$, respectively.
Here, the notation $\norm {\mathbf x}_p$ does not denote a norm, but is instead just a convenient notation similar to that of the $p$-norm, which is only defined when $p \ge 1$.
Let $\norm {\mathbf {x y} }_1$ denote the $1$-norm of $\mathbf {x y}$, if $\mathbf {x y}$ is in the Lebesgue space $\ell^1$.
Then $\norm {\mathbf {x y} }_1 \ge \norm {\mathbf x}_p \norm {\mathbf y}_{-q}$.
== Proof ==
, assume that $\mathbf x$ and $\mathbf y$ are non-zero.
Let:
:$\mathbf u = \sequence {u_n} = \dfrac {\mathbf x} {\norm {\mathbf x}_p}$
and:
:$\mathbf v = \sequence {v_n} = \dfrac {\mathbf y} {\norm {\mathbf y}_{-q} }$
Then:
:$\ds \norm {\mathbf u}_p = \dfrac 1 {\norm {\mathbf x}_p} \paren {\sum_{n \mathop = 1}^\infty \size {x_n}^p}^{1/p} = 1$
Similarly:
:$\norm {\mathbf v}_{-q} = 1$
It then suffices to prove that:
:$\norm {\mathbf {u v} }_1 = \dfrac {\norm {\mathbf {x y} }_1} {\norm {\mathbf x}_p \norm {\mathbf y}_{-q} } \ge 1$
By Reverse Young's Inequality for Products:
:$\size {u_n v_n} \ge \dfrac 1 p \size {u_n}^p - \dfrac 1 q \size {v_n}^{-q}$
Summing over all $n \in \N$ gives:
:$\norm {\mathbf {u v} }_1 \ge \dfrac 1 p \norm {\mathbf u}_p - \dfrac 1 q \norm {\mathbf v}_{-q} = 1$
as desired.
== Also see ==
* Hölder's Inequality for Sums
Category:Functional Analysis
Reverse Triangle Inequality

== Theorem ==
Let $M = \struct {X, d}$ be a metric space.
Then:
:$\forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$
=== Normed Division Ring ===
=== Normed Vector Space ===
=== Real and Complex Numbers ===
=== Seminormed Vector Space ===
== Proof ==
Let $M = \struct {X, d}$ be a metric space.
By , we have:
:$\forall x, y, z \in X: \map d {x, y} + \map d {y, z} \ge \map d {x, z}$
By subtracting $\map d {y, z}$ from both sides:
:$\map d {x, y} \ge \map d {x, z} - \map d {y, z}$
Now we consider 2 cases. 
;Case $1$: Suppose $\map d {x, z} - \map d {y, z} \ge 0$.
Then:
:$\map d {x, z} - \map d {y, z} = \size {\map d {x, z} - \map d {y, z} }$
and so:
:$\map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$
;Case 2: Suppose $\map d {x, z} - \map d {y, z} < 0$.
Applying again, we have:
:$\forall x, y, z \in X: \map d {y, x} + \map d {x, z} \ge \map d {y, z}$
Hence:
:$\map d {x, y} \ge \map d {y, z} - \map d {x, z}$
Since we assumed $\map d {x, z} - \map d {y, z} < 0$, we have that:
:$\map d {y, z} - \map d {x, z} > 0$
and so:
:$\map d {y, z} - \map d {x, z} = \size {\map d {y, z} - \map d {x, z} }$
Thus we obtain:
:$\map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$
Since these cases are exhaustive, we have shown that:
:$\forall x, y, z \in X: \map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 2$
Category:Triangle Inequality
Category:Named Theorems
Category:Inequalities
Reverse Triangle Inequality/Normed Division Ring

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Then:
:$\forall x, y \in R: \norm {x - y} \ge \bigsize {\norm x - \norm y}$
== Proof ==
Let $0$ be the zero of $\struct {R, \norm {\,\cdot\,} }$.
Let $d$ denote the metric induced by $\norm {\, \cdot \,}$, that is:
:$\map d {x, y} = \norm {x - y}$
From Metric Induced by Norm on Normed Division Ring is Metric we have that $d$ is indeed a metric.
Then, from the Reverse Triangle Inequality as applied to metric spaces:
:$\forall x, y, z \in R: \bigsize {\norm {x - z} - \norm {y - z} } \le \norm {x - y}$
Then:
:$\forall x, y \in R: \bigsize {\norm x - \norm y} = \bigsize {\norm{x - 0} - \norm{y - 0} } \le \norm {x - y}$
Category:Triangle Inequality
Category:Normed Division Rings
Reverse Triangle Inequality/Normed Vector Space

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Then:
:$\forall x, y \in X: \norm {x - y} \ge \size {\norm x - \norm y}$
== Proof ==
Let $d$ denote the metric induced by $\norm {\, \cdot \,}$, that is:
:$\map d {x, y} = \norm {x - y}$
From Metric Induced by Norm is Metric we have that $d$ is indeed a metric.
Then, from the Reverse Triangle Inequality as applied to metric spaces:
:$\forall x, y, z \in X: \size {\norm {x - z} - \norm {y - z} } \le \norm {x - y}$
Then:
:$\forall x, y \in X: \size {\norm x - \norm y} = \size {\norm {x - 0} - \norm {y - 0} } \le \norm {x - y}$.
== Sources ==
* : $\S 1.2$: Normed and Banach spaces. Normed spaces
Category:Triangle Inequality
Reverse Triangle Inequality/Real and Complex Fields

== Theorem ==
Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Then:
:$\cmod {x - y} \ge \size {\cmod x - \cmod y}$
where $\cmod x$ denotes either the absolute value of a real number or the complex modulus of a complex number.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 2$. Conjugate Complex Numbers
* : $\S 2$. Geometrical Representations: $(2.4)$
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Inequalities: $3.7.29$
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Corollary $1.1.10$
* : $1$: Review of some real analysis: Exercise $1.5: 8$
* : $\text {(iii)}$
* : $\text {(iii)}$
Category:Triangle Inequality
Reverse Triangle Inequality/Real and Complex Fields/Corollary 1

== Theorem ==
Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Then:
:$\size {x - y} \ge \size x - \size y$
where $\size x$ denotes either the absolute value of a real number or the complex modulus of a complex number.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $1$: Complex Numbers: Absolute Value: $4$
Category:Triangle Inequality
Reverse Triangle Inequality/Real and Complex Fields/Corollary 1/Proof 1

== Theorem ==
== Proof ==
From the Reverse Triangle Inequality:
:$\cmod {x - y} \ge \cmod {\cmod x - \cmod y}$
By the definition of both absolute value and complex modulus:
:$\cmod {\cmod x - \cmod y} \ge 0$
As:
:$\cmod x - \cmod y = \pm \cmod {\cmod x - \cmod y}$
it follows that:
:$\cmod {\cmod x - \cmod y} \ge \cmod x - \cmod y$
Hence the result.
Category:Triangle Inequality
Reverse Triangle Inequality/Real and Complex Fields/Corollary 1/Proof 2

== Theorem ==
== Proof ==
By the Triangle Inequality:
:$\cmod {x + y} - \cmod y \le \cmod x$
Let $z = x + y$.
Then $x = z - y$ and so:
:$\cmod z - \cmod y \le \cmod {z - y}$
Renaming variables as appropriate gives:
:$\cmod {x - y} \ge \cmod x - \cmod y$
== Sources ==
* : $\S 1$: Real Numbers: $\S 1.18$
* : $1$: Complex Numbers: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $7 \ \text{(c)}$
Category:Triangle Inequality
Reverse Triangle Inequality/Real and Complex Fields/Corollary 1/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 Subtraction, we can construct the parallelogram $OACB$ where:
:$OA$ and $OB$ represent $z_1$ and $z_2$ respectively
:$BA$ represents $z_1 - z_2$.
:
But $OA$, $OB$ and $BA$ form the sides of a triangle.
The result then follows directly from Sum of Two Sides of Triangle Greater than Third Side.
== Sources ==
* : $1$: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $7 \ \text{(c)}$
Category:Triangle Inequality
Reverse Triangle Inequality/Real and Complex Fields/Corollary 2

== Theorem ==
Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Then:
:$\size {x + y} \ge \size x - \size y$
where $\size x$ denotes either the absolute value of a real number or the complex modulus of a complex number.
== Proof ==
Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Let $z := -y$.
Then we have:
Hence the result.
== Sources ==
* : $3$: Elementary Analytic Methods: $3.2$ Inequalities: Triangle Inequalities: $3.2.5$
Category:Triangle Inequality
Reverse Triangle Inequality/Real and Complex Fields/Corollary 3

== Theorem ==
Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Then:
:$\size {x + y} \ge \size {\size x - \size y}$
where $\size x$ denotes either the absolute value of a real number or the complex modulus of a complex number.
== Proof ==
Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Let $z := -y$.
Then we have:
Hence the result.
== Sources ==
* : $\S 36$: Inequalities: Triangle Inequality: $36.1$
Category:Triangle Inequality
Reverse Triangle Inequality/Real and Complex Fields/Examples/6 - (-1)

== Example of Use of Reverse Triangle Inequality ==
:$7 = \size {6 - \paren {-1} } \ge \size 6 - \size {-1} = 6 - 1 = 5$
== Sources ==
* : $\S 1$: Real Numbers: Example $\S 1.19 \ \text{(ii)}$
Category:Examples of Triangle Inequality
Reverse Triangle Inequality/Real and Complex Fields/Proof 1

== Theorem ==
== Proof ==
Let $X$ denote either $\R$ or $\C$ as appropriate.
From Real Number Line is Metric Space and Complex Plane is Metric Space the distance function $d: X \times X \to \R$ can be defined as:
:$\map d {x, y} = \size {x - y}$
From the Reverse Triangle Inequality as applied to metric spaces:
:$(1): \quad \forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$
Let $z = 0$.
Then $(1)$ translates to:
:$\forall x, y, z \in X: \size {\size {x - 0} - \size {y - 0} } \le \size {x - y}$
Hence the result.
Category:Triangle Inequality
Reverse Triangle Inequality/Real and Complex Fields/Proof 2

== Theorem ==
== Proof ==
From proof $2$ of corollary $1$ to this result, which is derived independently:
:$\size {x - y} \ge \size x - \size y$
There are two cases:
$(1): \quad \size x \ge \size y$
We have :
:$\size {\size x - \size y} = \size x - \size y$
and the proof is finished.
$(2): \quad \size y \ge \size x$
We have:
:$\size {y - x} \ge \size y - \size x = \size {\size y - \size x}$
But:
:$\size {y - x} = \size {x - y}$
and:
:$\size {\size y - \size x} = \size {\size x - \size y}$
From this we have:
:$-\size {\size x - \size y} \ge -\size {x - y}$
Since, by Negative of Absolute Value, we have that:
:$\size x - \size y \ge -\size {\size x - \size y}$
it follows that:
:$-\size {x - y} \le \size x - \size y \le \size {x - y}$
The result follows.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.20 \ (2)$
Category:Triangle Inequality
Reverse Triangle Inequality/Seminormed Vector Space

== Theorem ==
Let $\struct {K, +, \circ}$ be a division ring with norm $\norm {\,\cdot\,}_K$.
Let $X$ be a vector space over $K$.
Let $p$ be a seminorm on $X$.
Then:
:$\forall x, y \in X : \size {\map p x - \map p y} \le \map p {x - y}$
== Proof ==
We have:
We also have:
Therefore:
:$-\map p {x - y} \le \map p x - \map p y \le \map p {x - y}$
In view of , the claim is proved.
Category:Triangle Inequality
Category:Seminorms
Reverse Young's Inequality for Products

== Theorem ==
Let $p, q \in \R_{> 0}$ be strictly positive real numbers satisfying:
:$\dfrac 1 p - \dfrac 1 q = 1$
Let $a \in \R_{\ge 0}$ be a positive real number.
Let $b \in \R_{> 0}$ be a strictly positive real number.
Then:
:$a b \ge \dfrac {a^p} p - \dfrac {b^{-q} } q$
== Proof ==
Define:
Then:
Thus Young's Inequality for Products can be applied:
== Geometric Proof ==
Since
:$\dfrac 1 p - \dfrac 1 q = 1$
we have $ p - 1 = \dfrac 1 {-q-1}$.
:
A graph $ y = x^{p - 1} $ on the $ xy $-plane is thus also a graph $ x = y^{-q - 1}$.
:
The above region corresponds to $\ds \int_0^a x^{p - 1} \rd x$
:
The above region corresponds to $\ds \int_b^\infty y^{-q - 1} \rd y$.
:
The above region corresponds to $\ds \int_0^a x^{p - 1} \rd x - \int_b^\infty y^{-q - 1} \rd y$.
Since
:$\dfrac 1 p = \dfrac 1 q + 1 > 1$
we have $0 < p < 1$.
Let $a, b$ be any positive real numbers.
Since $a b$ is the area of the rectangle in the given figure, we have:
:$\ds a b \ge \int_0^a x^{p - 1} \rd x - \int_b^\infty y^{-q - 1} \rd y = \frac {a^p} p - \frac {b^{-q}} q$
Category:Young's Inequality for Products
Reversed Complex Contour is Contour

== Theorem ==
Let $C$ be a contour in the complex plane $\C$ that is defined as a concatenation of a finite sequence $\sequence{ C_1, \ldots, C_n }$ of directed smooth curves in $\C$.
Then the finite sequence of reversed directed smooth curves:
:$\sequence{ -C_n, -C_{n - 1}, \ldots, -C_1 }$
defines a contour that is independent of the parameterization of $C_1, \ldots, C_n$.
== Proof ==
Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
From Reversed Directed Smooth Curve is Directed Smooth Curve, it follows that $-C_k$ is independent of the parameterization $\gamma_k$ of $C_k$.
We now prove that the end point of $-C_k$ is equal to the start point of $-C_{k - 1}$ for all $k \in \set {2, \ldots, n}$.
By definition of reversed directed smooth curve, $-C_k$ is parameterized by $\rho_k: \closedint {a_k} {b_k} \to \C$, defined by $\map {\rho_k} t = \map {\gamma_k} {a_k + b_k - t}$.
From Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints, it follows that the endpoints $\map {\rho_k} {a_k}$ and $\map {\rho_k} {b_k}$ are independent of the parameterization $\rho_k$.
Then:
By definition, it follows that $-C_n, \ldots, -C_1$ can be concatenated to form a contour.
== Sources ==
* $\S 2.2$
Category:Complex Contour Integrals
Reversed Contour Reverses Orientation

== Theorem ==
Let $C$ be a contour in the complex plane $\C$.
Let $D \subseteq \C$ be a connected domain.
Let $\Img C \subseteq \partial D$, where $\Img C$ denotes the image of $C$, and $\partial D$ denotes the boundary of $D$.
Let $-C$ be the reversed contour of $C$.
If $C$ is positively oriented with respect to $D$, then $-C$ is negatively oriented with respect to $D$.
If $C$ is negatively oriented with respect to $D$, then $-C$ is positively oriented with respect to $D$.
=== Corollary ===
== Proof ==
By definition of contour, $C$ is a concatenation of a finite sequence of directed smooth curves $C_1, \ldots, C_n$.
Let each directed smooth curve $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k}{b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
Let $\gamma: \closedint a b \to \C$ be the parameterization of $C$ defined by $\gamma_1, \ldots, \gamma_n$.
By definition of reversed contour, $-C$ is a concatenation of $-C_n , -C_{n-1}, \ldots, -C_1$.
By definition of reversed directed smooth curve, $-C_k$ is parameterized by the smooth path $\rho_k: \closedint {a_k}{b_k} \to \C$, defined by $\map {\rho_k} t = \map {\gamma_k}{a_k + b_k - t}$.
Let $\rho: \closedint c d \to \C$ be the parameterization of $-C$ defined by $\rho_n, \rho_{n-1}, \ldots, \rho_1$.
With $\psi_k: \closedint {a_k}{b_k} \to \closedint {a_k}{b_k}$ defined by $\map {\psi_k} t = a_k + b_k - t$, we have $\rho_k = \gamma_k \circ \psi_k$.
Suppose $C$ is positively oriented with respect to $D$.
Set $K_\rho := \set { u \in \closedint c d : \textrm{ $\rho$ is not differentiable at $u$ } }$.
If $u \in \openint c d \setminus K_\rho$, then there exists $k \in \set {1, \ldots, n}$ such that $t \in \openint {a_k}{b_k}$ with $\map \rho u = \map {\rho_k} t = \map {\gamma_k}{a_k + b_k - t}$.
Then:
By definition of positive orientation, there exists $r \in \R_{>0}$ such that:
:for all $\epsilon \in \openint 0 r$ : $\map {\gamma_k}{ a_k + b_k - t } + \epsilon i \map {\gamma_k'}{ a_k + b_k - t } \in D$, and $\map {\gamma_k}{ a_k + b_k - t } - \epsilon i \map {\gamma_k'}{ a_k + b_k - t } \notin D$
It follows that $-C$ is negatively oriented with respect to $D$, as:
:for all $\epsilon \in \openint 0 r$ : $\map \rho u - \epsilon i \map {\rho'} u \in D$, and $\map \rho u + \epsilon i \map {\rho'} u \notin D$
If $C$ is negatively oriented with respect to $D$, the proof is similar.
Category:Orientation of Complex Contour