avewright/proofwiki-math huggingface dataset:
Theorem: Evaluation Linear Transformation on Normed Vector Space is Linear Isometry

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 $J : X \to X^{\ast \ast}$ be the evaluation linear transformation for $X$. 
Then: 
:$J$ is a linear isometry.

Proof:
From Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual, we have that: 
:$J$ is a linear transformation.
It remains to show that:
:$\norm {\map J x}_{X^{\ast \ast} } = \norm x_X$
for each $x \in X$.
For each $x \in X$, we denote: 
:$\map J x = x^\wedge$
We then aim to show: 
:$\norm {x^\wedge}_{X^{\ast \ast} } = \norm x_X$
for each $x \in X$. 
As shown in Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual, we have:
:$\cmod {\map {x^\wedge} f} \le \norm f_{X^\ast} \norm x_X$
for each $f \in X^\ast$.
So, we have: 
:$\norm x_X \in \set {c > 0 : \cmod {\map {x^\wedge} f} \le c \norm f_{X^\ast} \text { for all } f \in X^\ast}$
So, from the definition of the norm on bounded functionals and the definition of infimum, we have: 
:$\norm {x^\wedge}_{X^{\ast \ast} } \le \norm x_X$
We now aim to show that: 
:$\norm x_X \le \norm {x^\wedge}_{X^{\ast \ast} }$
From Existence of Support Functional, there exists $f \in X^\ast$ such that: 
:$\map f x = \norm x_X$
and:
:$\norm f_{X^\ast} = 1$ 
Then we have: 
{{begin-eqn}}
{{eqn	| l = \cmod {\map {x^\wedge} f}
		| r = \cmod {\map f x}
}}
{{eqn	| r = \norm x_X
}}
{{eqn	| r = \norm f_{X^\ast} \norm x_X
}}
{{end-eqn}}
So we have: 
:$\norm x_X \in \set {\cmod {\map {x^\wedge} f} : \norm f_{X^\ast} = 1}$
So from the definition of the norm on bounded functionals and the definition of supremum, we have: 
:$\norm {x^\wedge}_{X^{\ast \ast} } \ge \norm x_X$
So we have: 
:$\norm {\map {x^\wedge} f} = \norm x_X$
That is: 
:$\norm {\map J x}_{X^{\ast \ast} } = \norm x_X$
{{qed}}
Category:Linear Isometries
Category:Evaluation Linear Transformations (Normed Vector Spaces)
9sef8hzqyxnwi5y4nsriwdbhh71efcu
Theorem: Pointwise Limit of Sequence of Bounded Linear Transformations from Banach Space to Normed Vector Space is Bounded Linear Transformation

Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot_X}$ be a Banach space over $\Bbb F$.
Let $\struct {Y, \norm \cdot_Y}$ be a normed vector space over $\Bbb F$.
Let $T : X \to Y$ be a function.
Let $\sequence {T_n}_{n \mathop \in \N}$ be a sequence of bounded linear transformations such that: 
:$\ds T x = \lim_{n \mathop \to \infty} T_n x$
for each $x \in X$.
Then $T$ is a bounded linear transformation.

Proof:
We first show that $T$ is a linear transformation.
Let $\alpha, \beta \in \Bbb F$ and $x, y \in X$. 
Then, we have: 
{{begin-eqn}}
{{eqn	| l = \map T {\alpha x + \beta y}
		| r = \lim_{n \mathop \to \infty} \map {T_n} {\alpha x + \beta y}
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \paren {\alpha T_n x + \beta T_n y}
		| c = {{Defof|Linear Transformation}}
}}
{{eqn	| r = \alpha \lim_{n \mathop \to \infty} T_n x + \beta \lim_{n \mathop \to \infty} T_n y
}}
{{eqn	| r = \alpha T x + \beta T y
}}
{{end-eqn}}
So $T$ is linear
Now we show that $T$ is bounded.
We use the Banach-Steinhaus Theorem.
From Modulus of Limit: Normed Vector Space, we have: 
:$\ds \norm {T x}_Y = \lim_{n \mathop \to \infty} \norm {T_n x}_Y$
From Convergent Real Sequence is Bounded, we have: 
:$\norm {T_n x}_Y$ is bounded in $n$ for each $x \in X$.
That is: 
:$\ds \sup_{n \in \N} \norm {T_n x}_Y$ is finite for each $x \in X$.
So, from the Banach-Steinhaus Theorem, we have: 
:$\ds \sup_{n \in \N} \norm {T_n}$ is finite, where $\norm \cdot$ is the norm on bounded linear transformation.
Pick $M > 0$ such that: 
:$\norm {T_n} \le M$
for each $n \in \N$.
Then: 
:$\norm {T_n x}_Y \le M \norm x_X$
for each $x \in X$ and $n \in \N$ from the definition of the norm on bounded linear transformation. 
Then for each $x \in X$ we have: 
{{begin-eqn}}
{{eqn	| l = \norm {T x}_Y
		| r = \lim_{n \mathop \to \infty} \norm {T_n x}_Y
}}
{{eqn	| o = \le
		| r = M \norm x_X
		| c = Limits Preserve Inequalities
}}
{{end-eqn}}
So $T$ is bounded.
{{qed}}
Theorem: Principle of Condensation of Singularities

Let $\struct {X, \norm {\,\cdot\,}_X}$ be a Banach space.
Let $\struct {Y, \norm {\,\cdot\,}_Y}$ be a normed vector space.
Let $\family {T_\alpha: X \to Y}_{\alpha \mathop \in A}$ be an $A$-indexed family of bounded linear transformations from $X$ to $Y$ such that: 
:$\ds \sup_{\alpha \in A} \norm {T_\alpha} = \infty$
Then: 
:$\ds \sup_{\alpha \in A} \norm {T_\alpha x} = \infty$ for some $x \in X$.

Proof:
{{AimForCont}} that: 
:$\ds \sup_{\alpha \in A} \norm {T_\alpha x} < \infty$ for each $x \in X$.
From the Banach-Steinhaus Theorem, we have: 
:$\ds \sup_{\alpha \in A} \norm {T_\alpha} < \infty$.
contradicting that: 
:$\ds \sup_{\alpha \in A} \norm {T_\alpha} = \infty$
So, we have: 
:$\ds \sup_{\alpha \in A} \norm {T_\alpha x} = \infty$ for some $x \in X$.
{{qed}}
Theorem: Banach-Schauder Theorem

Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be Banach spaces.
Let $T : X \to Y$ be a surjective bounded linear transformation.
Then $T$ is an open mapping.

Proof:
For each $x \in X$ and $r > 0$, let $\map {B_X} {x, r}$ be the open ball in $\struct {X, \norm \cdot_X}$ with centre $x$ and radius $r$.
For each $y \in Y$ and $r > 0$, let $\map {B_Y} {y, r}$ be the open ball in $\struct {Y, \norm \cdot_Y}$ with centre $y$ and radius $r$.
Note that we can write: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty \map {B_X} {0, n}$
From Image of Union under Mapping: General Result, we have: 
:$\ds T \sqbrk X = \bigcup_{n \mathop = 1}^\infty T \sqbrk {\map {B_X} {0, n} }$
Since $T$ is surjective, we have $T \sqbrk X = Y$, and so: 
:$\ds Y = \bigcup_{n \mathop = 1}^\infty T \sqbrk {\map {B_X} {0, n} }$
Since $\struct {Y, \norm \cdot_Y}$ is a Banach space, by the Baire Category Theorem it is also a Baire space.
From Baire Space is Non-Meager:
:$Y$ is non-meager.
So:
:$Y$ is not the countable union of nowhere dense subsets of $Y$.
So, we must have: 
:$T \sqbrk {\map {B_X} {0, m} }$ is not nowhere dense for some $m \in \N$.
So the topological closure of $T \sqbrk {\map {B_X} {0, m} }$ has non-empty interior.
So, there exists a open set $U \subseteq X$ such that: 
:$U \subseteq \paren {T \sqbrk {\map {B_X} {0, m} } }^-$
where $\paren {T \sqbrk {\map {B_X} {0, m} } }^-$ denotes the topological closure of $T \sqbrk {\map {B_X} {0, m} }$
Let $y \in U$.
Since $U$ is open set, there exists an open ball $\map {B_Y} {y, r}$ such that $r > 0$ and: 
:$\map {B_Y} {y, r} \subseteq U \subseteq \paren {T \sqbrk {\map {B_X} {0, m} } }^-$
Theorem: Sylvester's Law of Inertia

Let $\struct {V, q}$ be the scalar product space.
Let $V^*$ be the algebraic dual of $V$.
Suppose there exists a basis $\tuple {\beta^i}$ for $V^*$ such that $q$ is expressible as:
:$q = \paren {\beta^1}^2 + \ldots + \paren {\beta^r}^2 - \paren {\beta^{r + 1} }^2 - \ldots - \paren {\beta^{r + s} }^2$
where:
:$r, s \in \N : r + s = n$
Then $r$ is the maximum dimension among all subspaces on which the restriction of $q$ is positive definite.
Furthermore, $r$ and $s$ are independent of the choice of basis.

Proof:
{{ProofWanted}}
{{Namedfor|James Joseph Sylvester|cat = Sylvester}}
Theorem: Closed Graph Theorem/Banach Space

Let $\struct {X, \norm {\,\cdot\,}_X}$ and $\struct {Y, \norm {\,\cdot\,}_Y}$ be Banach spaces.
Let $T : X \to Y$ be a linear transformation.
Let $G_T \subseteq X \times Y$ be the graph of $T$. 
Suppose that: 
:$G_T$ is closed in the direct product $X \times Y$ equipped with the direct product norm $\norm \cdot_{X \times Y}$. 
Then: 
:$T$ is bounded.

Proof:
Let $\norm {\,\cdot\,}_{G_T}$ be the restriction of $\norm \cdot_{X \times Y}$ to $G_T$. 
By Closed Subspace of Banach Space forms Banach Space, $\struct {G_T, \norm \cdot_{G_T} }$ is a Banach space.
Define $\pi_X : G_T \to X$ by: 
:$\map {\pi_X} {x, y} = x$ for each $\tuple {x, y} \in X \times Y$. 
Note that by the definition of $G_T$, we have $\tuple {x, y} \in G_T$ {{iff}} $y = T x$.
So, we have that $\pi_X$ is a bijection with inverse $\pi_X^{-1} : X \to G_T$ with: 
:$\map {\pi_X^{-1} } x = \tuple {x, T x}$ for each $x \in X$. 
From Projections on Direct Product of Normed Vector Spaces define Bounded Linear Transformations, $\pi_X$ is a bounded linear transformation.
By the Banach Isomorphism Theorem: 
:since $\pi_X$ is a bijective bounded linear transformation between Banach spaces, its inverse $\pi_X^{-1}$ is a bounded linear transformation.
So there exists $C > 0$ such that: 
:$\norm {\pi_X^{-1} x}_{G_T} \le C \norm x_X$ 
for each $x \in X$.
That is: 
:$\norm {\tuple {x, T x} }_{X \times Y} \le C \norm x_X$ for each $x \in X$. 
From the definition of the direct product norm, we have: 
:$\max \set {\norm x_X, \norm {T x}_Y} \le C \norm x_X$ for each $x \in X$.
From the definition of the max operation, we have: 
:$\norm {T x}_Y \le \max \set {\norm x_X, \norm {T x}_Y}$
So:
:$\norm {T x}_Y \le C \norm x_X$ for all $x \in X$.
So:
:$T$ is bounded.
{{qed}}
Category:Linear Transformations on Banach Spaces
mfmqspe7lnceb625zxygfwu5uclkerv
Theorem: Hellinger-Toeplitz Theorem

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $T : \HH \to \HH$ be a Hermitian operator.
That is: 
:$\innerprod {T x} y = \innerprod x {T y}$ for each $x, y \in \HH$.
Then: 
:$T$ is bounded.

Proof:
Let $\struct {\HH \times \HH, \norm \cdot_{\HH \times \HH} }$ be the direct product of $\HH$ with itself, with the direct product norm.
From the Closed Graph Theorem, it suffices to show that: 
:$G_T = \set {\tuple {x, T x} \in \HH \times \HH : x \in \HH}$
is closed in $\struct {\HH \times \HH, \norm \cdot_{\HH \times \HH} }$.
Let $\sequence {\tuple {x_n, T x_n} }_{n \mathop \in \N}$ be a sequence in $G_T$ converging to $\tuple {x, y}$ in $\struct {\HH \times \HH, \norm \cdot_{\HH \times \HH} }$.
From Convergence in Direct Product Norm, we have: 
:$\sequence {x_n}_{n \mathop \in \N}$ converges to $x$ in $X$
and:
:$\sequence {y_n}_{n \mathop \in \N}$ converges to $y$ in $Y$. 
We show that $y = T x$.
Let $z \in \HH$. 
Then we have: 
:$\innerprod {T x_n} z = \innerprod {x_n} {T z}$
Taking $n \to \infty$, we have: 
:$\innerprod y z = \innerprod x {T z}$ 
from Inner Product is Continuous.
Since $T$ is Hermitian, we have: 
:$\innerprod x {T z} = \innerprod {T x} z$ 
Then we have: 
:$\innerprod y z = \innerprod {T x} z$ for all $z \in \HH$.
So from Inner Product is Sesquilinear, we have: 
:$\innerprod {y - T x} z = 0$ for all $z \in \HH$.
In particular: 
:$\innerprod {y - T x} {y - T x} = 0$
so: 
:$y = T x$
So $G_T$ is closed in $\struct {\HH \times \HH, \norm \cdot_{\HH \times \HH} }$.
So, by the Closed Graph Theorem, $T$ is bounded.
{{qed}}
Theorem: Hellinger-Toeplitz Theorem

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $T : \HH \to \HH$ be a Hermitian operator.
That is: 
:$\innerprod {T x} y = \innerprod x {T y}$ for each $x, y \in \HH$.
Then: 
:$T$ is bounded.

Proof:
{{AimForCont}} suppose that $T$ is not bounded.
Then there does not exist $C > 0$ such that: 
:$\norm {T x}_\HH \le C$ for each $x \in \HH$ with $\norm x_\HH = 1$.
That is, for each $n \in \N$ there exists $y_n \in \HH$ such that: 
:$\norm {T y_n}_\HH \ge n$ 
with $\norm {y_n}_\HH = 1$.
For each $n \in \N$, define the linear functional $f_n : \HH \to \HH$ by: 
:$\map {f_n} x = \innerprod x {T y_n}$
Then, from the Riesz Representation Theorem (Hilbert Spaces), we have that $f_n$ is a bounded linear functional with: 
:$\norm {f_n}_{\HH^\ast} = \norm {T y_n}_\HH \ge n$
So:
:$\ds \sup_n \norm {f_n}_{\HH^\ast} = \infty$ 
Then, from the Principle of Condensation of Singularities, there exists $x \in X$ such that: 
:$\ds \sup_n \cmod {\map {f_n} x} = \infty$
Since $T$ is Hermitian, we have: 
:$\map {f_n} x = \innerprod {T x} {y_n}$ for each $n \in \N$.
Then, we have:
{{begin-eqn}}
{{eqn	| l = \cmod {\map {f_n} x}
		| r = \cmod {\innerprod {T x} {y_n} }
}}
{{eqn	| o = \le
		| r = \norm {T x}_\HH \norm {y_n}_\HH
		| c = Cauchy-Bunyakovsky-Schwarz Inequality
}}
{{eqn	| r = \norm {T x}_\HH
}}
{{end-eqn}}
But then we have: 
:$\ds \sup_n \cmod {\map {f_n} x} \le \norm {T x}_\HH < \infty$
a contradiction.
So $T$ is bounded.
{{qed}}
Theorem: Characterisation of Terminal P-adic Expansion

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $x \in \Q_p$.
Then:
:the $p$-adic expansion of $x$ terminates 
{{iff}} 
:$\exists a \in \N : \exists k \in \Z : x = \dfrac a {p^k}$

Proof:
=== Necessary Condition ===
{{:Characterisation of Terminal P-adic Expansion/Necessary Condition}}{{qed|lemma}}
Theorem: Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual

Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\Bbb F$.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual of $\struct {X, \norm \cdot_X}$.
Let $\map L {X^\ast, \Bbb F}$ be the set of linear functionals on $X^\ast$. 
For each $x \in X$, define $x^\wedge : X^\ast \to \Bbb F$ by: 
:$\map {x^\wedge} f = \map f x$
Define $J : X \to \map L {X^\ast, \Bbb F}$ by: 
:$\map J x = x^\wedge$
for each $x \in X$.
Then: 
:$J$ is a linear transformation $X \to X^{\ast \ast}$. 
where $X^{\ast \ast}$ denotes the second normed dual.

Proof:
We first show that $J$ is a linear transformation.
Let $x, y \in X$ and $\alpha, \beta \in \Bbb F$.
Then, we have, for each $f \in X^\ast$:
{{begin-eqn}}
{{eqn	| l = \map {\paren {\map J {\alpha x + \beta y} } } f
		| r = \map {\paren {\alpha x + \beta y}^\wedge} f
}}
{{eqn	| r = \map f {\alpha x + \beta y}
}}
{{eqn	| r = \alpha \map f x + \beta \map f y
		| c = {{Defof|Linear Functional}}
}}
{{eqn	| r = \alpha \map {x^\wedge} f + \beta \map {y^\wedge} f
}}
{{eqn	| r = \alpha \map {\paren {\map J x} } f + \beta \map {\paren {\map J y} } f
}}
{{eqn	| r = \map {\paren {\alpha \map J x + \beta \map J y} } f
}}
{{end-eqn}}
So:
:$\map J {\alpha x + \beta y} = \alpha \map J x + \beta \map J y$
So $J$ is linear.
We now show that, for each $x \in X$, we have: 
:$\map J x = x^\wedge \in X^{\ast \ast}$
where $X^{\ast \ast}$ is the second normed dual.
We first show that $x^\wedge$ is a linear functional for each $x \in X$.
Let $x \in X$, $f, g \in X^\ast$ and $\alpha, \beta \in \Bbb F$.
Then: 
{{begin-eqn}}
{{eqn	| l = \map {x^\wedge} {\alpha f + \beta g}
		| r = \map {\paren {\alpha f + \beta g} } x
}}
{{eqn	| r = \alpha \map f x + \beta \map g x
}}
{{eqn	| r = \alpha \map {x^\wedge} f + \beta \map {x^\wedge} g
}}
{{end-eqn}}
so $x^\wedge$ is a linear functional for each $x \in X$.
We now show that $x^\wedge$ is bounded for each $x \in X$.
Let $x \in X$.
Then, for each $f \in X^\ast$, we have: 
{{begin-eqn}}
{{eqn	| l = \cmod {\map {x^\wedge} f}
		| r = \cmod {\map f x}
		| c = {{Defof|Evaluation Linear Transformation on Normed Vector Space}}
}}
{{eqn	| o = \le
		| r = \norm f_{X^\ast} \norm x_X
		| c = Fundamental Property of Norm on Bounded Linear Functional
}}
{{end-eqn}}
So: 
:$x^\wedge$ is a bounded linear functional.
That is:
:$\map J x \in X^{\ast \ast}$
{{qed}}
Theorem: Hensel's Lemma/P-adic Integers

Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\map F X \in \Z_p \sqbrk X$ be a polynomial.
Let $\map {F'} X$ be the (formal) derivative of $F$.
Let $p\Z_p$ denote the principal ideal of $\Z_p$ generated by $p$.
For all $x,y \in \Z_p$, let:
:$x \equiv y \pmod {p\Z_p}$
denote congruence modulo the principal ideal $p\Z_p$.
Let $\alpha_0 \in \Z_p$ be a $p$-adic integer:
{{begin-eqn}}
{{eqn | l = \map F {\alpha_0}
      | o = \equiv
      | r = 0
      | rr= \pmod {p\Z_p}
}}
{{eqn | l = \map {F'} {\alpha_0}
      | o = \not\equiv
      | r = 0
      | rr= \pmod {p\Z_p}
}}
{{end-eqn}}
Then there exists a unique $\alpha \in \Z_p$:
:$\alpha \equiv \alpha_0 \pmod {p\Z_p}$
:$\map F \alpha = 0$

Proof:
=== Lemma 1 ===
{{:Hensel's Lemma/P-adic Integers/Lemma 1}}{{qed|lemma}}
Let:
:$\ds \alpha = \sum_{n \mathop = 0}^\infty d_n p^n$
Theorem: Normed Vector Space is Reflexive iff Surjective Evaluation Linear Transformation

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 {{iff}}:
:$\iota$ is surjective.
That is: 
:$\iota X = X^{\ast \ast}$.

Proof:
From the definition of a reflexive space, we have that $X$ is reflexive {{iff}}:
:$\iota$ 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 {{iff}} $\iota$ is surjective.
That is: 
:$X$ is reflexive {{iff}} $\iota$ is surjective.
{{qed}}
Theorem: Image of Evaluation Linear Transformation on Banach Space is Closed

Let $\struct {X, \norm \cdot_X}$ be a Banach space.
Let $\struct {X^{\ast \ast}, \norm \cdot_{X^{\ast \ast} } }$ be the second normed dual of $\struct {X, \norm \cdot_X}$.
Let $J : X \to X^{\ast \ast}$ be the evaluation linear transformation.
Then: 
:$\map J X$ is closed in $X^{\ast \ast}$.

Proof:
Let $L$ be the limit of a convergent sequence in $\map J X$.
Let $\sequence {j_n}_{n \mathop \in \N}$ be a sequence in $\map J X$ such that: 
:$\sequence {j_n}_{n \mathop \in \N}$ converges to $L$.
Note that for each $n \in \N$ there exists $x_n \in X$ such that: 
:$j_n = J x_n$
From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, we have: 
:$J$ is a linear isometry.
Specifically:
:$J$ is a bounded linear transformation.
From Continuity of Linear Transformations, we have: 
:$J$ is continuous.
We show that $\sequence {x_n}_{n \mathop \in \N}$ converges to some $x \in X$.
We will then have:
:$L = J x \in \map J X$ 
from continuity.
Note that: 
{{begin-eqn}}
{{eqn	| l = \norm {j_n - j_m}_{X^{\ast \ast} }
		| r = \norm {J x_n - J x_m}_{X^{\ast \ast} }
}}
{{eqn	| r = \norm {\map J {x_n - x_m} }_{X^{\ast \ast} }
		| c = Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual
}}
{{eqn	| r = \norm {x_n - x_m}_X
		| c = Evaluation Linear Transformation on Normed Vector Space is Linear Isometry
}}
{{end-eqn}}
Let $\epsilon > 0$. 
Since: 
:$\sequence {j_n}_{n \mathop \in \N}$ is convergent
we have, from Convergent Sequence in Normed Vector Space is Cauchy Sequence:
:there exists $N \in \N$ such that for all $n, m \ge N$, we have $\norm {j_n - j_m}_X < \epsilon$.
So, for $n, m \ge N$, we have: 
:$\norm {x_n - x_m}_X < \epsilon$
Since $\epsilon$ was arbitrary, $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.
Since $X$ is a Banach space, we have: 
:$\sequence {x_n}_{n \mathop \in \N}$ converges to a limit $x \in X$.
Since $J$ is continuous, we obtain:
{{begin-eqn}}
{{eqn	| l = L
		| r = \lim_{n \mathop \to \infty} J x_n
}}
{{eqn	| r = \map J {\lim_{n \mathop \to \infty} x_n}
		| c = Continuous Mappings preserve Convergent Sequences
}}
{{eqn	| r = J x
}}
{{end-eqn}}
Since: 
:$J x \in \map J X$
we have: 
:$L \in \map J X$
Since $L$ was an arbitrary limit in $\map J X$, we have: 
:$\map J X$ contains each limit of a convergent sequence in $\map J X$
So, from Subset of Metric Space contains Limits of Sequences iff Closed:
:$\map J X$ is closed in $X^{\ast \ast}$.
{{qed}}
Theorem: Weakly Convergent Sequence in Normed Vector Space is Bounded

Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a weakly convergent sequence in $X$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is bounded.

Proof:
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual of $X$. 
Let $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$ be the second normed dual of $X$. 
Let $J : X \to X^{\ast \ast}$ be the evaluation linear transformation on $X$. 
Let: 
:$x^\wedge = \map J x$
for each $x \in X$. 
Consider the sequence $\sequence {x_n^\wedge}_{n \mathop \in \N}$ in $X^{\ast \ast}$. 
Then, for each $f \in X^\ast$, we have: 
:$\map {x_n^\wedge} f = \map f {x_n}$ 
Since $\sequence {x_n}_{n \mathop \in \N}$ converges weakly we have that: 
:$\sequence {\map f {x_n} }_{n \mathop \in \N}$ converges for each $f \in X^\ast$.
So, from Convergent Complex Sequence is Bounded, $\sequence {\map f {x_n} }_{n \mathop \in \N}$ is bounded for each $f \in X^\ast$. 
So for each $f \in X^\ast$ there exists a real number $M_f > 0$ such that: 
:$\cmod {\map f {x_n} } \le M_f$
for each $n \in \N$.
That is: 
:$\cmod {\map {x_n^\wedge} f} \le M_f < \infty$ 
so: 
:$\ds \sup_{n \mathop \in \N} \cmod {\map {x_n^\wedge} f}$ is finite for each $f \in X^\ast$. 
From Normed Dual Space is Banach Space, $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ is a Banach space, and so we may apply the Banach-Steinhaus Theorem.
From the Banach-Steinhaus Theorem, we then obtain: 
:$\ds \sup_{n \mathop \in \N} \norm {x_n^\wedge}_{X^{\ast \ast} }$ is finite. 
From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, we have: 
:$\norm {x_n^\wedge}_{X^{\ast \ast} } = \norm {x_n}_X$
So:
:$\ds \sup_{n \mathop \in \N} \norm {x_n}_X$ is finite.
So we may conclude that: 
:$\sequence {x_n}_{n \mathop \in \N}$ is bounded.
{{qed}}
Theorem: Equivalence of Definitions of Schauder Basis

Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.
Let $\set {e_n : n \in \N}$ be a countable subset of $X$. 
{{TFAE|def = Schauder Basis}}

Proof:
=== Definition 1 implies Definition 2 ===
Suppose that: 
:for each $x \in X$, there exists a unique sequence $\sequence {\map {\alpha_j} x}_{j \mathop \in \N}$ in $\Bbb F$ such that: 
::$\ds x = \sum_{j \mathop = 1}^\infty \map {\alpha_j} x e_j$
Clearly $\set {e_n : n \in \N}$ then satisfies $(1)$ of Definition 2.
Suppose the sequence $\sequence {\alpha_j}_{j \mathop \in \N}$ is such that: 
:$\ds 0 = \sum_{j \mathop = 1}^\infty \alpha_j e_j$
Note that: 
:$\ds 0 = \sum_{j \mathop = 1}^\infty 0 e_j$
By hypothesis, this expansion is unique, so we have: 
:$\alpha_j = 0$ 
for each $j \in \N$, as required. 
So $\set {e_n : n \in \N}$ also satisfies $(2)$ of Definition 2.
{{qed|lemma}}
Theorem: Resolvent Mapping Converges to 0 at Infinity

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 such that $\ds \lim_{z \mathop \to \infty} \norm {\map f z}_* = 0$.

Proof:
Pick $z \in \Bbb C$ such that $\size z > 2 \norm T_*$.
By Operator Norm is Norm:
:$\norm {\dfrac T z}_* = \dfrac {\norm T_*} {\size z} < \dfrac 1 2$
Hence:
{{begin-eqn}}
{{eqn | l = \norm {\map f z}_*
      | r = \norm {\paren {T - z I}^{-1} }_*
}}
{{eqn | r = \norm {\paren {z \paren {\dfrac T z - I} }^{-1} }_*
}}
{{eqn | r = \norm {\frac{-1} z \paren {I - \dfrac T z}^{-1} }_*
      | c = Norm of Inverse of Constant Times Operator
}}
{{eqn | r = \frac 1 {\size z} \norm {\paren {I - \dfrac T z}^{-1} }_*
      | c = Operator Norm is Norm
}}
{{eqn | r = \frac 1 {\size z} \norm {I + \dfrac T z + \paren {\dfrac T z}^2 + \cdots}_*
      | c = Invertibility of Identity Minus Operator since $\norm {\dfrac T z} < \dfrac 1 2 < 1$
}}
{{eqn | o = \le
      | r = \frac 1 {\size z} \sum_{k \mathop \ge 0} \norm {\paren {\dfrac T z}^k }_*
      | c = Triangle Inequality
}}
{{eqn | o = \le
      | r = \frac 1 {\size z} \sum_{k \mathop \ge 0} \paren {\dfrac {\norm T_*} {\size z} }^k
      | c = Operator Norm on Banach Space is Submultiplicative
}}
{{eqn | o = \le
      | r = \frac 1 {\size z} \sum_{k \mathop \ge 0} \paren {\dfrac 1 2}^k
      | c = since $\norm {\dfrac T z}_* < \dfrac 1 2$
}}
{{eqn | o = \le
      | r = \frac 2 {\size z}
}}
{{end-eqn}}
Taking limits of both sides as $\size z \to \infty$:
:$\norm {\map f z}_* \to 0$
{{qed}}
Category:Functional Analysis
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