nyuuzyou/wikis huggingface dataset:
Normed Dual Space Separates Points

== Theorem ==
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: 
:$x = y$
== Sources ==
* $20.1$: Existence of a Support Functional
Category:Normed Dual Spaces
Normed Dual Space is Banach Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\GF$. 
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual of $\struct {X, \norm {\, \cdot \,}_X}$.
Then $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ is a Banach space. 
== Proof ==
By definition, we have:
:$X^\ast = \map B {X, \GF}$
and:
:$\norm {\, \cdot \,}_{X^\ast} = \norm {\, \cdot \,}_{\map B {X, \GF} }$
From Real Number Line is Banach Space and Complex Plane is Banach Space, $\GF$ is a Banach space.
So from Space of Bounded Linear Transformations is Banach Space, $\struct {X, \norm {\, \cdot \,}_X}$ is a Banach space.
Category:Normed Dual Spaces
Category:Banach Spaces
Normed Dual Space of Infinite-Dimensional Normed Vector Space is Infinite-Dimensional

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be an infinite dimensional normed vector space.
Let $X^\ast$ be the normed dual space of $X$.
Then $X^\ast$ is infinite dimensional.
== Proof ==
 $X^\ast$ is finite dimensional.
Then from Normed Dual Space of Finite-Dimensional Vector Space is Isometrically Isomorphic to Original Space:
:$X^{\ast \ast} \cong X^{\ast}$
From Isometric Isomorphism between Normed Vector Spaces preserves Dimension:
:$\dim X^{\ast \ast} = \dim X^\ast$
Let $\iota : X \to X^{\ast \ast}$ be the evaluation linear transformation.
Then $\iota$ is a linear isometry from Evaluation Linear Transformation on Normed Vector Space is Linear Isometry.
Hence $\iota : X \to \iota X$ is an isometric isomorphism.
From Isometric Isomorphism between Normed Vector Spaces preserves Dimension:
:$\dim X = \dim \iota X$
From Image of Vector Subspace under Linear Transformation is Vector Subspace:
:$\iota X$ is a linear subspace of $X^{\ast \ast}$.
From Dimension of Proper Subspace is Less Than its Superspace:
:$\dim \iota X \le \dim X^{\ast \ast}$
So:
:$\dim X \le \dim X^{\ast \ast} = \dim X^\ast$
But $\dim X$ is not finite.
Hence $\dim X^\ast$ is not finite. 
This contradicts our assumption that $X^\ast$ is finite dimensional.
Hence by Proof by Contradiction $X^\ast$ is infinite dimensional.
Category:Normed Vector Spaces
Category:Normed Dual Spaces
Normed Dual Space of Normed Quotient Vector Space is Isometrically Isomorphic to Annihilator

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$.
Let $X^\ast$ be the normed dual space of $X$. 
Let $Y$ be a closed linear subspace of $X$. 
Let $\paren {X/Y}^\ast$ be the normed dual space of the normed quotient vector space $X/Y$.
Then: 
:$\paren {X/Y}^\ast$ is isometrically isomorphic to $Y^\bot$
where $Y^\bot$ is the annihilator of $Y$. 
== Proof ==
Let $\pi$ be the quotient mapping associated with $X/Y$. 
Note that for each $f \in \paren {X/Y}^\ast$ we have: 
:$f \circ \pi \in X^\ast$
by Composition of Bounded Linear Transformations is Bounded Linear Transformation.
Further, for $y \in Y$ we have $\map \pi y = 0_{X/Y}$ from Kernel of Quotient Mapping, and so: 
:$\map {\paren {f \circ \pi} } y = \map f {0_{X/Y} } = 0$ for all $y \in Y$. 
So we have $f \circ \pi \in Y^\bot$.
Now define $\phi : \paren {X/Y}^\ast \to Y^\bot$ by: 
:$\map \phi f = f \circ \pi$ 
From Linear Isometry is Injective: Corollary, it suffices to show that $\phi$ is a linear isometry and surjective.
First, take $f, g \in \paren {X/Y}^\ast$ and $\lambda, \mu \in \GF$.
We have: 
so $\phi$ is linear. 
Now, we have for $f \in \paren {X/Y}^\ast$: 
so $\phi$ is a linear isometry.
It remains to show that $\phi$ is surjective. 
Since we have shown that: 
:$\ds \sup_{\mathbf x \in \map \pi {B_X} } \cmod {\map f {\mathbf x} } = \sup_{\mathbf x \in B_{X/Y} } \cmod {\map f {\mathbf x} }$
we know that if $f \circ \pi \in X^\ast$, then $f \in \paren {X/Y}^\ast$.
From Condition for Mapping from Quotient Vector Space to be Well-Defined, we have that for each $g \in X^\ast$, there exists $f : X/Y \to \GF$ such that: 
:$g = f \circ \pi$ 
From the above, it follows that $f \in \paren {X/Y}^\ast$ and so $g = \map \phi f$. 
So $\phi$ is surjective, and we are done from Linear Isometry is Injective: Corollary.
== Sources ==
* : $4.9$: Theorem
Category:Normed Dual Spaces
Category:Normed Quotient Vector Spaces
Category:Annihilators of Subspaces of Banach Spaces
Normed Dual Space of Separable Normed Vector Space is Weak-* Separable

== Theorem ==
Let $X$ be a separable normed vector space.
Let $X^\ast$ be the normed dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Then $\struct {X^\ast, w^\ast}$ is separable. 
== Proof ==
Let $B^-_{X^\ast}$ be the closed unit ball of $X^\ast$.
From Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Separable, $\struct {B^-_{X^\ast}, w^\ast}$ is separable space.
Let $S$ be a countable dense subset of $B^-_{X^\ast}$. 
For $n \in \N$, we have: 
:$\map {\cl_{w^\ast} } {n S} = n B^-_{X^\ast}$ 
from Dilation of Closure of Set in Topological Vector Space is Closure of Dilation.
Now, note that: 
:$\ds \bigcup_{n \mathop = 1}^\infty n S$ is countable
from Countable Union of Countable Sets is Countable.
From Closure of Union contains Union of Closures, we therefore have: 
So:
:$\ds \map {\cl_{w^\ast} } {\bigcup_{n \mathop = 1}^\infty n S} = X^\ast$
So $\struct {X^\ast, w^\ast}$ is separable. 
== Sources ==
* : Proposition $3.25$
Category:Normed Dual Spaces
Category:Separable Spaces
Category:Weak-* Topologies
Normed Vector Space Requires Multiplicative Norm on Division Ring

== Theorem ==
Let $R$ be a normed division ring with a submultiplicative norm $\norm {\, \cdot \,}_R$.
Let $V$ be a vector space that is not a trivial vector space.
Let $\norm {\, \cdot \,}: V \to \R_{\ge 0}$ be a mapping from $V$ to the positive real numbers satisfying the vector space norm axioms.
Then $\norm {\, \cdot \,}_R$ is a multiplicative norm.
That is:
:$\forall r, s \in R: \norm {r s}_R = \norm r_R \norm s_R$
== Proof ==
Since $V$ is not a trivial vector space:
:$\exists \mathbf v \in V: \mathbf v \ne 0$
By :
:$\norm {\mathbf v} > 0$
Let $r, s \in R$:
By dividing both sides of the equation by $\norm {\mathbf v}$ then:
:$\norm {r s}_R = \norm r_R \norm s_R$
The result follows.
Category:Normed Division Rings
Category:Norm Theory
Normed Vector Space is Closed in Itself

== Theorem ==
Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Then $X$ is closed in $M$.
== Proof ==
From Empty Set is Open in Normed Vector Space, $\O$ is open in $M$.
But:
:$X = \relcomp X \O$
where $\complement_X$ denotes the set complement relative to $X$.
The result follows by definition of closed set.
== Also see ==
*Space is Closed in Itself
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Closed Sets (Normed Vector Spaces)
Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact

== Theorem ==
Let $X$ be a normed vector space.
Let $\Bbb S = \map {\Bbb S_1} 0$ be the unit sphere centred at $0$ in $X$.
Then $X$ is finite dimensional $\Bbb S$ is compact.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
Category:Compact Spaces
Category:Finite Dimensional Vector Spaces
Category:Normed Vector Spaces
Category:Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact
Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact/Necessary Condition

== Theorem ==
Let $X$ be a finite dimensional normed vector space.
Let $\Bbb S = \map {\Bbb S_1} 0$ be the unit sphere centred at $0$ in $X$.
Then $\Bbb S$ is compact.
== Proof ==
Let $X$ be a finite dimensional vector space $\R^d$.
We have that all norms on finite-dimensional vector space are equivalent.
Choose Euclidean norm $\norm {\, \cdot \,}_2$.
Let $\struct {\R^d, \norm {\, \cdot \,}_2}$ be the normed finite-dimensional real vector space with Euclidean norm.
Let $\map {\Bbb S^{d - 1}_1} 0$ be a unit sphere in $\struct {\R^d, \norm {\, \cdot \,}_2}$.
By definition, $\Bbb S^{d - 1}$ is bounded.
Let $\mathbf L := \tuple {L_1, \dots, L_d} \in \R^d$.
Let $\mathbf x_n := \tuple {x_n^{\paren 1}, \dots, x_n^{\paren d}}$.
Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ be a sequence in $\Bbb S^{d - 1}$.
Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converge to $\mathbf L$:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \norm {\mathbf x_n - \mathbf L}_2 < \epsilon$
Furthermore, for all $k \in \N : k \le d$ we have that:
Thus:
:$\forall k \in \N : k \le d : \forall \epsilon \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \size {x_n^{\paren k} - L_k} < \epsilon$
In other words, $\ds \lim_{n \mathop \to \infty} x_n^{\paren k} = L_k$.
We have that $\mathbf x_n \in \Bbb S^{d - 1}$.
Therefore:
:$\ds \norm {\mathbf x_n}_2^2 = \sum_{j \mathop = 1}^d \paren {x_n^{\paren j}}^2 = 1$
$\map f x = x^2$ is a continuous function in its whole range.
By Limit of Composite Function, taking the limit $n \to \infty$ results in:
:$\ds \norm {\mathbf L}_2^2 = \sum_{j \mathop = 1}^d \paren {L_j}^2 = 1$
Therefore, $\mathbf L \in \Bbb S^{d - 1}$.
Hence, $\Bbb S^{d - 1}$ is closed.
By Heine-Borel theorem, $\Bbb S^{d - 1}$ is compact.
== Sources ==
* : Chapter $\S 1.5$: Normed and Banach spaces. Compact sets
Category:Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact
Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact/Sufficient Condition

== Theorem ==
== Proof ==
Suppose that $X$ is not finite dimensional.
Then any finite dimensional subspace of $X$ is proper.
From Compact Subspace of Metric Space is Sequentially Compact in Itself, it suffices to show that $\Bbb S$ is not sequentially compact. 
For this, it suffices to show that there exists an $\epsilon > 0$ and sequence $\sequence {x_n}_{n \mathop \in \N}$ with $\norm {x_n} = 1$ for each $n$ and: 
:$\norm {x_i - x_j} \ge \epsilon$ for each $i \ne j$. 
Then no subsequence of $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy. 
We construct $\sequence {x_n}_{n \mathop \in \N}$ iteratively. 
From Convergent Sequence is Cauchy Sequence, we will have that no subsequence of $\sequence {x_n}_{n \mathop \in \N}$ can converge. 
Take $x_1 \in X$ with $\norm {x_1} = 1$.
Let: 
:$Y_1 = \span \set {x_1}$
From Finite Dimensional Subspace of Normed Vector Space is Closed, $Y_1$ is closed. 
From Riesz's Lemma, there exists $x_2 \in X$ with $\norm {x_2} = 1$ such that: 
:$\ds \norm {x_2 - x_1} > \frac 1 2$
Now suppose that we have constructed $x_1, x_2, \ldots, x_n$ such that: 
:$\ds \norm {x_i - x_j} > \frac 1 2$
for all $i \ne j$. 
Let: 
:$Y_n = \span \set {x_1, x_2, \ldots, x_n}$
From Finite Dimensional Subspace of Normed Vector Space is Closed, $Y_n$ is closed. 
From Riesz's Lemma, there exists $x_{n + 1} \in X$ with $\norm {x_{n + 1} } = 1$ such that: 
:$\ds \norm {x_{n + 1} - y} > \frac 1 2$
for all $y \in Y_n$. 
In particular: 
:$\ds \norm {x_{n + 1} - x_k} > \frac 1 2$
for all $1 \le k \le n$.
From , we have: 
:$\ds \norm {x_i - x_j} > \frac 1 2$
for each $1 \le i, j \le {n + 1}$ with $i \ne j$. 
Continuing this way, we have a sequence $\sequence {x_n}_{n \mathop \in \N}$ such that $\norm {x_n} = 1$ for each $n \in \N$ and: 
:$\ds \norm {x_i - x_j} > \frac 1 2$
for each $i \ne j$. 
So for any subsequence $\sequence {x_{n_j} }_{j \mathop \in \N}$ of $\sequence {x_n}_{n \mathop \in \N}$ we have: 
:$\ds \norm {x_{n_j} - x_{n_k} } > \frac 1 2$
for each $j \ne k$. 
So no subsequence $\sequence {x_{n_j} }_{j \mathop \in \N}$ can be Cauchy, and hence none can converge. 
Category:Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact
Normed Vector Space is Hausdorff

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X }$ be a normed vector space.
Let $\tau$ be the topology on $X$ that consists of all open sets in $X$.
That is, $U \in \tau$ :
:$\forall x \in U: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq U$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$.
Then $\struct {X, \tau}$ is a $T_2$ (Hausdorff) space.
== Proof ==
From Metric Induced by Norm is Metric, it follows that the norm $\norm {\,\cdot\,}$ induces a metric $d$ on $X$.
From the definition of open set in normed vector space, it follows that $\tau$ is the topology induced by the metric $d$.
The result now follows from Metric Space is Hausdorff.
Category:Normed Vector Spaces
Category:Hausdorff Spaces
Normed Vector Space is Hausdorff Locally Convex Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Then $\struct {X, \norm {\, \cdot \,} }$ is a Hausdorff locally convex space. 
 
== Proof ==
From Normed Vector Space is Locally Convex Space, $\struct {X, \norm {\, \cdot \,} }$ can be viewed as the locally convex space $\struct {X, \norm {\, \cdot \,} }$.
Now for $x \ne \mathbf 0_X$, we have: 
:$\norm x \ne 0$ 
from . 
So $\struct {X, \norm {\, \cdot \,} }$ is a Hausdorff locally convex space.
Category:Locally Convex Spaces
Category:Normed Vector Spaces
Normed Vector Space is Hausdorff Topological Vector Space

== Theorem ==
Let $\struct { K, +_K, \circ_K }$ be a valued field with norm $\norm {\,\cdot\,}_K$.
Let $\struct {X, \norm {\, \cdot \,}_X }$ be a normed vector space over $K$.
Let $\tau$ be the topology on $X$ that consists of all open sets in $X$.
That is, $U \in \tau$ :
:$\forall x \in U: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq U$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$.
Then $\struct { X, \tau }$ is a Hausdorff topological vector space.
== Proof ==
By its definition, a valued field is a normed division ring.
Let $\tau_K$ be the topology on $K$ induced by the metric induced by the norm $\norm {\,\cdot\,}_K$
From Normed Division Ring Operations are Continuous:Corollary, it follows that $\struct {K, \tau_K}$ is a topological field.
Let $d_X : X \times X \to \R_{\ge 0}$ be the metric induced by the inner product norm $\norm {\,\cdot\,}_X$ on $X$.
By definition of open set in normed vector space, it follows that $\tau$ is the topology on $X$ induced by the metric $d_X$.
To show that $\tau$ is a topology on $X$, we examine each of the open set axioms in turn:
=== ===
Follows from Union of Open Sets of Normed Vector Space is Open.
=== ===
Follows from Finite Intersection of Open Sets of Normed Vector Space is Open.
=== ===
Follows from Normed Vector Space is Open in Itself.
Let $+_{\scriptscriptstyle X}$ denote the vector addition on $X$ , and let $\circ_{\scriptscriptstyle X}$ denote 
the scalar multiplication on $X$.
To show that $\struct { X, \tau }$ is a topological vector space, we examine each of the axioms from the definition of topological vector space in turn:
=== $\tau$ is a Hausdorff topology ===
Follows from Normed Vector Space is Hausdorff.
=== $+_{\scriptscriptstyle X}: X \times X \to X$ is continuous with respect to $\tau$ ===
Follows from Vector Addition on Normed Vector Space is Continuous.
=== $\circ_{\scriptscriptstyle X}: K \times X \to X$ is continuous with respect to $\tau$ ===
Follows from Scalar Multiplication on Normed Vector Space is Continuous.
Category:Normed Vector Spaces
Category:Hausdorff Topological Vector Spaces
Normed Vector Space is Locally Connected

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$. 
Then $\struct {X, \norm {\, \cdot \,} }$ is locally connected.
== Proof ==
Let $x \in X$.
Let $\epsilon > 0$. 
Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball around $x$ in $\struct {X, \norm {\, \cdot \,} }$. 
From Open Balls form Local Basis of Metric Space, the set $\BB_x = \set {\map {B_\epsilon} x : \epsilon > 0}$ is a local basis at $x$.
From Open Ball in Normed Vector Space is Connected, $\map {B_\epsilon} x$ is connected.
Hence $x$ has a local basis of connected sets.
Since $x$ was arbitrary, $\struct {X, \norm {\, \cdot \,} }$ is locally connected.
Category:Locally Connected Spaces
Category:Normed Vector Spaces
Normed Vector Space is Locally Convex Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\tau$ be the standard topology on the locally convex space $\struct {X, \set {\norm {\, \cdot \,} } }$.
Then $\tau$ consists precisely of the open sets of $\struct {X, \norm {\, \cdot \,} }$.
== Proof ==
Let $U \subseteq X$. 
From Open Sets in Standard Topology of Locally Convex Space, we have $U \in \tau$ :
:for each $x \in U$ there exists $\epsilon > 0$ such that: 
::$\set {y \in X : \norm {y - x} < \epsilon} \subseteq U$
That is for each $x \in U$ there exists $\epsilon > 0$ such that: 
:$\map {B_\epsilon} x \subseteq U$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$.
So from the definition of an open set of $\struct {X, \norm {\, \cdot \,} }$, this is equivalent to $U$ being an open set of $\struct {X, \norm {\, \cdot \,} }$.
Category:Normed Vector Spaces
Category:Locally Convex Spaces
Normed Vector Space is Open in Itself

== Theorem ==
Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Then the set $X$ is an open set of $M$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Open Sets (Normed Vector Spaces)
Category:Normed Vector Space is Open in Itself
Normed Vector Space is Open in Itself/Proof 1

== Theorem ==
Let $M = \struct{X, \norm {\, \cdot \,}}$ be a normed vector space.
Then the set $X$ is an open set of $M$.
== Proof ==
By definition, an open set $S \subseteq A$ is one where every point inside it is an element of an open ball contained entirely within that set.
Let $x \in X$.
An open ball of $x$ in $M$ is by definition a subset of $X$.
Hence the result.
== Sources ==
* : Chapter $1$: Normed and Banach spaces
Category:Normed Vector Space is Open in Itself
Normed Vector Space is Open in Itself/Proof 2

== Theorem ==
Let $M = \struct{X, \norm {\, \cdot \,}}$ be a normed vector space.
Then the set $X$ is an open set of $M$.
== Proof ==
By definition, a subset $S \subseteq X$ is open if:
:$\forall x \in X : \exists \epsilon \in \R_{>0} : \map {B_\epsilon} x \subseteq S$
Let $S = X$.
 $X$ is not open.
By De Morgan's laws:
:$\exists x \in X : \forall \epsilon \in \R_{>0} : \map {B_\epsilon} x \cap \paren {X \setminus S} \ne \O$
Note that:
:$X \setminus S = X \setminus X = \O$.
By Intersection with Empty Set:
:$\map {B_\epsilon} x \cap \O = \O$
Hence:
:$\exists x \in X : \forall \epsilon \in \R_{>0} : \O \ne \O$.
Since Empty Set is Unique, we have a contradiction.
Category:Normed Vector Space is Open in Itself
Normed Vector Space is Reflexive iff Closed Unit Ball in Original Space is Mapped to Closed Unit Ball in Second Dual

== Theorem ==
Let $X$ be a normed vector space. 
Let $X^\ast$ be the normed dual of $X$.
Let $X^{\ast \ast}$ be the second norm dual.
Let $\iota : X \to X^{\ast \ast}$ be the evaluation linear transformation.
Let $B_X^-$ be the closed unit ball of $X$.
Let $B_{X^{\ast \ast} }^-$ be the closed unit ball of $X^{\ast \ast}$. 
Then $X$ is reflexive $\iota B_X^- = B_{X^{\ast \ast} }^-$.
== Proof ==
=== Necessary Condition ===
Suppose that $X$ is reflexive.
Then $\iota : X \to X^{\ast \ast}$ is an isometric isomorphism.
So from Injective Linear Transformation between Normed Vector Spaces sends Closed Unit Ball to Closed Unit Ball iff Isometric Isomorphism, we have that $\iota B_X^- = B_{X^{\ast \ast} }^-$. 
=== Sufficient Condition ===
Conversely suppose that $\iota B_X^- = B_{X^{\ast \ast} }^-$. 
From Linear Isometry is Injective, $\iota$ is injective.
Then we see that $\iota : X \to X^{\ast \ast}$ is an injective linear transformation with:
:$\iota B_X^- = B_{X^{\ast \ast} }^-$
It follows that $\iota$ is an isometric isomorphism from Injective Linear Transformation between Normed Vector Spaces sends Closed Unit Ball to Closed Unit Ball iff Isometric Isomorphism. 
Category:Reflexive Spaces
Normed Vector Space is Reflexive iff Surjective Evaluation Linear Transformation

== Theorem ==
Let $\struct {X, \norm \cdot_X}$ be a normed vector space.
Let $\struct {X^{\ast \ast}, \norm \cdot_{X^{\ast \ast} } }$ be the second normed dual of $\struct {X, \norm \cdot_X}$.
Let $\iota : X \to X^{\ast \ast}$ be the evaluation linear transformation.
Then $X$ is reflexive :
:$\iota$ is surjective.
That is: 
:$\iota X = X^{\ast \ast}$. 
== Proof ==
From the definition of a reflexive space, we have that $X$ is reflexive :
:$\iota$ is an isometric isomorphism.
From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, we have: 
:$\iota$ is a linear isometry.
From Linear Isometry is Injective: Corollary, we then have: 
:$\iota$ is an isometric isomorphism $\iota$ is surjective.
That is: 
:$X$ is reflexive $\iota$ is surjective.
== Sources ==
* $26.1$: The Second Dual
Category:Reflexive Spaces
Category:Evaluation Linear Transformations (Normed Vector Spaces)
Normed Vector Space is Reflexive iff Weak and Weak-* Topologies on Normed Dual coincide

== Theorem ==
Let $X$ be a normed vector space. 
Let $X^\ast$ be the normed dual of $X$.
Let $w$ be the weak topology on $X^\ast$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$. 
Then $X$ is reflexive $\struct {X^\ast, w} = \struct {X^\ast, w^\ast}$. 
== Proof ==
=== Necessary Condition ===
Suppose that $X$ is reflexive.
Then for each $\Phi \in X^{\ast \ast}$ there exists $x \in X$ such that $\Phi = x^\wedge$.
Conversely, by Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual, we have $x^\wedge \in X^{\ast \ast}$ for each $x \in X$. 
Hence, we have: 
:$\set {x^\wedge : x \in X} = X^{\ast \ast}$ 
By definition, $w$ is the initial topology on $X^\ast$ generated by $X^{\ast \ast}$, and $w^\ast$ is the initial topology on $X^\ast$ generated by the family $\set {x^\wedge : x \in X}$.
It follows that $w = w^\ast$. 
=== Sufficient Condition ===
Suppose that $\struct {X^\ast, w} = \struct {X^\ast, w^\ast}$.
Then $\struct {X^\ast, w}^\ast = \struct {X^\ast, w^\ast}^\ast$. 
From Characterization of Continuity of Linear Functional in Weak Topology, we have $\struct {X^\ast, w}^\ast = X^{\ast \ast}$.
From Characterization of Continuity of Linear Functional in Weak-* Topology , we have $\struct {X^\ast, w^\ast}^\ast = \iota X$.
So we have: 
:$\iota X = X^{\ast \ast}$
From Normed Vector Space is Reflexive iff Surjective Evaluation Linear Transformation, we have that $X$ is reflexive.
Category:Weak Topologies on Topological Vector Spaces
Category:Weak-* Topologies
Category:Reflexive Spaces
Category:Normed Dual Spaces