nyuuzyou/wikis huggingface dataset:
Naturality of Yoneda Lemma for Covariant Functors

== Theorem ==
Let $C$ be a locally small category.
Let $\mathbf {Set}$ be the category of sets.
Let $\sqbrk {C, \mathbf {Set} }$ be the covariant functor category.
Let $C \times \sqbrk {C, \mathbf {Set} }$ be the product category.
Let $C \times \sqbrk {C, \mathbf {Set} } \to \mathbf {Set}: \tuple {A, F} \mapsto \map {\operatorname {Nat} } {h^A, F}$ be the covariant functor defined as the composition of the hom bifunctor and the product of the contravariant Yoneda functor $h^-$ and the identity functor $\operatorname{id}_{\sqbrk {C, \mathbf {Set} } }$.
Let $\operatorname{ev} : C \times \sqbrk {C, \mathbf {Set} } \to \mathbf {Set}: \tuple {A, F} \mapsto \map F A$ be the functor evaluation functor.
Then $\Phi_{\tuple {A, F} } : \map {\operatorname {Nat} } {h^A, F} \to \map F A: \eta \mapsto \map {\eta_A} {\operatorname {id}_A}$ defines a natural isomorphism, where $\operatorname{id}_A$ is the identity morphism of $A$.
== Proof ==
By the Bijection in Yoneda Lemma for Covariant Functors, $\Phi_{\tuple {A, F} }$ is a bijection for all $\tuple {A, F}$.
Let $\tuple {f, \xi}: \tuple {A, F} \to \tuple {B, G} $ be a morphism in $C \times \sqbrk {C, \mathbf {Set} }$.
To prove that $\Phi$ is a natural isomorphism, it remains to prove that the following diagram commutes:
:$\xymatrix{
\map {\operatorname {Nat} } {h^A, F} \ar[d]^{\Phi_{\tuple {A, F} } } \ar[r]^{\map {\operatorname {Nat} } {h^f, \xi} } & \map {\operatorname {Nat} } {h^B, G} \ar[d]^{\Phi_{\tuple {B, G} } } \\
F(A) \ar[r]^{\map {\operatorname {ev} } {f, \xi} } & \map G B}$
Let $\eta \in \map {\operatorname {Nat} } {h^A, F}$.
We have 
and:
== Also see ==
* Naturality of Yoneda Lemma for Contravariant Functors
Category:Category Theory
Naturally Ordered Semigroup Axioms are Independent

== Theorem ==
Consider the naturally ordered semigroup axioms:
Each of the naturally ordered semigroup axioms is independent of all the others.
That is, you cannot drop any one of them and still uniquely define a naturally ordered semigroup.
== Proof ==
This will be proved by demonstrating that for each of the naturally ordered semigroup axioms, it is possible to create an algebraic structure which fulfils all the others, but is not a naturally ordered semigroup.
=== Positive Rational Numbers under Addition fulfil Naturally Ordered Semigroup Axioms 2 to 4 ===
=== Natural Numbers with Extension fulfil Naturally Ordered Semigroup Axioms 1, 3 and 4 ===
=== Natural Numbers without 1 fulfil Naturally Ordered Semigroup Axioms 1, 2 and 4 ===
=== Singleton fulfils Naturally Ordered Semigroup Axioms 1 to 3 ===
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers
Category:Naturally Ordered Semigroup
Naturally Ordered Semigroup Axioms imply Commutativity

== Theorem ==
Consider the naturally ordered semigroup axioms:
Axioms $\text {NO} 1$, $\text {NO} 2$ and $\text {NO} 3$ together imply the commutativity of the naturally ordered semigroup $\struct {S, \circ, \preceq}$.
== Proof ==
From , $\struct {S, \circ, \preceq}$ has a smallest element.
This is identified as zero: $0$.
From Zero is Identity in Naturally Ordered Semigroup, $0$ is the identity element of $\struct {S, \circ, \preceq}$,
It may be the case that $S$ is a singleton such that $S = \set 0$ .
Then $\struct {S, \circ}$ degenerates to the trivial group.
From Trivial Group is Abelian, it follows that $\circ$ is commutative.
Let $S^* := S \setminus \set 0$ denote the complement of $\set 0$ in $S$.
From , $S^*$ also has a smallest element.
This we will call $1$.
It will be shown that $1$ commutes with every element of $S$.
Let $T \subseteq S$ be the set of element of $S$ which commute with $1$.
We have that $0$ is the identity element of $\struct {S, \circ, \preceq}$.
Hence:
:$1 \circ 0 = 1 = 0 \circ 1$
and it is seen that $0 \in T$.
Now suppose $n \in T$.
That is:
:$1 \circ n = n \circ 1$
We have:
So
:$n \in T \implies n \circ 1 \in T$.
It follows from Principle of Mathematical Induction for Naturally Ordered Semigroup that:
:$T = S$
That is, all the element of $S$ commute with $1$.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.5$
Category:Naturally Ordered Semigroup
Naturally Ordered Semigroup Exists

== Theorem ==
Let the Zermelo-Fraenkel axioms be accepted as axiomatic.
Then there exists a naturally ordered semigroup.
== Proof ==
We take as axiomatic the Zermelo-Fraenkel axioms.
From these, Minimally Inductive Set Exists is demonstrated.
This proves the existence of a minimally inductive set.
Then we have that the Minimally Inductive Set forms Peano Structure.
It follows that the existence of a Peano structure depends upon the existence of such a minimally inductive set.
Then we have that a Naturally Ordered Semigroup forms Peano Structure.
Hence the result.
== Also defined as ==
Some sources accept as axiomatic the naturally ordered semigroup itself.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers
Category:Naturally Ordered Semigroup
Naturally Ordered Semigroup forms Peano Structure

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Let $0 \in S$ be the zero of $S$.
Let $1 \in S$ be the one of $S$.
Let $s: S \to S$ be the mapping defined as:
:$\map s n := n \circ 1$
Then $\struct {S, 0, s}$ is a Peano structure.
== Proof ==
We verify Peano's axioms in turn.
First, suppose that $\map s m = \map s n$ for some $m, n \in S$.
That is:
:$m \circ 1 = n \circ 1$
By Axiom $(\text {NO} 2)$, it follows that $m = n$.
Hence Axiom $(\text P 3)$ holds.
 that $\map s n = 0$ for some $n \in S$.
That is:
:$n \circ 1 = 0$
By Sum with One is Immediate Successor in Naturally Ordered Semigroup, this would imply:
:$n \prec 0$
However, this contradicts the definition of $0$.
Consequently, Axiom $(\text P 4)$ is satisfied.
Finally, let $A \subseteq S$ be such that $0 \in A$ and, for all $n \in S$:
:$n \in A \implies \map s n \in A$
Define $T := S \setminus A$.
 $T \ne \O$.
Then by Axiom $(\text {NO} 3)$, $T$ has a minimal element $n$.
By assumption, $0 \in A$, so that $n \ne 0$.
Therefore, by definition of $1$:
:$1 \preceq n$
By Ordering in terms of Addition, there exists a $p \in S$ such that:
:$p \circ 1 = n$
since $\circ$ is commutative.
By Sum with One is Immediate Successor in Naturally Ordered Semigroup:
:$p \prec n$
Since $n$ is the minimal element of $T$, it follows that $p \notin T$.
Therefore, by definition, $p \in A$.
But by assumption on $A$, this means that $n = p \circ 1 \in A$.
This is a contradiction, hence $T = \O$.
That is to say, $A = S$, and Axiom $(\text P 5)$ holds as well.
Having verified all the axioms, we conclude that $\struct {S, 0, s}$ is a Peano structure.
Category:Naturally Ordered Semigroup
Category:Peano's Axioms
Naturally Ordered Semigroup is Unique

== Theorem ==
Let $\struct {S, \circ, \preceq}$ and $\struct {S', \circ', \preceq'}$ be naturally ordered semigroups.
Let:
:$0'$ be the smallest element of $S'$
:$1'$ be the smallest element of $S' \setminus \set {0'} = S'^*$.
Then the mapping $g: S \to S'$ defined as:
:$\forall a \in S: \map g a = \circ'^a 1'$
is an isomorphism from $\struct {S, \circ, \preceq}$ to $\struct {S', \circ', \preceq'}$.
This isomorphism is unique.
Thus, up to isomorphism, there is only one naturally ordered semigroup.
== Proof ==
=== Proof that Mapping is Isomorphism ===
=== Proof that Isomorphism is Unique ===
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.15$
Category:Naturally Ordered Semigroup is Unique
Category:Naturally Ordered Semigroup
Category:Isomorphisms (Abstract Algebra)
Naturally Ordered Semigroup is Unique/Existence of Isomorphism

== Theorem ==
Let $\struct {S, \circ, \preceq}$ and $\struct {S', \circ', \preceq'}$ be naturally ordered semigroups.
Let:
:$0'$ be the smallest element of $S'$
:$1'$ be the smallest element of $S' \setminus \set {0'} = S'^*$.
Then the mapping $g: S \to S'$ defined as:
:$\forall a \in S: \map g a = \circ'^a 1'$
is an isomorphism from $\struct {S, \circ, \preceq}$ to $\struct {S', \circ', \preceq'}$.
== Proof ==
Let $T' = \Cdm g$, that is, the codomain of $g$.
By Zero is Identity in Naturally Ordered Semigroup, $0'$ is the identity for $\circ'$.
Thus:
:$\map g 0 = \circ'^0 1' = 0'$
and so $0' \in T'$
Suppose $x' \in T'$.
Then:
:$\map g n = x'$
and so:
So:
:$x' \in T' \implies x' \circ' 1' \in T'$
Thus, by the Principle of Mathematical Induction applied to $S'$:
:$T' = S'$
So:
:$\forall a' \in S': \exists a \in S: \map g a = a'$
and so $g$ is surjective by definition.
From Index Laws for Semigroup: Sum of Indices:
:$\map g {a \circ b} = \map g a \circ' \map g b$
and therefore $g$ is a homomorphism from $\struct {S, \circ}$ to $\struct {S', \circ'}$.
Now:
For $p \in S$, let $S_p$ be the initial segment of $S$:
:$S_p = \set {x \in S: x \prec p}$
Let:
:$T = \set {p \in S: \forall a \in S_p: \circ'^a 1' \prec' \circ'^p 1'}$
Now $S_0 = \O \implies 0 \in T$.
Suppose $p \in T$.
Then:
:$a \prec p \circ 1 \implies a \preceq p$
By Strict Lower Closure of Sum with One, either of these is the case:
:$(1): \quad a \prec p: p \in T \implies \circ'^a 1' \prec' \circ'^p 1' \prec' \circ'^{\paren {p \circ 1} } 1'$
:$(2): \quad a = p: \circ'^a 1' = \circ'^p 1' \prec' \circ'^{\paren {p \circ 1} } 1'$
In either case, we have:
:$p \in T \implies p \circ 1 \in T$, and by the Principle of Mathematical Induction:
:$T = S$
So $n \prec p \implies \circ'^n 1' \prec' \circ'^p 1'$.
Thus $g$ is a surjective monomorphism and therefore is an isomorphism from $\struct {S, \circ, \preceq}$ to $\struct {S', \circ', \preceq'}$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.15$
Category:Naturally Ordered Semigroup is Unique
Naturally Ordered Semigroup is Unique/Isomorphism is Unique

== Theorem ==
Let $\struct {S, \circ, \preceq}$ and $\struct {S', \circ', \preceq'}$ be naturally ordered semigroups.
Let:
:$0'$ be the smallest element of $S'$
:$1'$ be the smallest element of $S' \setminus \set {0'} = S'^*$.
Then the isomorphism $g: \struct {S, \circ, \preceq} \to \struct {S', \circ', \preceq'}$ defined as:
:$\forall a \in S: \map g a = \circ'^a 1'$
is unique.
== Proof ==
Let $f: S \to S'$ be another isomorphism different from $g$.
 $\map f 1 \ne 1'$.
We show by induction that $1' \notin \Cdm f$.
... Thus $1' \notin \Cdm f$ which is a contradiction.
Thus $\map f 1 = 1$ and it follows
that $f = g$.
Thus the isomorphism $g$ is unique.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.28 \ \text {(b)}$
Category:Naturally Ordered Semigroup is Unique
Nature of Logistic Map for Various Parameter Values

== Theorem ==
Let $f$ denote the logistic map on the closed unit interval $\closedint 0 1$:
:$x_{n + 1} = a x_n \paren {1 - x_n}$
where:
:$x_0$ can be anywhere in $\closedint 0 1$
:$a$ is a real constant in the interval $\hointl 0 4$.
The nature of $f$ depends on the parameter as follows:
=== Between $0$ and $1$ ===
=== Between $1$ and $3$ ===
=== Between $3$ and $1 + \sqrt 6$ ===
=== Greater than $1 + \sqrt 6$ ===
== Sources ==
* 
* 
Category:Nature of Logistic Map for Various Parameter Values
Category:Logistic Map
Nature of Logistic Map for Various Parameter Values/Between 0 and 1

== Theorem ==
Let $f$ denote the logistic map on the closed unit interval $\closedint 0 1$:
:$x_{n + 1} = a x_n \paren {1 - x_n}$
where:
:$x_0$ can be anywhere in $\closedint 0 1$
:$a$ is a real constant in the interval $\hointl 0 4$.
Let $0 < a < 1$.
Then the point $x = 0$ is an attracting fixed point for all $x_0 \in \closedint 0 1$.
== Proof ==
== Sources ==
* 
* 
Category:Nature of Logistic Map for Various Parameter Values
Nature of Logistic Map for Various Parameter Values/Between 1 and 3

== Theorem ==
Let $f$ denote the logistic map on the closed unit interval $\closedint 0 1$:
:$x_{n + 1} = a x_n \paren {1 - x_n}$
where:
:$x_0$ can be anywhere in $\closedint 0 1$
:$a$ is a real constant in the interval $\hointl 0 4$.
Let $1 < a < 3$.
Then:
:$f$ has a repelling fixed point at $x = 0$ for all $x_0 \in \closedint 0 1$
:$f$ has an attracting fixed point at $x = 1 - \dfrac 1 a$ for all $x_0 \in \closedint 0 1$.
== Proof ==
== Sources ==
* 
* 
Category:Nature of Logistic Map for Various Parameter Values
Nature of Logistic Map for Various Parameter Values/Between 3 and 1 + sqrt 6

== Theorem ==
Let $f$ denote the logistic map on the closed unit interval $\closedint 0 1$:
:$x_{n + 1} = a x_n \paren {1 - x_n}$
where:
:$x_0$ can be anywhere in $\closedint 0 1$
:$a$ is a real constant in the interval $\hointl 0 4$.
Let $3 < a < 1 + \sqrt 6$.
Then:
:$f$ has a repelling fixed point at $x = 0$ for all $x_0 \in \closedint 0 1$
:$f$ has $2$ repelling periodic points of period $2$ for all $x_0 \in \closedint 0 1$.
== Proof ==
== Sources ==
* 
* 
Category:Nature of Logistic Map for Various Parameter Values
Nature of Logistic Map for Various Parameter Values/Greater than 1 + sqrt 6

== Theorem ==
Let $f$ denote the logistic map on the closed unit interval $\closedint 0 1$:
:$x_{n + 1} = a x_n \paren {1 - x_n}$
where:
:$x_0$ can be anywhere in $\closedint 0 1$
:$a$ is a real constant in the interval $\hointl 0 4$.
Let $a > 1 + \sqrt 6$.
Then:
:$f$ has a repelling fixed point at $x = 0$ for all $x_0 \in \closedint 0 1$
:$f$ has repelling periodic points which bifurcate into orbits of periods $2, 4, 8 \ldots$ for all $x_0 \in \closedint 0 1$.
== Proof ==
== Sources ==
* 
* 
Category:Nature of Logistic Map for Various Parameter Values
Navier-Stokes Existence and Smoothness

== Unsolved Problem ==
It has not yet been proven that the Navier-Stokes equations:
:always exist in ordinary $3$-dimensional space
:if they do exist, they do not contain any singular points.
== Progress ==
== Sources ==
* 
* 
* 
* : Appendix $18$: Millennium Prize problems: $5$.
* : Appendix $23$: Millennium Prize problems: $5$.
Category:Navier-Stokes Equations
Category:Millennium Problems
Category:Unsolved Problems
Near-Repdigit Prime/Sequence

== Sequence of Near-Repdigit Primes ==
The sequence of near-repdigit primes begins:
:$101$, $113$, $131$, $151$, $181$, $191$, $199$, $211$, $223$, $227$, $229$, $233$, $277$, $311$, $313$, $331$, $337$, $\ldots$
Category:Near-Repdigit Primes
Necessary Condition for Autoregressive Process to be Stationary

== Theorem ==
Let $S$ be a stochastic process based on an equispaced time series.
Let the values of $S$ at timestamps $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be deviations from a constant mean level $\mu$:
:$\tilde z_t = z_t - \mu$
Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a sequence of independent shocks at timestamps $t, t - 1, t - 2, \dotsc$
Let $M$ be an autoregessive model on $S$ of order $p$:
:$\map \phi B \tilde z_t = a_t$
where $\map \phi B := 1 - \phi_1 B - \phi_2 B^2 - \dotsb - \phi_p B^p$ is the autoregressive operator of order $p$.
Consider the polynomial equation in $B$ of degree $p$:
:$(1): \quad \map \phi B = 0$
Let $\map R \phi \subseteq \C$ denote the set of roots of $(1)$, considered as a polynomial of degree $p$.
It is noted that the elements of $\map R \phi$ may be real or complex.
For $S$ modelled by $M$ to be a stationary process, it is necessary that the elements of $\map R \phi$ have a complex modulus greater than $1$:
:$\forall z \in \map R \phi: \size z > 1$
== Proof ==
== Sources ==
* :
::$1$: Introduction:
:::$1.2$ Stochastic and Deterministic Dynamic Mathematical Models
::::$1.2.1$ Stationary and Nonstationary Stochastic Models for Forecasting and Control: Autoregressive models
Category:Autoregressive Models
Necessary Condition for Existence of BIBD

== Theorem ==
Let there exist a BIBD with parameters $v, b, r, k, \lambda$.
Then the following are true:
:$(1): \quad b k = r v$
:$(2): \quad \lambda \paren {v - 1} = r \paren {k - 1}$
:$(3): \quad b \dbinom k 2 = \lambda \dbinom v 2$
:$(4): \quad k < v$
:$(5): \quad r > \lambda$
All of $v, b, r, k, \lambda$ are integers.
Some sources prefer to report $(3)$ as:
:$b = \dfrac {\dbinom v 2} {\dbinom k 2} \lambda$
which is less appealing visually, and typographically horrendous.
== Proof ==
:$(1)$: We have by definition of balanced incomplete block design that:
:each treatment is in exactly $r$ blocks
:each block is of size $k$.
We have that $b k$ is the number of blocks times the size of each block.
We also have that $r v$ is the number of treatments times the number of blocks each treatment is in.
The two must clearly be equal.
:$(2)$: Comparing the and of the equation we can see that:
: An arbitrary treatment must be paired with $v - 1$ other treatments.
If $\lambda > 1$ then every treatment is paired $\lambda \paren {v - 1}$ times.
: An arbitrary treatment is paired with $k - 1$ other treatments for each of the $r$ blocks it is in.
Therefore it is paired $r \paren {k - 1}$ times.
Both values give the number of times an arbitrary treatment is paired, therefore the equals the .
:$(3)$: From equation $(1)$, we have that $r = \dfrac {b k} v$
From $(2)$ we have that:
:$r = \dfrac {v - 1} {k - 1} \lambda$
So:
Category:Balanced Incomplete Block Designs
Necessary Condition for Integral Functional to have Extremum/Two Variables

== Theorem ==
Let $D \subset \R^2$.
Let $\Gamma$ be the boundary of $D$.
Let $S$ be a set of real mappings such that:
:$S = \set {\map z {x, y}: \paren {z: S_1 \subseteq \R^2 \to S_2 \subseteq \R}, \paren {\map z {x, y} \in \map {C^2}D}, \paren {\map z \Gamma = 0} }$
Let $J \sqbrk z: S \to S_3 \subseteq \R$ be a functional of the form:
:$\ds \iint_D \map F {x, y, z, z_x, z_y} \rd x \rd y$ 
Then a necessary condition for $J \sqbrk y$ to have an extremum (strong or weak) for a given function $\map z {x, y}$ is that $\map z {x, y}$ satisfy Euler's equation:
:$F_z - \dfrac \partial {\partial x} F_{z_x} - \dfrac \partial {\partial y} F_{z_y} = 0$
== Proof ==
From Condition for Differentiable Functional to have Extremum we have
:$\bigvalueat {\delta J \sqbrk {z; h} } {z \mathop = \hat z} = 0$
The variation exists if $J$ is a differentiable functional.
Since $\map z \Gamma = 0$, $\map h {x, y}$ vanishes on the boundary $\Gamma$:
:$\bigvalueat h \Gamma = 0$
From the definition of increment of a functional:
:
Using multivariate Taylor's theorem, expand $\map F {x, y, z + h, z_x + h_x, z_y + h_y}$ $h$, $h_x$ and $h_y$.
Denote the ordered tuples $\tuple {h, h_x, h_y}$ as $\mathbf h$ and $\tuple {z, z_x, z_y}$ as $\mathbf z$ respectively.
Then:
:$\ds \map F {x, y, \mathbf z + \mathbf h} = \valueat {\map F {x, y, \mathbf z + \mathbf h} } {\mathbf h \mathop = 0} + \valueat {\frac {\partial {\map F {x, y, \mathbf z + \mathbf h} } } {\partial z} } {\mathbf h \mathop = 0} h + \valueat {\frac {\partial {\map F {x, y, \mathbf z + \mathbf h} } } {\partial z_x} } {\mathbf h \mathop = 0} z_x + \valueat {\frac {\partial {\map F {x, y, \mathbf z + \mathbf h} } } {\partial z_y} } {\mathbf h \mathop = 0} z_y + \map \OO {\mathbf h^2}$
where the last term includes all terms of the order not lesser than 2 the elements of $\mathbf h$.
Substitute this back into the integral: 
:$\ds \Delta J \sqbrk {y; h} = \iint_D \paren {\map F {x, y, \mathbf h}_z h + \map F {x, y, \mathbf z}_{z_x} h_x + \map F {x, y, \mathbf z}_{z_y} h_y + \map \OO {\mathbf h^2} } \rd x \rd y$
Terms in $\map \OO {\mathbf h^2}$ represent terms of order higher than 1 with respect to elements of $\mathbf h$.
By definition, the integral not counting in $\map \OO {\mathbf h^2}$ is a variation of functional:
Then, by the use of the lemma:
:$F_z - \dfrac \partial {\partial x} F_{z_x} - \dfrac \partial {\partial y} F_{z_y} = 0$
== Sources ==
* : $\S 1.5$: The Case of Several Variables
Category:Calculus of Variations
Necessary Condition for Integral Functional to have Extremum/Two Variables/Lemma

== Theorem ==
Let $D \subset \R^2$.
Let $\Gamma$ be the boundary of $D$.
Let $\alpha : D \to \R$ be a continuous mapping.
Let $h : D \to \R$ be a twice differentiable mapping such that $\map h \Gamma = 0$.
Suppose for every $h$ we have that:
:$\ds \iint_D \map \alpha {x, y} \map h {x,y} \rd x \rd y = 0$.
Then:
:$\ds \forall x, y \in D : \map \alpha {x, y} = 0$
== Proof ==
 that:
:$\ds \exists x_0,y_0 \in D : \map \alpha {x_0,y_0} > 0$
$\alpha$ is continuous in $D$. 
Therefore, there exists a closed ball $B^-_{\epsilon}$ defined by:
:$\map {B^-_{\epsilon}} {x_0, y_0} := \set {\tuple{x,y} \in D : \paren {x - x_0}^2 + \paren {y - y_0}^2 \le \epsilon^2}$
such that:
:$\forall x,y \in B^-_{\epsilon} : \map \alpha {x,y} > 0$
Choose $\map h {x, y}$ in the following way:
:$\map h {x,y} = \begin {cases} 0, \forall x, y \notin B^-_{\epsilon} \\ \sqbrk {\epsilon^2 - \paren {x - x_0}^2 - \paren {y - y_0}^2}^3, \forall x,y \in B^-_{\epsilon} \end{cases}$
Such a choice for $\map h {x,y}$ satisfies the conditions of the lemma.
But then both $\alpha$ and $h$ are positive inside this ball.
Hence, the integral is positive.
This contradicts assumptions of the lemma.
Hence:
:$\ds \forall x, y \in D : \map \alpha {x,y} = 0$
== Mistake ==
The author uses a slightly different form of the function $h$:
:$\map h {x,y} = \begin{cases} 0, \forall x, y \notin B^-_{\epsilon} \\ \sqbrk {\paren {x - x_0}^2 + \paren {y - y_0}^2 - \epsilon^2}^3, \forall x,y \in B^-_{\epsilon} \end{cases}$
Then he concludes that for points inside the circle the integrand is positive.
$\alpha$ is positive by assumption.
However, $h$ is not positive inside since $\paren {x - x_0}^2 + \paren {y - y_0}^2 < \epsilon^2$.
The result is negative, and stays negative after being raised to the $3$rd power.
== Sources ==
* : $\S 1.5$: The Case of Several Variables
Category:Calculus of Variations
Necessary Condition for Integral Functional to have Extremum for given Function/Non-differentiable at Intermediate Point

== Theorem ==
Let $y, F$ be real functions.
Let $y$ be continuously differentiable for $x \in \hointr a c \cap \hointl c b$ and satisfy:
:$\map y a = A$
:$\map y b = B$
Let $J\sqbrk y$ be a functional of the form
:$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
Then the functional $J$ has a weak extremum if $y$ satisfies the following system of equations:
where, by the use of limit from the left and limit from the right, the following abbreviations are denoted as follows:
:$\ds \bigvalueat {\map y x} {x \mathop = c \mathop + 0} = \lim_{x \mathop \to c^+} \map y x$
:$\ds \bigvalueat {\map y x} {x \to x \mathop = c \mathop - 0} = \lim_{x \mathop \to c^-} \map y x$
The last two equations are known as the Weierstrass-Erdmann corner conditions.
== Proof ==
Rewrite $J \sqbrk y$ as a sum of two functionals:
Recall that end points $x = a,x = b$ are fixed.
The function $\map y x$ has to be $C^0$ at $x = c$, but otherwise this point can move freely.
From general variation of functional, and noting that $y = \map y x$ is an extremal, write down variations for $J_1 \sqbrk y$ and $J_2 \sqbrk y$ separately:
$\ds \delta J_1 = \bigvalueat {F_{y'} } {x \to c \mathop - 0} \delta y_1 + \bigvalueat {\paren {F - y' F_{y'} } } {x \to c \mathop - 0} \delta x_1$
$\ds \delta J_2 = \bigvalueat {-F_{y'} } {x \to c \mathop + 0} \delta y_1 - \bigvalueat {\paren {F - y' F_{y'} } } {x \to c \mathop + 0} \delta x_1$
Note that $\delta J_1$ and $\delta J_2$ involve the same increments $\delta x_1$ and $\delta y_1$.
Since $y = \map y x$ is an extremum of $J$, we have:
Since $ \delta x_1$ and $ \delta y_1$ are arbitrary, both collections of terms have to vanish independently.
== Sources ==
* 
* : $\S 3.15$: Broken Extremals. The Weierstrass-Erdmann Conditions
Category:Calculus of Variations