avewright/proofwiki-math huggingface dataset:
Theorem: Weak Limit in Normed Vector Space is Unique

Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $x, y \in X$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ such that: 
:$x_n \weakconv x$
and:
:$x_n \weakconv y$
where $\weakconv$ denotes weak convergence.
Then: 
:$x = y$

Proof:
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.
Since: 
:$x_n \weakconv x$
we have: 
:$\map f {x_n} \to \map f x$ for each $f \in X^\ast$.
Since: 
:$x_n \weakconv y$
we have: 
:$\map f {x_n} \to \map f y$ for each $f \in X^\ast$.
From Convergent Complex Sequence has Unique Limit, we have: 
:$\map f x = \map f y$ for each $f \in X^\ast$. 
From Normed Dual Space Separates Points, we therefore have: 
:$x = y$
{{qed}}
Theorem: Weak Convergence in Hilbert Space

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\HH$. 
Let $x \in X$.
Then:
:$\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$
{{iff}}:
:$\innerprod {x_n} y \to \innerprod x y$ for each $y \in \HH$.

Proof:
Let $\struct {\HH^\ast, \norm \cdot_{\HH^\ast} }$ be the normed dual space of $\HH$.
Theorem: Orthonormal Sequence in Hilbert Space Converges Weakly to Zero

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\sequence {e_n}_{n \mathop \in \N}$ be a sequence in $\HH$ such that: 
:$\innerprod {e_n} {e_m} = 0$ if $n \ne m$
and:
:$\norm {e_n} = 1$ for each $n \in \N$. 
Then: 
:$e_n \weakconv 0$
where $\rightharpoonup$ denotes weak convergence.

Proof:
From Bessel's Inequality, we have: 
:$\ds \sum_{n \mathop = 1}^\infty \size {\innerprod y {e_n} }^2$ converges for each $y \in \HH$.
So from Terms in Convergent Series Converge to Zero, we have: 
:$\cmod {\innerprod y {e_n} }^2 \to 0$ for each $y \in \HH$. 
We then have, from Complex Sequence is Null iff Positive Integer Powers of Sequence are Null: 
:$\cmod {\innerprod y {e_n} } \to 0$ for each $y \in \HH$
and so, from Complex Sequence is Null iff Modulus of Sequence is Null: 
:$\innerprod y {e_n} \to 0$ for each $y \in \HH$.
From Weak Convergence in Hilbert Space: Corollary, we therefore have:
:$e_n \weakconv 0$
{{qed}}
Theorem: Complex Sequence is Null iff Positive Integer Powers of Sequence are Null

Let $\sequence {z_n}_{n \mathop \in \N}$ be a complex sequence.
Let $k \in \N$. 
Then: 
:$z_n \to 0$
{{iff}}:
:${z_n}^k \to 0$

Proof:
=== Necessary Condition ===
Suppose that: 
:$z_n \to 0$
Let $\epsilon > 0$. 
Then from the definition of convergent sequence we can find $N \in \N$ such that for $n \ge N$ we have: 
:$\cmod {z_n} < \epsilon^{1/k}$
then: 
:$\cmod {z_n}^k < \epsilon$ 
From Power of Complex Modulus equals Complex Modulus of Power, this gives: 
:$\cmod { {z_n}^k} < \epsilon$
for all $n \ge N$. 
So:
:${z_n}^k \to 0$ 
{{qed|lemma}}
Theorem: Convergence of Complex Conjugate of Convergent Complex Sequence

Let $z \in \C$. 
Let $\sequence {z_n}_{n \mathop \in \N}$ be a complex sequence converging to $z$. 
Then: 
:$\overline {z_n} \to \overline z$

Proof:
Let $\epsilon > 0$.
Since $z_n \to z$, from the definition of convergence, we can find $N \in \N$ such that: 
:$\cmod {z_n - z} < \epsilon$ 
From Complex Modulus equals Complex Modulus of Conjugate, we have: 
:$\cmod {\overline {z_n - z} } = \cmod {z_n - z}$
From Difference of Complex Conjugates, we have: 
:$\cmod {z_n - z} = \cmod {\overline {z_n} - \overline z}$
So we have: 
:$\cmod {\overline {z_n} - \overline z} < \epsilon$ 
for each $n \ge N$.
Since $\epsilon$ was arbitrary, we have: 
:$\overline {z_n} \to \overline z$
{{qed}}
Category:Complex Conjugates
Category:Convergent Complex Sequences
3ftcx5xqnh3a00r394hbpealwtkccvr
Theorem: Rectangular Delta Sequence

thumb300pxThe graph of the rectangular delta sequence. As $n$ grows, the rectangle becomes thinner and taller. The area of each rectangle is equal to $1$.
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
:<nowiki>$\map {\delta_n} x := \begin {cases}
0 & : x < - \dfrac 1 {2 n} \\ \\
n & : - \dfrac 1 {2 n} \le x \le \dfrac 1 {2 n} \\ \\
0 & : x > \dfrac 1 {2 n} \end {cases}$</nowiki>
Then $\sequence {\map {\delta_n} x}_{n \mathop \in {\N_{>0} } }$ is a delta sequence.
That is, in the distributional sense it holds that:
:$\ds \lim_{n \mathop \to \infty} \map {\delta_n} x = \map \delta x$
or
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \delta \phi$
where:
:$\phi \in \map \DD \R$ is a test function
:$\delta$ is the Dirac delta distribution
:$\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.

Proof:
Let $\phi \in \map \DD \R$ be a test function.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x
      | r = \lim_{n \mathop \to \infty} n \int_{-\frac 1 {2 n} }^{\frac 1 {2 n} } \map \phi x \rd x
}}
{{eqn | r = \lim_{n \mathop \to \infty} n \map \phi {\xi_n} \paren {\frac 1 {2 n} - \paren {-\frac 1 {2 n} } }
      | c = Mean Value Theorem for Integrals, $\xi_n \in \closedint {-\frac 1 {2 n} } {\frac 1 {2 n} }$
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map \phi {\xi_n}
}}
{{eqn | r = \map \phi {\lim_{n \mathop \to \infty} \xi_n}
      | c = Limit of Image of Sequence on Real Numbers
}}
{{eqn | r = \map \phi 0
      | c = $\ds \lim_{n \mathop \to \infty} \frac 1 {2 n} = \lim_{n \mathop \to \infty} \paren {-\frac 1 {2 n} } = 0$, Squeeze Theorem for Real Sequences
}}
{{eqn | r = \map \delta \phi
      | c = {{Defof|Dirac Delta Distribution}}
}}
{{end-eqn}}
{{qed}}
Theorem: Minkowski Functional of Open Convex Set in Normed Vector Space is Well-Defined

Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.
Let $C$ be an open convex subset of $X$ with $0 \in C$.
Then, for each $x \in X$:
:$\set {t > 0 : t^{-1} x \in C} \ne \O$
and so the Minkowski functional of $C$ is well-defined.

Proof:
If $x = 0$, then: 
:$t^{-1} x \in C$
for all $t > 0$, so:
:$\set {t > 0 : t^{-1} x \in C} = \openint 0 \infty$
Now take $x \ne 0$. 
Since $C$ is open, there exists $\delta > 0$ such that for all $x \in X$ with: 
:$\norm x < \delta$
we have $x \in C$.
Note that we have, from positive homogeneity: 
:$\ds \norm {\frac \delta {2 \norm x} x} = \frac \delta 2$ 
so:
:$\dfrac \delta {2 \norm x} x \in C$
We therefore have:
:$\dfrac {2 \norm x} \delta \in \set {t > 0 : t^{-1} x \in C}$
so in particular: 
:$\set {t > 0 : t^{-1} x \in C} \ne \O$
We also have that:
:$t \ge 0$ for all $t \in \set {t > 0 : t^{-1} x \in C}$
So, from the Continuum Property:
:$\set {t > 0 : t^{-1} x \in C}$ has an infimum.
So:
:$\ds \inf \set {t > 0 : t^{-1} x \in C}$ is well-defined.
Since $x \in X$ was arbitrary, we have: 
:the Minkowski functional of $C$ is well-defined
{{qed}}
Category:Minkowski Functionals in Normed Vector Spaces
h7wwaat04yx6pr7yvq04und5wmb6jy6
Theorem: Minkowski Functional of Open Convex Set in Normed Vector Space is Bounded

Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $C$ be an open convex subset of $X$ with $0 \in C$.
Let $p_C$ be the Minkowski functional for $C$. 
Then there exists a real number $c > 0$ such that: 
:$0 \le \map {p_C} x \le c \norm x$
for each $x \in X$.

Proof:
From the definition of the Minkowski functional, we have: 
:$\map {p_C} x = \inf \set {t > 0 : t^{-1} x \in C}$
We have:
:$t \ge 0$ for every $t \in \set {t > 0 : t^{-1} x \in C}$
so:
:$\inf \set {t > 0 : t^{-1} x \in C} \ge 0$
from the definition of infimum. 
If $x = 0$, we have: 
:$t^{-1} x \in C$
for each $t > 0$, so:
:$\set {t > 0 : t^{-1} x \in C} = \openint 0 \infty$
so:
:$\inf \set {t > 0 : t^{-1} x \in C} = 0$ 
So the desired inequality holds for $x = 0$ for all real numbers $c > 0$. 
Now take $x \ne 0$.
Since $C$ is open, there exists $\delta > 0$ such that for all $x \in X$ with: 
:$\norm x < \delta$
we have $x \in C$.
In Minkowski Functional of Open Convex Set in Normed Vector Space is Well-Defined, it is shown that: 
:$\dfrac {2 \norm x} \delta \in \set {t > 0 : t^{-1} x \in C}$
So from the definition of infimum, we have: 
:$\ds \inf \set {t > 0 : t^{-1} x \in C} \le \frac {2 \norm x} \delta$
That is: 
:$\ds \map {p_C} x \le \frac 2 \delta \norm x$
Note that $\delta > 0$ depended only on $C$, so we have: 
:$\ds 0 \le \map {p_C} x \le \frac 2 \delta \norm x$ 
for all $x \in X$, hence the result.
{{qed}}
Theorem: Minkowski Functional of Open Convex Set in Normed Vector Space is Sublinear Functional

Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$.
Let $C$ be an open convex subset of $X$ with $0 \in C$.
Let $p_C$ be the Minkowski functional for $C$. 
Then $p_C$ is a sublinear functional.

Proof:
We will show that: 
:$(1): \quad \map {p_C} {\lambda x} = \lambda \map {p_C} x$ for each $x \in X$ and $\lambda \in \R_{\ge 0}$
:$(2): \quad \map {p_C} {x + y} \le \map {p_C} x + \map {p_C} y$ for each $x, y \in X$.
Theorem: Minkowski Functional of Open Convex Set in Normed Vector Space recovers Set

Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $C$ be an open convex subset of $X$ with $0 \in C$.
Let $p_C$ be the Minkowski functional for $C$. 
Then: 
:$C = \set {x \in X : \map {p_C} x < 1}$

Proof:
We first show that: 
:$\set {x \in X : \map {p_C} x < 1} \subseteq C$
Suppose that: 
:$x \in \set {x \in X : \map {p_C} x < 1}$
Then: 
:$\inf \set {t > 0 : t^{-1} x \in C} < 1$
from the definition of a Minkowski functional.
So, there exists $0 < \alpha < 1$ such that: 
:$\alpha^{-1} x \in C$ 
Since $C$ is convex and $0 \in C$, we have: 
:$\alpha \paren {\alpha^{-1} x} + \paren {1 - \alpha} \times 0 \in C$
so:
:$x \in C$
So we have: 
:$\set {x \in X : \map {p_C} x < 1} \subseteq C$
Now we show that: 
:$C \subseteq \set {x \in X : \map {p_C} x < 1}$
Let:
:$x \in C$ 
Since $C$ is open, there exists $\epsilon > 0$ such that for all $y \in X$ with: 
:$\norm {y - x} < \epsilon$
we have $y \in C$. 
So, we have: 
:$\ds \paren {1 + \frac \epsilon 2} x \in C$
So:
:$\ds \frac 2 {2 + \epsilon} \in \set {t > 0 : t^{-1} x \in C}$
So from the definition of infimum, we have: 
:$\ds \map {p_C} x \le \frac 2 {2 + \epsilon}$
So:
:$\map {p_C} x < 1$ 
We therefore have: 
:$C \subseteq \set {x \in X : \map {p_C} x < 1}$
So, from the definition of set equality, we have: 
:$C = \set {x \in X : \map {p_C} x < 1}$
{{qed}}
Theorem: True Weight from False Balance/Unequal Arms

Let $B$ be a body whose weight is $W$.
Let $B$ be weighed in a false balance with unequal arms.
Let the readings of the weight of $B$ be $a$ and $b$ when placed in opposite pans.
Then:
:$W = \sqrt {a b}$

Proof:
We have that the false balance has arms of different lengths.
Let the lengths of the arms of the false balance be $x$ and $y$.
{{WLOG}}, placing $B$ in the pan at the end of $x$ gives:
:$W x = a y$
and placing $B$ in the pan at the end of $y$ gives:
:$W y = b x$
{{TheoremWanted|We need to invoke the physics of couples to justify the above statements, but we haven't done the work yet to cover it.}}
Then:
{{begin-eqn}}
{{eqn | l = W x
      | r = a y
      | c = 
}}
{{eqn | l = W y
      | r = a x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac x y
      | r = \dfrac W a
      | c = 
}}
{{eqn | r = \dfrac b W
      | c = 
}}
{{eqn | ll= \leadsto
      | l = W^2
      | r = a b
      | c = multiplying both sides by $W a$
}}
{{eqn | ll= \leadsto
      | l = W
      | r = \sqrt {a b}
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:False Balances
taqb7lrw1xov7rbzj8d06uu3rz06k4n
Theorem: True Weight from False Balance/Imbalanced Pans

Let $B$ be a body whose weight is $W$.
Let $B$ be weighed in a false balance with imbalanced pans.
Let the readings of the weight of $B$ be $a$ and $b$ when placed in opposite pans.
Then:
:$W = \dfrac {a + b} 2$

Proof:
We have that the false balance has pans such that one weighs more than the other.
Let the lengths of the arms of the false balance be $x$.
Let one of the pans weigh $m$ more than the other.
Placing $B$ in the lighter pan gives:
:$W x = \paren {a + m} x$
and placing $B$ in the heavier pan gives:
:$\paren {W + m} x = b x$
{{TheoremWanted|We need to invoke the physics of couples to justify the above statements, but we haven't done the work yet to cover it.}}
Then:
{{begin-eqn}}
{{eqn | l = W
      | r = a + m
      | c = 
}}
{{eqn | l = W + m
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 W
      | r = a + b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = W
      | r = \dfrac {a + b} 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:False Balances
tu0h6svf22ebve0cfkj9gqfqqwucisg
Theorem: Seminorm is Sublinear Functional

Let $X$ be a vector space over $\R$.
Let $p : X \to \R$ be a seminorm on $X$. 
Then: 
:$p$ is a sublinear functional.

Proof:
Since $p$ is a seminorm, we have: 
:$\map p {x + y} \le \map p x + \map p y$ for each $x, y \in X$
We also have:
:$\map p {\lambda x} = \cmod \lambda \map p x$ for each $\lambda \in \R$ and $x \in X$.
and in particular: 
:$\map p {\lambda x} = \lambda \map p x$ for each $\lambda \in \R_{\ge 0}$ and $x \in X$.
So:
:$p$ is a sublinear functional.
{{qed}}
Category:Seminorms
Category:Sublinear Functionals
mmwj9b8t78cklil1f1t0as675m02m6i
Theorem: Weak-* Limit in Normed Dual Space is Unique

Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$. 
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space for $\struct {X, \norm \cdot}$.
Let $f, g \in X^\ast$.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence in $X^\ast$ such that: 
:$f_n \weakstarconv f$
and:
:$f_n \weakstarconv g$
where $\weakstarconv$ denotes weak-$\ast$ convergence.
Then: 
:$f = g$

Proof:
By the definition of weak-$\ast$ convergence, we have: 
:$\map {f_n} x \to \map f x$ for each $x \in X$
and:
:$\map {f_n} x \to \map g x$ for each $x \in X$.
Then, from Convergent Complex Sequence has Unique Limit we have: 
:$\map f x = \map g x$ for each $x \in X$.
That is: 
:$f = g$
{{qed}}
Theorem: Limit Inferior of Norm of Weakly Convergent Sequence is Bounded Below by Norm of Weak Limit

Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $x \in X$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ converging weakly to $x$. 
Then, we have: 
:$\ds \norm x \le \liminf_{n \mathop \to \infty} \norm {x_n}$

Proof:
From Existence of Support Functional, there exists $f \in X^\ast$ such that: 
:$\map f x = \norm x$
and:
:$\norm f_{X^\ast} = 1$
Since:
:$x_n \weakconv x$
we have: 
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f x$ 
That is:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = \norm x$
From Absolute Value of Limit, we have: 
:$\ds \lim_{n \mathop \to \infty} \cmod {\map f {x_n} } = \norm x$ 
From Convergence of Limsup and Liminf, we have: 
:$\ds \liminf_{n \mathop \to \infty} \cmod {\map f {x_n} } = \norm x$
From Fundamental Property of Norm on Bounded Linear Functional, we have: 
:$\cmod {\map f {x_n} } \le \norm f_{X^\ast} \norm {x_n}$
Since:
:$\norm f_{X^\ast} = 1$
we have: 
:$\cmod {\map f {x_n} } \le \norm {x_n}$
So, from Inequality Rule for Real Sequences: 
:$\ds \liminf_{n \mathop \to \infty} \cmod {\map f {x_n} } \le \liminf_{n \mathop \to \infty} \norm {x_n}$
That is: 
:$\ds \norm x \le \liminf_{n \mathop \to \infty} \norm {x_n}$
{{qed}}
Theorem: Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence

Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $x \in X$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$.
Then $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$ {{iff}}:
:$\norm {x_n - x} \to 0$

Proof:
From the definition of a convergent sequence in a normed vector space, we have that: 
:$x_n$ converges to $x$
{{iff}}:
:for each $\epsilon > 0$ there exists $N \in \N$ such that $\norm {x_n - x} < \epsilon$.
From the definition of a convergent real sequence, we have that: 
:$\norm {x_n - x} \to 0$
{{iff}}:
:for each $\epsilon > 0$ there exists $N \in \N$ such that $\size {\norm {x_n - x} - 0} < \epsilon$.
Since the norm is non-negative, we have that:
:$\norm {x_n - x} \to 0$
{{iff}}:
:for each $\epsilon > 0$ there exists $N \in \N$ such that $\norm {x_n - x} < \epsilon$.
We can therefore immediately deduce the result.
{{qed}}
Category:Convergent Sequences (Normed Vector Spaces)
gwmirajdy2oztkn0xvc8cvdda1bah8y
Theorem: Weakly Convergent Sequence in Hilbert Space with Convergent Norm is Convergent

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\HH$. 
Let $x \in X$ be such that: 
:$x_n \weakconv x$
and:
:$\norm {x_n} \to \norm x$
where $\weakconv$ denotes weak convergence.
Then: 
:$x_n \to x$

Proof:
Let $\norm \cdot$ be the inner product norm for $\struct {\HH, \innerprod \cdot \cdot}$.
From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we have: 
:$x_n \to x$
{{iff}}:
:$\norm {x_n - x} \to 0$
From Complex Sequence is Null iff Positive Integer Powers of Sequence are Null, it suffices to show that: 
:$\norm {x_n - x}^2 \to 0$
We have: 
{{begin-eqn}}
{{eqn	| l = \norm {x_n - x}^2
		| r = \innerprod {x_n - x} {x_n - x}
		| c = {{Defof|Inner Product Norm}}
}}
{{eqn	| r = \innerprod {x_n} {x_n} - \innerprod x {x_n} - \innerprod {x_n} x + \innerprod x x
		| c = Inner Product is Sesquilinear
}}
{{eqn	| r = \norm {x_n}^2 - \innerprod x {x_n} - \innerprod {x_n} x + \norm x^2
		| c = {{Defof|Inner Product Norm}}
}}
{{end-eqn}}
From Weak Convergence in Hilbert Space, we have: 
:$\innerprod {x_n} x \to \innerprod x x = \norm x^2$
since $x_n$ converges weakly to $x$.
From Weak Convergence in Hilbert Space: Corollary, we also have: 
:$\innerprod x {x_n} \to \innerprod x x = \norm x^2$
From hypothesis, we have: 
:$\norm {x_n}^2 \to \norm x$ 
Then, from Sum Rule for Real Sequences, we have: 
:$\norm {x_n - x}^2 \to 0$
so:
:$x_n \to x$
{{qed}}
Theorem: Convergent Sequence in Normed Vector Space is Weakly Convergent

Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $x \in X$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ converging to $x$. 
Then $\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$.

Proof:
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.
Let $f \in X^\ast$. 
If $\norm f_{X^\ast} = 0$, then $f = 0$ and: 
:$\map f {x_n} \to \map f x$
Take $f \ne 0$, so that $\norm f_{X^\ast} \ne 0$.
Let $\epsilon > 0$. 
We then have: 
{{begin-eqn}}
{{eqn	| l = \cmod {\map f {x_n} - \map f x} 
		| r = \cmod {\map f {x_n - x} }
		| c = {{Defof|Linear Functional}}
}}
{{eqn	| o = \le
		| r = \norm f_{X^\ast} \norm {x_n - x}
		| c = Fundamental Property of Norm on Bounded Linear Functional
}}
{{end-eqn}}
Since $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$:
:there exists $N \in \N$ such that $\ds \norm {x_n - x} < \frac \epsilon {\norm f_{X^\ast} }$ for all $n \ge N$.
Then, for $n \ge N$, we have: 
{{begin-eqn}}
{{eqn	| l = \cmod {\map f {x_n} - \map f x}
		| o = \le
		| r = \norm f_{X^\ast} \norm {x_n - x}
}}
{{eqn	| o = <
		| r = \norm f_{X^\ast} \paren {\frac \epsilon {\norm f_{X^\ast} } }
}}
{{eqn	| r = \epsilon
}}
{{end-eqn}}
Since $\epsilon > 0$ was arbitrary, we obtain: 
:$\map f {x_n} \to \map f x$
Since $f \in X^\ast$ was arbitrary, we have: 
:$\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$.
{{qed}}
Theorem: Convergent Sequence in Normed Vector Space is Weakly Convergent

Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $x \in X$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ converging to $x$. 
Then $\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$.

Proof:
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.
Then, for each $f \in X^\ast$:
{{begin-eqn}}
{{eqn | l = \size {\map f {x_n} - \map f x}
      | o = \le
      | r = \norm f_{X^\ast} \norm {x_n - x}_X
      | c = Fundamental Property of Norm on Bounded Linear Functional
}}
{{eqn | o = 
      | r = \stackrel{n \to \infty}{\longrightarrow} 0
}}
{{end-eqn}}
{{qed}}
Theorem: Linear Combination of Weakly Convergent Sequences is Weakly Convergent

Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\Bbb F$.
Let $x, y \in X$.
Let $\alpha, \beta \in \Bbb F$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ converging weakly to $x$. 
Let $\sequence {y_n}_{n \mathop \in \N}$ be a sequence in $X$ converging weakly to $y$. 
Then: 
:$\alpha x_n + \beta y_n \weakconv \alpha x + \beta y$
where $\weakconv$ denotes weak convergence.

Proof:
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,} }$. 
Since: 
:$x_n \weakconv x$
We have: 
:$\map f {x_n} \to \map f x$ 
for each $f \in X^\ast$.
Similarly, since: 
:$y_n \weakconv y$
we have: 
:$\map f {y_n} \to \map f y$ 
From Combined Sum Rule for Real Sequences, we therefore have: 
:$\alpha \map f {x_n} + \beta \map f {y_n} \to \alpha \map f x + \beta \map f {y_n}$
Since $f$ is linear, we have: 
:$\alpha \map f {x_n} + \beta \map f {y_n} = \map f {\alpha x_n + \beta y_n}$ for each $n \in \N$.
and:
:$\alpha \map f x + \beta \map f y = \map f {\alpha x + \beta y}$
So we have: 
:$\map f {\alpha x_n + \beta y_n} \to \map f {\alpha x + \beta y}$
for each $f \in X^\ast$. 
So, from the definition of weak convergence, we have: 
:$\alpha x_n + \beta y_n \weakconv \alpha x + \beta y$
{{qed}}
Category:Weak Convergence (Normed Vector Spaces)
rn3cfmb48cp04qwyb9b0hwpncaacsiq
Theorem: Largest Rectangle Contained in Triangle

Let $T$ be a triangle.
Let $R$ be a rectangle contained within $T$.
Let $R$ have the largest area possible for the conditions given.
Then:
:$(1): \quad$ One side of $R$ is coincident with part of one side of $T$, and hence two vertices lie on that side of $T$
:$(2): \quad$ The other two vertices of $R$ bisect the other two sides of $T$
:$(3): \quad$ The area of $R$ is equal to half the area of $T$.

Proof:
Note that a rectangle is a parallelogram.
By Largest Parallelogram Contained in Triangle, the area of $R$ cannot exceed half the area of $T$.
Hence we only need to show that when the first two conditions above are satisfied, the area of $R$ is exactly half the area of $T$.
Consider the diagram below.
:400px
Since $AD = DC$ and $CE = EB$:
{{begin-eqn}}
{{eqn | l = AF
      | r = FH
}}
{{eqn | l = HG
      | r = GB
}}
{{eqn | l = DF
      | r = \frac 1 2 GH
}}
{{eqn | l = \triangle CDE
      | o = \cong
      | r = \triangle HDE
}}
{{eqn | l = \triangle ADF
      | o = \cong
      | r = \triangle HDF
}}
{{eqn | l = \triangle BGE
      | o = \cong
      | r = \triangle HGE
}}
{{end-eqn}}
and so the area of $R$ is equal to the area of the parts of $T$ not included in $R$.
That is, the area of $R$ is exactly half the area of $T$.
{{qed}}
Category:Triangles
Category:Rectangles
5qjvgathfp1gk7u331za8bcm49f7cg3
Theorem: Characterization of Hyperplanes

Let $\Bbb F$ be a field.
Let $X$ be a vector space over $\Bbb F$.
Let $U$ be a subspace of $X$. 
{{TFAE}}
:$(1): \quad$ $U$ is a hyperplane
:$(2): \quad$ $U \ne X$, and for any $x \in X \setminus U$ we have $\map \span {U \cup \set x} = X$
:$(3): \quad$ there exists a non-zero linear functional $\phi : X \to \Bbb F$ such that $U = \map \ker \phi$.

Proof:
=== $(1)$ implies $(2)$ ===
Suppose that: 
:$U$ is a hyperplane.
Let $x \in X \setminus U$. 
Then from Linear Span is Linear Subspace, we have: 
:$\map \span {U \cup \set x}$ is a subspace of $X$.
We have that: 
:$U \subseteq \map \span {U \cup \set x}$
So either: 
:$U = \map \span {U \cup \set x}$
or: 
:$X = \map \span {U \cup \set x}$ 
Since $x \in \map \span {U \cup \set x}$ but $x \not \in U$, we must have: 
:$X = \map \span {U \cup \set x}$ 
{{qed|lemma}}