avewright/proofwiki-math huggingface dataset:
Theorem: Existence of Support Functional

Let $\mathbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\mathbb F$. 
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $X$. 
Let $x \in X$. 
Then there exists $f \in X^\ast$ such that:
:$(1): \quad$ $\norm f_{X^\ast} = 1$
:$(2): \quad$ $\map f x = \norm x_X$.
That is: 
:there exists a support functional at $x$.

Proof:
Let: 
:$U = \span {\set x}$ 
Then $U$ consists precisely of the $u \in X$ of the form: 
:$u = \alpha x$
for $\alpha \in \mathbb F$.
From Linear Span is Linear Subspace, we have: 
:$U$ is a linear subspace of $X$.
Let $\struct {U^\ast, \norm \cdot_{U^\ast} }$ be the normed dual space of $U$.
Define $f_0 : U \to \R$ by: 
:$\map {f_0} {\alpha x} = \alpha \norm x_X$
for each $\alpha \in \mathbb F$.
In particular, we have: 
:$\map {f_0} x = \norm x$ 
We show that this is a linear functional.
Let $u, v \in U$ and $k, l \in \mathbb F$. 
Then there exists $\alpha, \beta \in \mathbb F$ such that: 
:$u = \alpha x$
and:
:$v = \beta x$
We then have:
{{begin-eqn}}
{{eqn	| l = \map {f_0} {k u + l v}
		| r = \map {f_0} {k \alpha x + l \beta x}
}}
{{eqn	| r = \map {f_0} {\paren {k \alpha + l \beta} x}
}}
{{eqn	| r = \paren {k \alpha + l \beta} \norm x_X
}}
{{eqn	| r = k \alpha \norm x_X + l \beta \norm x_X
}}
{{eqn	| r = k \map {f_0} u + l \map {f_0} v
}}
{{end-eqn}}
so $f_0$ is a linear functional.
Now we show that $f_0 \in X^\ast$ and: 
:$\norm {f_0}_{U^\ast} = 1$
Let $u \in U$ and write: 
:$u = \alpha x$
for $\alpha \in \mathbb F$. 
We then have:
{{begin-eqn}}
{{eqn	| l = \size {\map {f_0} u}
		| r = \size {\map {f_0} {\alpha x} }
}}
{{eqn	| r = \size {\alpha \norm x_X}
}}
{{eqn	| r = \size \alpha \norm x_X
}}
{{eqn	| r = \norm {\alpha x}_X
		| c = {{Defof|Norm}}
}}
{{eqn	| r = \norm u_X
}}
{{end-eqn}}
so we have that $f_0$ is bounded.
That is, $f \in U^\ast$.  
We then have: 
:$\ds \sup_{\norm u_X = 1} \size {\map {f_0} u} = 1$ 
That is, from the definition of dual norm, we have: 
:$\norm {f_0}_{U^\ast} = 1$
We apply: 
:{{Corollary|Hahn-Banach Theorem/Real Vector Space|2|disp = Hahn-Banach Theorem: Real Vector Space}} if $\mathbb F = \R$
:{{Corollary|Hahn-Banach Theorem/Complex Vector Space|disp = Hahn-Banach Theorem: Complex Vector Space}} if $\mathbb F = \C$
to find that there exists $f \in X^\ast$ such that:
:$f$ extends $f_0$ to $X$
and:
:$\ds \norm f_{X^\ast} = \norm {f_0}_{U^\ast} = 1$
Since $f$ extends $f_0$, we have: 
:$\map f x = \map {f_0} x = \norm x$
So $f$ is the required linear functional.
{{qed}}
Theorem: Normed Dual Space Separates Points

Let $\struct {X, \norm \cdot_X}$ be a normed vector space. 
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $X$. 
Then $X^\ast$ separates points.
That is, suppose that $x, y \in X$ are such that:
:$\map f x = \map f y$ for each $f \in X^\ast$.
Then $x = y$.

Proof:
From Existence of Support Functional, there exists a $\phi \in X^\ast$ such that: 
:$\map \phi {x - y} = \norm {x - y}$
Since $\phi$ is linear, we then have: 
:$\map \phi x - \map \phi y = \norm {x - y}$
By hypothesis, we have: 
:$\map \phi x = \map \phi y$
so:
:$\norm {x - y} = 0$
Since a norm is positive definite, we then have: 
:$x - y = 0$
so that: 
:$x = y$
{{qed}}
Theorem: Scheffé's Lemma

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f_n$ be a sequence of $\mu$-integrable functions that converge almost everywhere to another $\mu$-integrable function $f$.
Then $f_n$ converges to $f$ in $L^1$ {{iff}} $\ds \int_X \size {f_n} \rd \mu$ converges to $\ds \int_X \size f \rd \mu$.

Proof:
=== Sufficient Condition ===
Let $f_n \to f$ in $L^1$.
Then:
{{begin-eqn}}
{{eqn | l = \size {\int_X \size f \rd \mu - \int_X \size {f_n} \rd \mu}
      | o = \le
      | r = \int_X \size {f - f_n} d\mu
      | c = Reverse Triangle Inequality on $L^1$
}}
{{end-eqn}}
and so $f_n \to f$ in $L^1$ implies the {{RHS}} of this inequality goes to $0$ as $n$ grows to infinity.
Since the {{LHS}} of the inequality is non-negative, it also goes to $0$ as $n$ grows to infinity.
{{qed|lemma}}
Theorem: Positive Definite and Positive Homogeneous Map with Convex Closed Unit Ball is Norm

Let $\mathbb F$ be a subfield of $\C$. 
Let $X$ be a vector space over $\mathbb F$. 
Let $N : X \to \R_{\ge 0}$ be a positive definite and positive homogeneous function.
That is: 
:$(1): \quad$ $\map N x = 0$ {{iff}} $x = 0$
:$(2): \quad$ $\map N {\lambda x} = \cmod \lambda \map N x$ for each $x \in X$ and $\lambda \in \mathbb F$.
Suppose further that: 
:$B = \set {x \in X : \map N x \le 1}$ is convex.
Then: 
:$N$ is a norm on $X$.

Proof:
Note that $N$ satisfies {{NormAxiomVector|1}} and {{NormAxiomVector|2}}.
So we only need to verify {{NormAxiomVector|3}}.
That is, we want to show that: 
:$\map N {x + y} \le \map N x + \map N y$ 
for each $x, y \in X$. 
Fix $x, y \in X$. 
If $\map N x = 0$, then $x = 0$ and we have: 
{{begin-eqn}}
{{eqn	| l = \map N {x + y}
		| r = \map N y
}}
{{eqn	| r = \map N x + \map N y
}}
{{end-eqn}}
Similarly, if $\map N y = 0$, then $y = 0$ and: 
{{begin-eqn}}
{{eqn	| l = \map N {x + y}
		| r = \map N x
}}
{{eqn	| r = \map N x + \map N y
}}
{{end-eqn}}
In both cases we clearly have: 
:$\map N {x + y} \le \map N x + \map N y$
So suppose that $\map N x > 0$ and $\map N y > 0$.
We then have: 
{{begin-eqn}}
{{eqn	| l = \map N {\frac x {\map N x} }
		| r = \frac {\map N x} {\map N x}
		| c = since $N$ is positive homogeneous
}}
{{eqn	| r = 1
}}
{{end-eqn}}
so:
:$\dfrac x {\map N x} \in B$
Similarly: 
{{begin-eqn}}
{{eqn	| l = \map N {\frac y {\map N y} }
		| r = \frac {\map N y} {\map N y}
		| c = since $N$ is positive homogeneous
}}
{{eqn	| r = 1
}}
{{end-eqn}}
so:
:$\dfrac y {\map N y} \in B$
Now note that we can write: 
:$1 = \dfrac {\map N x} {\map N x + \map N y} + \dfrac {\map N y} {\map N x + \map N y}$
Since $B$ is convex, we have: 
:$\dfrac {\map N x} {\map N x + \map N y} \paren {\dfrac x {\map N x} } + \dfrac {\map N y} {\map N x + \map N y} \paren {\dfrac y {\map N y} } \in B$
That is: 
{{begin-eqn}}
{{eqn	| l = \frac {\map N x} {\map N x + \map N y} \paren {\frac x {\map N x} } + \frac {\map N y} {\map N x + \map N y} \paren {\frac y {\map N y} }
		| r = \frac x {\map N x + \map N y} + \frac y {\map N x + \map N y}
}}
{{eqn	| r = \frac {x + y} {\map N x + \map N y}
}}
{{eqn	| o = \in
		| r = B
}}
{{end-eqn}}
That is: 
:$\map N {\dfrac {x + y} {\map N x + \map N y} } \le 1$
Using positive homogeneity, we then obtain: 
:$\dfrac {\map N {x + y} } {\map N x + \map N y} \le 1$
so:
:$\map N {x + y} \le \map N x + \map N y$
as was required.
{{qed}}
Theorem: Characterization of Unit Open Balls of Norms of Euclidean Space

Let $\struct {\R^n, \norm \cdot}$ be the Euclidean $n$-space. 
Let $K \subseteq \R^n$ be a non-empty open set of $\struct {\R^n, \norm \cdot}$.
Then there exists a norm $\norm \cdot_* : \R^n \to \R$ such that: 
:$\set {x \in \R^n : \norm x_* < 1} = K$
{{iff}}:
:$(1): \quad$ $K$ is bounded in $\struct {\R^n, \norm \cdot}$
:$(2): \quad$ $K$ is convex
:$(3): \quad$ $K$ is symmetric.
That is: 
:an open set $K \subseteq \R^n$ is the unit ball of some norm on $\R^n$ {{iff}} $K$ is bounded, (with respect to the Euclidean norm), convex and symmetric.

Proof:
=== Necessary Condition ===
Theorem: Polarization Identity/Real Vector Space

Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space over $\R$. 
Let $\norm \cdot$ be the inner product norm for $V$. 
Then we have: 
:$4 \innerprod x y = \norm {x + y}^2 - \norm {x - y}^2$
for all $x, y \in V$.

Proof:
We have: 
{{begin-eqn}}
{{eqn	| l = \norm {x + y}^2 - \norm {x - y}^2
		| r = \innerprod {x + y} {x + y} - \innerprod {x - y} {x - y}
		| c = {{Defof|Inner Product Norm}}
}}
{{eqn	| r = \paren {\innerprod x {x + y} + \innerprod y {x + y} } - \paren {\innerprod x {x - y} - \innerprod y {x - y} }
		| c = since an inner product is linear in the first argument
}}
{{eqn	| r = \paren {\innerprod {x + y} x + \innerprod {x + y} y} - \paren {\innerprod {x - y} x - \innerprod {x - y} y}
		| c = since a real inner product is symmetric
}}
{{eqn	| r = \paren {\innerprod x x + \innerprod y x + \innerprod x y + \innerprod y y} - \paren {\innerprod x x - \innerprod y x - \innerprod x y + \innerprod y y}
		| c = using linearity in the first argument
}}
{{eqn	| r = 2 \innerprod x y + 2 \innerprod y x
}}
{{eqn	| r = 4 \innerprod x y
		| c = since a real inner product is symmetric
}}
{{end-eqn}}
{{qed}}
Theorem: Polarization Identity/Complex Vector Space

Let $V$ be a vector space over $\C$. 
Let $q : V \times V \to \C$ be a sesquilinear form.
Define $Q : V \to \C$ by:
:$\map Q x = \map q {x, x}$
Then, we have: 
:$4 \map q {x, y} = \map Q {x + y} - \map Q {x - y} + i \map Q {x + i y} - i \map Q {x - i y}$
for each $x, y \in V$.

Proof:
We write: 
{{begin-eqn}}
{{eqn	| l = \map Q {x + y} - \map Q {x - y} + i \map Q {x + i y} - i \map Q {x - i y}
		| r = \sum_{n \mathop = 0}^3 i^n \map Q {x + i^n y}
}}
{{eqn	| r = \sum_{n \mathop = 0}^3 i^n \map q {x + i^n y, x + i^n y}
		| c = {{Defof|Inner Product Norm}}
}}
{{end-eqn}}
We then compute: 
{{begin-eqn}}
{{eqn	| l = i^n \map q {x + i^n y, x + i^n y}
		| r = i^n \paren {\map q {x, x + i^n y} + i^n \map q {y, x + i^n y} }
		| c = Inner Product is Sesquilinear
}}
{{eqn	| r = i^n \map q {x, x + i^n y} + i^{2 n} \map q {y, x + i^n y}
}}
{{eqn	| r = i^n \paren {\map q {x, x} + \overline {i^n} \map q {x, y} } + i^{2 n} \paren {\map q {y, x} + \overline {i^n} \map q {y, y} }
		| c = Inner Product is Sesquilinear
}}
{{eqn	| r = i^n \map q {x, x} + i^n \paren {-i}^n \map q {x, y} + i^{2 n} \map q {y, x} + i^{2 n} \paren {-i}^n \map q {y, y}
		| c = Power of Complex Conjugate is Complex Conjugate of Power
}}
{{eqn	| r = i^n \map q {x, x} + \map q {x, y} + \paren {-1}^n \map q {y, x} + i^n \map q {y, y}
}}
{{end-eqn}}
We then have: 
{{begin-eqn}}
{{eqn	| l = \sum_{n \mathop = 0}^3 i^n
		| r = 1 + i + i^2 + i^3
}}
{{eqn	| r = 1 + i - 1 - i
}}
{{eqn	| r = 0
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn	| l = \sum_{n \mathop = 0}^3 \paren {-1}^n
		| r = \paren {-1}^0 + \paren {-1}^1 + \paren {-1}^2 + \paren {-1}^3
}}
{{eqn	| r = 1 - 1 + 1 - 1
}}
{{eqn	| r = 0
}}
{{end-eqn}}
Then we have: 
{{begin-eqn}}
{{eqn	| l = \sum_{n \mathop = 0}^3 i^n \map Q {x + i^n y}
		| r = \sum_{n \mathop = 0}^3 \paren {i^n \map q {x, x} + \map q {x, y} + \paren {-1}^n \map q {y, x} + i^n \map q {y, y} }
}}
{{eqn	| r = \sum_{n \mathop = 0}^3 \map q {x, y} + \paren {\map q {x, x} + \map q {y, y} } \sum_{n \mathop = 0}^3 i^n + \map q {y, x} \sum_{n \mathop = 0}^3 \paren {-1}^n
}}
{{eqn	| r = 4 \map q {x, y}
}}
{{end-eqn}}
{{qed}}
Theorem: Existence of Distance Functional

Let $\mathbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\mathbb F$.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $X$. 
Let $Y$ be a proper closed linear subspace of $X$. 
Let $x \in X \setminus Y$.
Let:
:$d = \map {\operatorname {dist} } {x, Y}$
where $\map {\operatorname {dist} } {x, Y}$ denotes the distance between $x$ and $Y$.
Then there exists $f \in X^\ast$ such that:
:$(1): \quad$ $\norm f_{X^\ast} = 1$
:$(2): \quad$ $\map f y = 0$ for each $y \in Y$
:$(3): \quad$ $\map f x = d$.
That is: 
:there exists a distance functional for $x$.

Proof:
Consider the normed quotient vector space $X/Y$ with quotient mapping $\pi$.
From Kernel of Quotient Mapping, we have $\map \pi x \ne 0$. 
So, from Existence of Support Functional, there exists $f \in \paren {X/Y}^\ast$ such that: 
:$\norm f_{\paren {X/Y}^\ast} = 1$
and:
:$\map f {\map \pi x} = \norm {\map \pi x}_{X/Y}$
From the definition of the quotient norm, we have: 
:$\norm {\map \pi x}_{X/Y} = \map {\operatorname {dist} } {x, Y}$
From Normed Dual Space of Normed Quotient Vector Space is Isometrically Isomorphic to Annihilator, $g = f \circ \pi \in X^\ast$ and: 
:$\norm g_{X^\ast} = \norm f_{\paren {X/Y}^\ast} = 1$ 
with: 
:$\map g x = \map {\operatorname {dist} } {x, Y}$
So $g$ is a linear functional satisfying our requirements.
{{qed}}
Theorem: Characterization of Separable Normed Vector Space

Let $\mathbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.
{{TFAE}}
:$(1): \quad$ $X$ is separable
:$(2): \quad$ $S_X = \set {x \in X : \norm x = 1}$ is separable
:$(3): \quad$ there exists a countable set $\set {x_n : n \in \N} \subseteq X$ such that the closed linear span of $\set {x_n : n \in \N}$ is $X$.

Proof:
=== $(1)$ implies $(2)$ ===
This is immediate from Subspace of Separable Metric Space is Separable.
{{qed|lemma}}
Theorem: Subspace of Separable Metric Space is Separable

Let $\struct {X, d}$ be a separable metric space.
Let $\struct {Y, d_Y}$ be a metric subspace of $\struct {X, d}$.
Then $Y$ is separable.

Proof:
If $Y = X$, we are done immediately.
Now consider the case $Y \ne X$. 
Pick $x \in X \setminus Y$. 
Let $\set {x_n : n \in \N}$ be a countable everywhere dense subset of $X$.
Iterate over the pairs $\struct {n, k}$ where $n, k \in \N$. 
Check whether:
:$\map {B_{1/k} } {x_n} \cap Y \ne \O$
where $\map {B_{1/k} } {x_n}$ is the open ball of radius $r$, centered at $x_n$. 
If the intersection is non-empty, let $y_{n, k}$ be any element of $\map {B_{1/k} } {x_n} \cap Y$. 
Otherwise, let $y_{n, k} = x$.
Now consider the set $\mathcal S = \set {y_{n, k} : y_{n, k} \ne x}$.
Note that it is not yet known that $\mathcal S \ne \O$.
We show that $\mathcal S$. is everywhere dense.
Let $y \in Y$ and let $\epsilon > 0$ be arbitrary.
Let $k$ be such that $1/k < \epsilon/2$. 
Since $\set {x_n : n \in \N}$ is everywhere dense, there exists $x_n \in \set {x_n : n \in \N}$ such that: 
:$\ds \map d {y, x_n} < 1/k$
Now, we have that:
:$\map {B_{1/k} } {x_n} \cap Y \supseteq \set {y_{n, k} } \ne \O$
We are now assured that $\mathcal S \ne \O$. 
So, we have: 
{{begin-eqn}}
{{eqn	| l = \map {d_Y} {y, y_{n, k} }
		| r = \map {d_Y} {y, x_n} + \map {d_Y} {x_n, y_{n, k} }
		| c = {{Defof|Metric}}
}}
{{eqn	| o = <
		| r = \frac \epsilon 2 + \frac 1 k
}}
{{eqn	| o = <
		| r = \epsilon
}}
{{end-eqn}}
So, from Condition for Point being in Closure, we have: 
:$y \in \mathcal S^-$
But $y \in Y$ was arbitrary, so we have: 
:$Y \subseteq \mathcal S^-$
So:
:$Y = \mathcal S^-$ 
That is: 
:$\mathcal S$ is everywhere dense in $Y$.
We now show that $\mathcal S$ is countable.
From Lexicographic Order on Pair of Well-Ordered Sets is Well-Ordering and the Well-Ordering Principle, there exists a well-order $\sqsubseteq$ on $\N \times \N$.
For each $y \in \mathcal S$, define a map $f : \mathcal S \to \N \times \N$ by picking $\map f y$ to be the $\sqsubseteq$-least element of $\set {y_{n, k} \in \mathcal S : y_{n, k} = y}$. 
Then $f$ is an injection.
From Cartesian Product of Countable Sets is Countable, $\N \times \N$ is countable.
So, from Domain of Injection to Countable Set is Countable, we have that: 
:$\mathcal S$ is countable.
So $\mathcal S$ is a countable everywhere dense subset of $Y$.
{{qed}}
Category:Metric Spaces
Category:Separable Spaces
fdlkbz226j05b17snm8ytxn5uzk7l94
Theorem: Normed Vector Space with Separable Dual is Separable

Let $\struct {X, \norm \cdot_X}$ be a normed vector space.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $X$. 
Then, if $X^\ast$ is separable: 
:$X$ is separable.

Proof:
From Characterization of Separable Normed Vector Space, we have that: 
:$S_{X^\ast} = \set {f \in X^\ast : \norm f_{X^\ast} = 1}$ is separable.
Let $\mathcal S_{X^\ast} = \set {f_n : n \in \N}$ be a countable everywhere dense subset of $S_{X^\ast}$.
For each $n \in \N$, pick $x_n \in X$ such that $\norm {x_n}_X = 1$ and:
:$\ds \size {\map {f_n} {x_n} } \ge \frac 1 2$
Let: 
:$M = \span \set {x_1, x_2, \ldots}$
From Characterization of Separable Normed Vector Space, it suffices to show that: 
:$M^- = X$
Suppose that $M^- \ne X$. 
From Closed Linear Span is Closed Vector Subspace, we have that: 
:$M^-$ is a closed vector subspace.
So, from Existence of Distance Functional, there exists $f \in X^\ast$ such that: 
:$\map f x = 0$ for each $x \in M^-$ and $\norm f_{X^\ast} = 1$.
That is, $f \in S_{X^\ast}$.
Then we have: 
:$\map f {x_n} = 0$ for each $n \in \N$
and so:
{{begin-eqn}}
{{eqn	| l = \size {\map {f_n} {x_n} }
		| r = \size {\map {f_n} {x_n} - \map f {x_n} }
}}
{{eqn	| o = \le
		| r = \norm {f_n - f}_{X^\ast} \norm {x_n}_X
		| c = {{Defof|Norm on Bounded Linear Functional}}
}}
{{eqn	| r = \norm {f_n - f}_{X^\ast}
}}
{{end-eqn}}
However since: 
:$\ds \size {\map {f_n} {x_n} } \ge \frac 1 2$
for each $n \in \N$, we have: 
:$\ds \norm {f_n - f}_{X^\ast} \ge \frac 1 2$
for each $n \in \N$. 
So there does not exist $g \in \mathcal S_{X^\ast}$ such that: 
:$\ds \norm {g - f}_{X^\ast} < \frac 1 3$
while $f \in S_{X^\ast}$. 
This contradicts that $\mathcal S_{X^\ast}$ is everywhere dense in $S_{X^\ast}$. 
So we must have $M^- = X$. 
So, from Characterization of Separable Normed Vector Space, we have: 
:$X$ is separable.
{{qed}}
Category:Separable Spaces
Category:Normed Vector Spaces
2z87m6546mn3gczsj4y27a3lv1xfys1
Theorem: Pell's Equation/Examples/2/-1

:$x^2 - 2 y^2 = -1$
has the positive integral solutions:
:<nowiki>$\begin {array} {r|r} x & y \\ \hline
1 & 1 \\
7 & 5 \\
41 & 29 \\
239 & 169 \\
1393 & 985 \\
\end {array}$</nowiki>
and so on.

Proof:
From Continued Fraction Expansion of $\sqrt 2$:
:$\sqrt 2 = \sqbrk {1, \sequence 2}$
The cycle is of length is $1$.
By Solution of Pell's Equation, the only solutions of $x^2 - 2 y^2 = -1$ are:
:${p_r}^2 - 2 {q_r}^2 = \paren {-1}^r$
for $r = 1, 2, 3, \ldots$
From Convergents to Continued Fraction Expansion of $\sqrt 2$:
{{:Continued Fraction Expansion of Irrational Square Root/Examples/2/Convergents}}
from which the solutions are obtained by taking the convergents with odd indices.
{{qed}}
Category:Pell's Equation
Category:2
8glj5lxd14v6el15zliv94mmxvty8u4
Theorem: Pell's Equation/Examples/2

:$x^2 - 2 y^2 = 1$
has the positive integral solutions:
:<nowiki>$\begin {array} {r|r} x & y \\ \hline
3 & 2 \\
17 & 12 \\
99 & 70 \\
577 & 408 \\
3363 & 2378 \\
\end {array}$</nowiki>
and so on.

Proof:
From Continued Fraction Expansion of $\sqrt 2$:
:$\sqrt 2 = \sqbrk {1, \sequence 2}$
The cycle is of length is $1$.
By Solution of Pell's Equation, the only solutions of $x^2 - 2 y^2 = -1$ are:
:${p_r}^2 - 2 {q_r}^2 = \paren {-1}^r$
for $r = 1, 2, 3, \ldots$
From Convergents to Continued Fraction Expansion of $\sqrt 2$:
{{:Continued Fraction Expansion of Irrational Square Root/Examples/2/Convergents}}
from which the solutions are obtained by taking the convergents with even indices.
{{qed}}
Category:Pell's Equation
Category:2
469xkp1p3g0bwblqo6pq1ak28ges9rp
Theorem: Convergence in Measure Implies Convergence a.e. of Subsequence

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $D \in \Sigma$. 
Let $f : D \to \R$ be a $\Sigma$-measurable function.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of $\Sigma$-measurable functions $f_n : D \to \R$ such that: 
:$\sequence {f_n}_{n \mathop \in \N}$ converges in measure to $f$. 
Then there is a subsequence $\sequence {f_{n_k} }_{k \mathop \in \N}$ of $\sequence {f_n}_{n \mathop \in \N}$ that converges a.e. to $f$.

Proof:
For each $n, k \ge 1$, define:
:$\ds B_{n, k} = \set {x \in X : \size {\map {f_n} x - \map f x} > \frac 1 k}$
Since $\sequence {f_n}_{n \mathop \in \N}$ converges in measure to $f$, we have: 
:$\ds \lim_{n \mathop \to \infty} \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x} > \frac 1 k} } = 0$
Then from the definition of convergence, for each $k$ we can select $n_k$ such that: 
:$\ds \map \mu {\set {x \in X : \size {\map {f_{n_k} } x - \map f x} > \frac 1 k} } < \frac 1 {2^k}$ 
That is: 
:$\ds \map \mu {B_{n_k, k} } < \frac 1 {2^k}$ 
We then have: 
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^\infty \map \mu {B_{n_k, k} } 
      | r = \sum_{k \mathop = 1}^\infty \frac 1 {2^k}
}}
{{eqn | r = 1
      | c = Sum of Infinite Geometric Progression
}}
{{eqn | o = <
      | r = \infty
}}
{{end-eqn}}
By the Borel-Cantelli Lemma, we therefore have: 
:$\ds \map \mu {\limsup_{k \mathop \to \infty}  B_{n_k, k} } = 0$
By Equivalence of Definitions of Limit Superior of Sequence of Sets:
:$\map \mu {\set {x \in X : \size {\map {f_{n_k} } x - \map f x} > \dfrac 1 k \text { for infinitely many } k} } = 0$
Let: 
:$N = \set {x \in X : \size {\map {f_{n_k} } x - \map f x} > \dfrac 1 k \text { for infinitely many } k}$ 
Then, for $x \not \in N$, we have: 
:$\size {\map {f_{n_k} } x - \map f x} > \dfrac 1 k$ holds for only finitely many $k$.
That is, there exists $K \in \N$ such that: 
:$\size {\map {f_{n_k} } x - \map f x} \le \dfrac 1 k$ for $k \ge K$.
Then from the Squeeze Theorem, we have: 
:$\size {\map {f_{n_k} } x - \map f x} \to 0$
so:
:$\map {f_{n_k} } x \to \map f x$ 
So if $x \notin N$: 
:$\map {f_{n_k} } x \to \map f x$
Since $\map \mu N = 0$: 
:$\sequence {f_{n_k}}_{n_k \mathop \in \N}$ converges a.e. to $f$.
{{qed}}
Theorem: Convergence of Product of Convergent Scalar Sequence and Convergent Vector Sequence in Normed Vector Space

Let $\Bbb F \in \set {\R, \C}$.
Let $\struct {V, \norm \cdot}$ be a normed vector space on $\Bbb F$.
Let $\alpha \in \R$.
Let $x \in V$.
Let $\sequence {\alpha_n}_{n \mathop \in \N}$ be a real sequence in $\Bbb F$ such that: 
:$\alpha_n \to \alpha$
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $V$ such that:
:$x_n \to x$
in $\struct {V, \norm \cdot}$. 
Then we have: 
:$\alpha_n x_n \to \alpha x$
in $\struct {V, \norm \cdot}$.

Proof:
Let $\epsilon > 0$.
We have: 
{{begin-eqn}}
{{eqn	| l = \norm {\alpha_n x_n - \alpha x}
		| r = \norm {\alpha_n x_n - \alpha x + \alpha_n x - \alpha_n x}
}}
{{eqn	| o = \le
		| r = \norm {\alpha_n x_n - \alpha_n x} + \norm {\alpha_n x - \alpha x}
		| c = using the triangle inequality for the norm $\norm \cdot$
}}
{{eqn	| r = \size {\alpha_n} \norm {x_n - x} + \norm x \size {\alpha_n - \alpha}
		| c = using positive homogeneity for the norm $\norm \cdot$
}}
{{end-eqn}}
From Convergent Real Sequence is Bounded, there exists $M > 0$ such that we have: 
:$\size {\alpha_n} \le M$
for each $n \in \N$. 
Then we have: 
:$\norm {\alpha_n x_n - \alpha x} \le M \norm {x_n - x} + \norm x \size {\alpha_n - \alpha}$
Suppose that: 
:$\norm x = 0$
then, from positive definiteness of $\norm \cdot$: 
:$x = 0$ 
In this case:
:$x_n \to 0$
and:
:$\norm {\alpha_n x_n} = \norm {\alpha_n x_n - \alpha x} \le M \norm {x_n - x} = M \norm {x_n}$
Then we can choose $N \in \N$ such that: 
:$\ds \norm {x_n} < \frac \epsilon M$
for $n > N$.
From which we have: 
:$\norm {\alpha_n x_n} < M$ 
for $n > N$. 
So, from the definition of a convergent sequence, we have: 
:$\alpha_n x_n \to 0 = \alpha x$
in the case $\norm x = 0$.
Now suppose that: 
:$\norm x \ne 0$
Now suppose that: 
:$\norm x \ne 0$
Since: 
:$\alpha_n \to \alpha$
we can pick $N_1 \in \N$ such that: 
:$\ds \size {\alpha_n - \alpha} < \frac \epsilon {2 \norm x}$
for $n \ge N_1$, from the definition of convergence.
Since we also have: 
:$x_n \to x$
we can pick $N_2 \in \N$ such that: 
:$\ds \norm {x_n - x} < \frac \epsilon {2 M}$
from the definition of a convergent sequence.
Let:
:$N = \max \set {N_1, N_2}$
We then have, for $n > N$: 
{{begin-eqn}}
{{eqn	| l = \norm {\alpha_n x_n - \alpha x}
		| o = <
		| r = M \paren {\frac \epsilon {2 M} } + \norm x \paren {\frac \epsilon {2 \norm x} }
}}
{{eqn	| r = \frac \epsilon 2 + \frac \epsilon 2
}}
{{eqn	| r = \epsilon
}}
{{end-eqn}}
So, from the definition of a convergent sequence, we have: 
:$\alpha_n x_n \to \alpha x$
in the case $x \ne 0$ also, hence the result.
{{qed}}
Category:Convergent Sequences (Normed Vector Spaces)
Category:Normed Vector Spaces
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