avewright/proofwiki-math huggingface dataset:
Theorem: Hyperplane in Normed Vector Space generated by Unbounded Linear Functional is Everywhere Dense

Let $\Bbb F \in \set {\R, \C}$.
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.
Let $\phi : X \to \Bbb F$ be a linear functional that is not bounded.
Let $U$ be a hyperplane in $X$ given by: 
:$U = \map \ker \phi$
Then: 
:$U$ is everywhere dense in $X$.

Proof:
Since $\phi$ is not bounded, for each $n \in \N$ we can pick $v_n \ne 0$ such that:
:$\cmod {\map \phi {v_n} } \ge n \norm {v_n}$
Set:
:$\ds x_n = \frac {v_n} {\norm {v_n} }$
for each $n \in \N$.
Then for each $n \in \N$, we have:
:$\cmod {\map \phi {x_n} } \ge n$
from linearity with: 
:$\norm {x_n} = 1$
Fix $x \in X$ and $\epsilon > 0$.
We aim to show that there exists $y \in U$ such that: 
:$\norm {x - y} < \epsilon$
It suffices to find a sequence $\sequence {y_n}_{n \mathop \in \N}$ in $U$ such that: 
:$y_n \to x$ 
Then, from the definition of convergent sequence, we can find $N \in \N$ such that: 
:$\norm {x - y_n} < \epsilon$
For each $n \in \N$, let: 
:$\ds y_n = x - \frac {\map \phi x} {\map \phi {x_n} } x_n$
Then we have: 
{{begin-eqn}}
{{eqn	| l = \map \phi {y_n}
		| r = \map \phi x - \map \phi {\frac {\map \phi x} {\map \phi {x_n} } x_n}
		| c = {{Defof|Linear Functional}}
}}
{{eqn	| r = \map \phi x - \frac {\map \phi x} {\map \phi {x_n} } \map \phi {x_n}
		| c = {{Defof|Linear Functional}}
}}
{{eqn	| r = \map \phi x - \map \phi x
}}
{{eqn	| r = 0
}}
{{end-eqn}}
So, we have: 
:$y_n \in \map \ker \phi$
and so:
:$y_n \in U$
We also have: 
{{begin-eqn}}
{{eqn	| l = \norm {x - y_n}
		| r = \norm {\frac {\map \phi x} {\map \phi {x_n} } x_n}
}}
{{eqn	| r = \frac {\cmod {\map \phi x} } {\cmod {\map \phi {x_n} } } \norm {x_n}
		| c = positive homogeneity of the norm.
}}
{{eqn	| r = \frac {\cmod {\map \phi x} } {\cmod {\map \phi {x_n} } }
		| c = since $\norm {x_n} = 1$
}}
{{end-eqn}}
Since: 
:$\cmod {\map \phi {x_n} } \ge n$
We have: 
:$\ds \frac {\cmod {\map \phi x} } {\cmod {\map \phi {x_n} } } \le \frac {\cmod {\map \phi x} } n$
So, we have: 
:$\ds \norm {x - y_n} \le \frac {\cmod {\map \phi x} } n$
We therefore have: 
:$\ds \norm {x - y_n} \to 0$ as $n \to \infty$.
From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we have: 
:$\ds y_n \to x$
So $\sequence {y_n}_{n \mathop \in \N}$ is the desired sequence converging to $x$.
Since $x$ was arbitrary, we have that: 
:$U$ is everywhere dense in $X$.
{{qed}}
Category:Hyperplanes
Category:Normed Vector Spaces
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Theorem: Preimage of Maximum of Real Part of Continuous Linear Functional on Extreme Set in Convex Compact Set is Extreme Set

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \PP}$ be a locally convex space over $\GF$ with its standard topology.
Let $X^\ast$ be the topological dual space of $X$.
Let $K$ be a compact convex subset of $X$. 
Let $M \subseteq K$ be an extreme set in $K$. 
Let $f \in X^\ast$. 
Let:
:$\ds M^f = \set {x \in M : \map \Re {\map f x} = \max_{y \in M} \map \Re {\map f y} }$
Then $M^f$ is an extreme set in $K$.

Proof:
Note that from Continuous Function on Compact Space is Bounded we have that: 
:$\map \Re f$ attains its supremum
and so $M^f$ is well-defined.
From Mapping is Continuous iff Inverse Images of Open Sets are Open: Corollary, we have: 
:$M^f$ is closed.
Let $x, y \in K$ and $t \in \openint 0 1$ be such that: 
:$t x + \paren {1 - t} y \in M^f$
We show that $x, y \in M^f$. 
Since $M^f \subseteq M$, we have: 
:$t x + \paren {1 - t} y \in M$
and so $x, y \in M$ since $M$ is an extreme set in $K$. 
Since $t x + \paren {1 - t} y \in M^f$, we have: 
:$\ds \map \Re {\map f {t x + \paren {1 - t} y} } = \max_{z \in M} \map \Re {\map f z}$
Since $f \in X^\ast$, we have: 
:$\ds t \map \Re {\map f x} + \paren {1 - t} \map \Re {\map f y} = \max_{z \in M} \map \Re {\map f z}$
On the other hand, since $x, y \in M$, we have: 
:$\ds \map \Re {\map f x} \le \max_{z \in M} \map \Re {\map f z}$
and:
:$\ds \map \Re {\map f y} \le \max_{z \in M} \map \Re {\map f z}$ 
from the definition of maximum.
If: 
:$\ds \map \Re {\map f x} < \max_{z \in M} \map \Re {\map f z}$
or: 
:$\ds \map \Re {\map f y} < \max_{z \in M} \map \Re {\map f z}$
we have: 
:$\ds t \map \Re {\map f x} + \paren {1 - t} \map \Re {\map f y} < \max_{z \in M} \map \Re {\map f z}$
So, we must have: 
:$\ds \map \Re {\map f x} = \map \Re {\map f y} = \max_{z \in M} \map \Re {\map f z}$
So:
:$x, y \in M^f$
So:
:$M^f$ is an extreme set in $K$.
{{qed}}
Theorem: Point in Convex Set is Extreme Point iff Singleton is Extreme Set

Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$. 
Let $K$ be a convex subset of $X$. 
Let $a \in K$.
Then $a$ is an extreme point of $K$ {{iff}}:
:$\set a$ is an extreme set in $K$.

Proof:
We have: 
:$a$ is an extreme point of $K$.
{{iff}}: 
:whenever $a = t x + \paren {1 - t} y$ for some $x, y \in K$ and $t \in \openint 0 1$, we have $x = y = a$.
We can rewrite this: 
:whenever $t x + \paren {1 - t} y \in \set a$ for some $x, y \in K$ and $t \in \openint 0 1$, we have $x, y \in \set a$.
This is precisely the criteria for $\set a$ being an extreme set in $K$.
{{qed}}
Theorem: Convex Hull is Smallest Convex Set containing Set

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $U \subseteq X$ be non-empty.
Let $\map {\operatorname {conv} } U$ be the convex hull of $U$.
Then $\map {\operatorname {conv} } U$ is the smallest convex subset of $X$ containing $U$ in the sense that: 
:$\map {\operatorname {conv} } U$ is convex and if $K \subseteq X$ is a convex subset with $U \subseteq K$, we have that $\map {\operatorname {conv} } U \subseteq K$.

Proof:
We have: 
:$\ds \map {\operatorname {conv} } U = \set {\sum_{j \mathop = 1}^n \lambda_j u_j : n \in \N, \, u_j \in U \text { and } \lambda_j \in \R_{> 0} \text { for each } j, \, \sum_{j \mathop = 1}^n \lambda_j = 1}$
Considering $n = 1$ and $\lambda_1 = 1$, we obtain: 
:$u \in \map {\operatorname {conv} } U$ for each $u \in U$. 
So:
:$U \subseteq \map {\operatorname {conv} } U$
Now we prove that $\map {\operatorname {conv} } U$ is convex.
Let $u, v \in \map {\operatorname {conv} } U$.
Then there exists $u_1, u_2, \ldots, u_n \in U$, $v_1, v_2, \ldots, v_m$ and $\lambda_1, \lambda_2, \ldots, \lambda_n, \mu_1, \ldots, \mu_m \in \R_{> 0}$ such that: 
:$\ds u = \sum_{i \mathop = 1}^n \lambda_i u_i$
and:
:$\ds v = \sum_{j \mathop = 1}^n \mu_j v_j$
with:
:$\ds \sum_{i \mathop = 1}^n \lambda_i = 1$
and:
:$\ds \sum_{j \mathop = 1}^n \mu_j = 1$ 
Take:
:$u_{n + j} = v_j$ for each $1 \le j \le m$
and:
:$\lambda_{n + j} = \mu_j$ for each $1 \le j \le m$ 
Then:
:$\ds u = \sum_{j \mathop = 1}^n \lambda_j u_j$
and:
:$\ds v = \sum_{j \mathop = n + 1}^{n + m} \lambda_j u_j$
Now let $t \in \closedint 0 1$.
We have: 
:$\ds t u + \paren {1 - t} v = \sum_{j \mathop = 1}^n t \lambda_j u_j + \sum_{j \mathop = n + 1}^{n + m} \paren {1 - t} \mu_j u_j$ 
We have: 
{{begin-eqn}}
{{eqn	| l = \sum_{i \mathop = 1}^n t \lambda_i + \sum_{j \mathop = n + 1}^{n + m} \paren {1 - t} \lambda_j
		| r = t \sum_{i \mathop = 1}^n \lambda_i + \paren {1 - t} \sum_{j \mathop = 1}^m \mu_j
}}
{{eqn	| r = t + \paren {1 - t}
}}
{{eqn	| r = 1
}}
{{end-eqn}}
with $t \lambda_i \ge 0$ and $\paren {1 - t} \mu_j \ge 0$ for each $i, j$.
Let:
:$s_j = t \lambda_i$ for each $1 \le i \le n$
:$s_{n + j} = \paren {1 - t} \mu_j$ for each $1 \le j \le m$. 
We have:
:$\ds t x + \paren {1 - t} y = \sum_{j \mathop = 1}^{n + m} s_j x_j$
with:
:$\ds \sum_{j \mathop = 1}^{n + m} s_j = 1$ 
Since we have $u_i \in U$ for each $i$ we have: 
:$t u + \paren {1 - t} v \in \map {\mathrm {conv} } U$
So $\map {\mathrm {conv} } U$ is convex.
From Convex Combination contained in Convex Set, we have $\map {\mathrm {conv} } U \subseteq K$.
Hence the result.
{{qed}}
Theorem: Krein-Milman Theorem

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\GF$.
Let $K$ be a non-empty compact convex subset of $X$.
Let $\map E K$ be the set of extreme points in $K$. 
Then $K$ is the closed convex hull of $\map E K$.

Proof:
Let $K'$ be the closed convex hull of $\map E K$.
We show that $K' = K$.
From Convex Hull is Smallest Convex Set containing Set, we have: 
:$\map E K \subseteq \map {\operatorname {conv} } {\map E K}$
From Set is Subset of its Topological Closure, we therefore have: 
:$\map E K \subseteq K'$
We also have, from Convex Hull preserves Subsets:
:$\map {\operatorname {conv} } {\map E K} \subseteq \map {\operatorname {conv} } K$ 
From Convex Hull is Smallest Convex Set containing Set: Corollary, we have: 
:$\map {\operatorname {conv} } {\map E K} \subseteq K$ 
Then, from Set Closure Preserves Set Inclusion, we have: 
:$K' \subseteq K^-$
where $K^-$ is the closure of $K$.
From Compact Subspace of Hausdorff Space is Closed, we have that: 
:$K$ is closed.
So we have: 
:$K' \subseteq K$ 
{{AimForCont}} suppose that $K' \ne K$.
Then there exists $a \in K \setminus K'$. 
Note that $K'$ is closed from Topological Closure is Closed.
Since $K$ is compact and $K' \subseteq K$, we have: 
:$K'$ is compact
from Closed Subspace of Compact Space is Compact. 
From Closed Convex Hull in Normed Vector Space is Convex, we have that: 
:$K'$ is convex.
Finally:
:$\set a$ is closed
from Singleton in Normed Vector Space is Closed, and:
:$\set a$ is convex
from Singleton is Convex Set.
So, from the:
:real Hahn-Banach Separation Theorem in the case of a compact convex set and a closed convex set if $\GF = \R$
:real Hahn-Banach Separation Theorem in the case of a compact convex set and a closed convex set if $\GF = \C$
there exists $f \in X^\ast$ such that: 
:$\map \Re {\map f x} < \map \Re {\map f a}$
for each $x \in K'$. 
Now let: 
:$\ds K^f = \set {x \in K : \map \Re {\map f x} = \max_{y \mathop \in K} \map \Re {\map f y} }$ 
From Preimage of Maximum of Real Part of Continuous Linear Functional on Extreme Set in Convex Compact Set is Extreme Set:
:$K^f$ is an extreme set in $K$. 
Since: 
:$\map \Re {\map f x} < \map \Re {\map f a}$ for each $x \in K'$
and $a \in K$, we have: 
:$\ds \map \Re {\map f x} < \max_{y \mathop \in K} \map \Re {\map f y}$ for each $x \in K'$
from the definition of maximum.
So, we have: 
:$K^f \cap K' = \O$ 
However, from Extreme Set in Compact Convex Set contains Extreme Point, we also have: 
:$K^f$ contains an extreme point of $K$, say $x \in \map E K$
while: 
:$\map E K \subseteq K$  
So we have: 
:$x \in K^f \cap K'$
contradicting that $K^f \cap K$ is empty.
So, there exists no such $a \in K \setminus K'$, and we have: 
:$K = K'$
{{qed}}
{{Namedfor|Mark Grigorievich Krein|name2 = David Pinhusovich Milman|cat = Krein|cat2 = Milman, D.}}
Theorem: Extreme Set in Compact Convex Set contains Extreme Point

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \PP}$ be a locally convex space over $\GF$ with its standard topology.
Let $K$ be a non-empty compact convex subset of $X$.
Let $E$ be an extreme set of $K$. 
Then $E$ contains an extreme point of $K$.

Proof:
Let $P$ be the set of extreme sets in $K$ that are contained in $E$. 
Since $E \in P$, $P$ is certainly non-empty.
Define a relation $\preceq$ on $P$ by $A \preceq B$ {{iff}} $B \subseteq A$. 
From Subset Relation is Ordering and Dual Ordering is Ordering, we have:
:$\struct {P, \preceq}$ is an ordered set.
We show that every non-empty chain in $\struct {P, \preceq}$ has an upper bound. 
We will then invoke Zorn's Lemma.
Theorem: Resolvent Mapping is Continuous/Bounded Linear Operator

Let $B$ be a Banach space.
Let $\mathfrak{L}(B, B)$ be the set of bounded linear operators from $B$ to itself.
Let $T \in \mathfrak{L}(B, B)$.
Let $\rho(T)$ be the resolvent set of $T$ in the complex plane.
Then the resolvent mapping $f : \rho(T) \to \mathfrak{L}(B,B)$ given by:
:$f(z) = (T - zI)^{-1}$
is continuous in the operator norm $\|\cdot\|_*$.

Proof:
Pick $z\in\rho(T)$. Since $z\in\rho(T)$, the operator $R_z = (T - zI)^{-1}$ exists and has finite norm $C \geq 0$.
Since Resolvent Set is Open, $z+h\in\rho(T)$ for any $h\in \Bbb C$ smaller than some $\delta > 0$. For such $h$,
{{begin-eqn}}
{{eqn | l = \norm{ f(z+h) - f(z) }_*
      | r = \norm{ (T-(z+h)I)^{-1} - (T-zI)^{-1}  }_*
}}
{{eqn | r = \norm{ ((T-zI)-hI)^{-1} - (T-zI)^{-1}  }_*
}}
{{eqn | r =  \norm{ [(T-zI)(I-h(T-zI)^{-1})]^{-1} - (T-zI)^{-1}  }_*
      | c = by factoring out $T-zI$ from inside the first term
}}
{{eqn | r =  \norm{ (I-h(T-zI)^{-1})^{-1}(T-zI)^{-1} - (T-zI)^{-1}  }_*
}}
{{eqn | r =  \norm{ \big[(I-h(T-zI)^{-1})^{-1} - I\big] (T-zI)^{-1}  }_*
}}
{{eqn | o = \leq
      | r =  \norm{ (I-h(T-zI)^{-1})^{-1} - I }_* \norm{(T-zI)^{-1}  }_*
      | c = by Operator Norm on Banach Space is Submultiplicative.
}}
{{eqn | r =  C\norm{ (I-hR_z)^{-1} - I }_*.
      | c = (1)
}}
{{end-eqn}}
Consider:
:$(I - hR_z)^{-1}$
Either:
:$\|R_z\|_* = 0$ 
so that:
:$\forall h \in \Bbb C: \|hR_z\|_* = |h|\|R_z\|_* = 0$ (by Operator Norm is Norm)
or else:
:$\|hR_z\|_* < 1$ for $h < 1/\|R_z\|_*$
In either case:
:$\|hR_z\|_* < 1$ 
for all sufficiently small $h$.
Therefore, by Invertibility of Identity Minus Operator, for sufficiently small $h$ we have $(I - hR_z)^{-1} = I + hR_z + h^2R_z^2 + \ldots$. Substituting this into $(1)$, we get
{{begin-eqn}}
{{eqn | l = \norm{ f(z+h) - f(z) }_*
      | r = C\norm{ hR_z + h^2R_z^2 + \ldots }_*
      | o = \leq
}}
{{eqn | r = C\norm{hR_z (I + hR_z + h^2R_z^2 + \ldots) }_*
}}
{{eqn | r = C\size{h}\norm{R_z}_*\norm{I + hR_z + h^2R_z^2 + \ldots }_*
      | o = \leq
      | c = by Operator Norm on Banach Space is Submultiplicative
}}
{{eqn | r = C^2\size{h} \sum_{n\in N} (\size{h}\norm{R_z})^n
      | o = \leq
      | c = by Triangle Inequality and Operator Norm on Banach Space is Submultiplicative again on each term
}}
{{eqn | r = 2C^2\size{h}
      | o = \leq
}}
{{end-eqn}}
as long as:
:$|h| \leq \frac{1}{2\norm{R_z}}$
The expression on the right-hand side goes to zero as $h$ gets small. Therefore, taking limits, we get
$$
\lim_{h\to0} \norm{ f(z+h) - f(z) }_* = 0.
$$
This establishes continuity of $f$ at arbitrary $z\in\rho(T)$.
{{qed}}
Category:Bounded Linear Operators
Category:Resolvent Mapping is Continuous
ee2c0e1804wct5w0tfwtlssd4wshrbn
Theorem: Resolvent Mapping is Analytic/Bounded Linear Operator

Let $B$ be a Banach space.
Let $\map \LL {B, B}$ be the set of bounded linear operators from $B$ to itself.
Let $T \in \map \LL {B, B}$.
Let $\map \rho T$ be the resolvent set of $T$ in the complex plane.
Then the resolvent mapping $f : \map \rho T \to \map \LL {B, B}$ given by $\map f z = \paren {T - z I}^{-1}$ is analytic, and:
:$\map {f'} z = \paren {T - z I}^{-2}$
where $f'$ denotes the derivative of $f$ {{WRT|Differentiation}} $z$.

Proof:
For $a \in \map \rho T$, define:
:$R_a = \paren {T - a I}^{-1}$
Then we have:
{{begin-eqn}}
{{eqn | l = \frac {\norm {\map f {z + h} - \map f z - \paren {T - z I}^{-2} h }_*} {\size h}
      | r = \frac {\norm {R_{z + h} - R_z - R_z^2 h}_*} {\size h}
}}
{{eqn | r = \frac {\norm {h R_{z + h} R_z - R_z^2 h }_*} {\size h}
      | c = Resolvent Identity
}}
{{eqn | r = \frac {\size h \norm {R_{z + h} R_z - R_z^2 }_*} {\size h}
      | c = Operator Norm is Norm
}}
{{eqn | r = \norm {R_{z + h} R_z - R_z^2 }_*
}}
{{end-eqn}}
From Resolvent Mapping is Continuous we have:
:$R_{z + h} \to R_z$ as $h \to 0$
Taking limits of both sides and using Norm is Continuous, we get:
:$\ds \lim_{h \mathop \to 0} \dfrac {\norm {\map f {z + h} - \map f z - \paren {T - z I}^{-2} h }_*} {\size h} = \norm {R_z^2 - R_z^2}_* = 0$
which is the result.
{{qed}}
Theorem: Resolvent Mapping is Analytic/Bounded Linear Operator

Let $B$ be a Banach space.
Let $\map \LL {B, B}$ be the set of bounded linear operators from $B$ to itself.
Let $T \in \map \LL {B, B}$.
Let $\map \rho T$ be the resolvent set of $T$ in the complex plane.
Then the resolvent mapping $f : \map \rho T \to \map \LL {B, B}$ given by $\map f z = \paren {T - z I}^{-1}$ is analytic, and:
:$\map {f'} z = \paren {T - z I}^{-2}$
where $f'$ denotes the derivative of $f$ {{WRT|Differentiation}} $z$.

Proof:
If $B=\set {\mathbf 0_B}$, the statement is trivial, since $\map f z = \mathbf 0_{\map \LL {B, B}}$ for all $z \in \C$.
We assume that $B \ne \set {\mathbf 0_B}$.
Especially, for all $z \in \map \rho T$:
:$\map f z \ne \mathbf 0_{\map \LL {B, B}}$
since:
:$\paren {T - z I} \map f z = I \ne \mathbf 0_{\map \LL {B, B}}$
Let $a \in \map \rho T$.
Then for each $z \in \C$:
{{begin-eqn}}
{{eqn | l = T - z I
      | r = \paren {T - a I - \paren {z - a} I}
}}
{{eqn | r = \paren {T - a I} \paren {I - \paren {z - a} \paren {T - a I}^{-1} }
}}
{{eqn | r = \paren {T - a I} \paren {I - \paren {z - a} \map f a}
      | n = 1
}}
{{end-eqn}}
Recall that $\map B {c, r}$ denotes the open disc of center $c \in \C$ and radius $r>0$.
For all $z \in \map B {a, \frac{1}{\norm {\map f a} } }$, we have:
{{begin-eqn}}
{{eqn | l = \paren {I - \paren {z - a} \map f a}^{-1}
      | r = \sum_{n \mathop \ge 0} \paren {\paren {z - a} \map f a}^n
      | c = Neumann Series, as $\norm {\paren {z - a} \map f a} < 1$
}}
{{eqn | r = \sum_{n \mathop \ge 0} \paren {\map f a}^n \paren {z - a}^n
      | n = 2
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \paren {T - zI}^{-1}
      | r = \paren {\paren {T - a I} \paren {I - \paren {z - a} \map f a} }^{-1}
      | c = by $(1)$
}}
{{eqn | r = \paren {I - \paren {z - a} \map f a}^{-1} \paren {T - a I}^{-1}
      | c = Inverse of Product
}}
{{eqn | r = \paren {I - \paren {z - a} \map f a}^{-1} \map f a
}}
{{eqn | r = \paren {\sum_{n \mathop \ge 0} \paren {\map f a}^n \paren {z - a}^n} \map f a
      | c = by $(2)$
}}
{{eqn | r = \sum_{n \mathop \ge 0} \paren {\map f a}^{n+1} \paren {z - a}^n
}}
{{end-eqn}}
That is, in a neighborhood of each $a \in \map \rho T$, $f$ can be written as:
:$\ds \map f z = \sum_{n \mathop \ge 0} \paren {\map f a}^{n+1} \paren {z - a}^n$
This means that $f$ is analytic on $\map \rho T$.
Finally, by Derivative of Power Series, in the neighborhood of each $a \in \map \rho T$:
:$\ds \map {f'} z = \sum_{n \mathop \ge 1} \paren {\map f a}^{n+1} n \paren {z - a}^{n-1}$
Choosing $z = a$, we obtain:
:$\map {f'} a = \paren {\map f a}^2 = \paren {T - aI}^{-2}$
{{qed}}
Theorem: Numbers for which Euler Phi Function equals Product of Digits

The sequence of positive integers $n$ for which $\map \phi n$ is equal to the product of the digits of $n$ begins:
:$1, 24, 26, 87, 168, 388, 594, 666, 1998, 2688, 5698, 5978, 6786, 7888, 68 \, 796$
{{OEIS|A058627}}
It is known that this sequence is finite, but it is unknown whether a $16^{\text {th}}$ term exists.

Proof:
{{begin-eqn}}
{{eqn | l = \map \phi 1
      | r = 1
      | c = {{EulerPhiLink|1}}
}}
{{eqn | l = \map \phi {24}
      | r = 8
      | c = {{EulerPhiLink|24}}
}}
{{eqn | r = 2 \times 4
}}
{{eqn | l = \map \phi {26}
      | r = 12
      | c = {{EulerPhiLink|26}}
}}
{{eqn | r = 2 \times 6
}}
{{eqn | l = \map \phi {87}
      | r = 56
      | c = {{EulerPhiLink|87}}
}}
{{eqn | r = 8 \times 7
}}
{{eqn | l = \map \phi {168}
      | r = 48
      | c = {{EulerPhiLink|168}}
}}
{{eqn | r = 1 \times 6 \times 8
}}
{{end-eqn}}
and so on.
{{ProofWanted}}
Category:Euler Phi Function
2cpi7xdok7v21u18ob5smwr4zewu65f
Theorem: Linear Isometry is Injective

Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be normed vector spaces.
Let $T : X \to Y$ be a linear isometry.
Then $T$ is injective.

Proof:
Let $x, y \in X$. 
We have: 
:$\norm {\map T {x - y} }_Y = \norm {T x - T y}_Y$
from the definition of a linear transformation.
Since $T$ is a linear isometry, we have: 
:$\norm {\map T {x - y} }_Y = \norm {x - y}_X$
So:
:$\norm {T x - T y}_Y = 0$
{{iff}}:
:$\norm {x - y}_X = 0$
Since the norm is positive definite, this gives:
:$T x = T y$
{{iff}}:
:$x = y$
So $T$ is injective.
{{qed}}
Category:Linear Isometries
Category:Linear Isometry is Injective
l6rw5cw9gzz4387ybtjc6acvxga4uej
Theorem: Hahn-Banach Separation Theorem/Normed Vector Space/Real Case/Open Convex Set and Convex Set

Let $A \subseteq X$ be an open convex set.
Let $B \subseteq X$ be a convex set disjoint from $A$. 
Then there exists $f \in X^\ast$ and $c \in \R$ such that: 
:$A \subseteq \set {x \in X : \map f x < c}$
and:
:$B \subseteq \set {x \in X : \map f x \ge c}$
That is: 
:there exists $f \in X^\ast$ and $c \in \R$ such that $\map f a < c \le \map f b$ for each $a \in A$ and $b \in B$.

Proof:
Let $a_0 \in A$ and $b_0 \in B$.
Let:
:$v_0 = b_0 - a_0$
Let:
:$C = v_0 + A - B = \set {v_0 + a - b : a \in A, \, b \in B}$
Theorem: Translation of Open Set in Normed Vector Space is Open

Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $U \subseteq X$ be an open set.
Let $x \in X$.
Then:
:$U + x$ is open.

Proof:
Let: 
:$v \in U + x$
then:
:$v = u + x$
for some $u \in U$.
So:
:$v - x \in U$ 
Since $U$ is open, there exists $\epsilon > 0$ such that whenever $v' \in X$ and: 
:$\norm {\paren {v - x} - \paren {v' - x} } < \epsilon$
we have $v' - x \in U$.
That is: 
:$v' \in U + x$
Note that: 
:$\norm {\paren {v - x} - \paren {v' - x} } = \norm {v - v'}$
So, whenever $v \in U + x$ and $v' \in X$ is such that: 
:$\norm {v - v'} < \epsilon$
we have: 
:$v' \in U + x$
Since $v$ was arbitrary: 
:$U + x$ is open.
{{qed}}
Category:Open Sets (Normed Vector Spaces)
Category:Translation of Subsets of Vector Spaces
5s04728p618v13aoc3ze436dqvhkxn1
Theorem: Element of Leibniz Harmonic Triangle as Sum of Elements on Diagonal from Below/Lemma 2

:$\ds \forall r \in \N_{>0}: \tuple {n, m} = \tuple {n + r, m + r} + \sum_{k \mathop = 1}^r \tuple {n + k, m + k - 1}$
That is, each number in the Leibniz harmonic triangle is equal to the sum of the number below it, $\paren {r - 1}$ numbers diagonally below that number, and the number to the right of the last number.

Proof:
Proof by induction:
For all $r \in \N_{>0}$, let $\map P r$ be the proposition:
:$\ds \tuple {n, m} = \tuple {n + r, m + r} + \sum_{k \mathop = 1}^r \tuple {n + k, m + k - 1}$
Theorem: Convex Hull is Smallest Convex Set containing Set/Corollary

Let $K \subseteq X$ be non-empty.
Then: 
:$K$ is convex {{iff}} $\map {\operatorname {conv} } K = K$

Proof:
=== Sufficient Condition ===
Suppose that: 
:$\map {\operatorname {conv} } K = K$
From Convex Hull is Smallest Convex Set containing Set, we have: 
:$\map {\operatorname {conv} } K$ is convex.
So:
:$K$ is convex.
{{qed|lemma}}
Theorem: Convex Hull preserves Subsets

Let $X$ be a vector space over $\R$.
Let $U, V \subseteq X$ be non-empty with $U \subseteq V$. 
Then: 
:$\map {\operatorname {conv} } U \subseteq \map {\operatorname {conv} } V$
where $\operatorname {conv}$ denotes the convex hull.

Proof:
Let: 
:$u \in \map {\operatorname {conv} } U$
From the definition of the convex hull, there exists $u_1, u_2, \ldots, u_n \in U$ and $\lambda_1, \lambda_2, \ldots, \lambda_n \in \R_{> 0}$ with: 
:$\ds \sum_{i \mathop = 1}^n \lambda_i = 1$
such that: 
:$\ds u = \sum_{i \mathop = 1}^n \lambda_i u_i$
Since $U \subseteq V$, we have: 
:$u_i \in V$ for each $i$.
So, we have: 
:$u \in \map {\operatorname {conv} } V$ 
So:
:if $u \in \map {\operatorname {conv} } U$ then $u \in \map {\operatorname {conv} } V$.
So, from the definition of subset, we have: 
:$\map {\operatorname {conv} } U \subseteq \map {\operatorname {conv} } V$
{{qed}}
Category:Convex Hulls
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