avewright/proofwiki-math huggingface dataset:
Theorem: Number of form 28000...0007 is Divisible by 7

Let $x$ be a natural number in the form:
:$\sqbrk {28 \underbrace {000 \cdots 0}_n 7}_{10}$ 
Then $x$ is divisible by $7$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \sqbrk {28 \underbrace {000 \cdots 0}_n 7}_{10}
      | r = 7 \times \sqbrk {4 \underbrace {000 \cdots 0}_n 1}_{10}
      | c = 
}}
{{end-eqn}}
and the result follows by definition of divisibility.
{{qed}}
Theorem: Derivative of Hyperbolic Tangent/Corollary

:$\map {\dfrac \d {\d x} } {\tanh x} = 1 - \tanh^2 x$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\tanh x}
      | r = \sech^2 x
      | c = Derivative of Hyperbolic Tangent
}}
{{eqn | r = 1 - \tanh^2 x
      | c = Sum of Squares of Hyperbolic Secant and Tangent
}}
{{end-eqn}}
{{qed}}
Theorem: Derivative of Hyperbolic Cotangent/Corollary

:$\map {\dfrac \d {\d x} } {\coth x} = 1 - \coth^2 x$

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac \d {\d x} \coth x
      | r = -\csch^2 x
      | c = Derivative of Hyperbolic Cotangent
}}
{{eqn | r = 1 - \coth^2 x
      | c = Difference of Squares of Hyperbolic Cotangent and Cosecant
}}
{{end-eqn}}
{{qed}}
Theorem: Interior of Proper Subspace of Normed Vector Space is Empty

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$. 
Let $U$ be a proper vector subspace of $X$. 
Let $U^\circ$ be the interior of $U$. 
Then $U^\circ = \O$.

Proof:
{{AimForCont}} $U^\circ \ne \O$.
Take $x \in U^\circ$.
Then there exists $\epsilon > 0$ such that $\map {B_\epsilon} x \subseteq U$, where $\map {B_\epsilon} x$ is the open ball centered at $x$ with radius $\epsilon$.
Since $U$ is a vector subspace, we have:
:$\map {B_\epsilon} 0 = \map {B_\epsilon} x - x \subseteq U$
So if $x \in X$ has $\norm x < \epsilon$, then $x \in U$.
Take $x \in U \setminus \set {\mathbf 0_X}$. 
Then, we have: 
:$\norm {\dfrac \epsilon {2 \norm x} x} = \dfrac \epsilon 2 < \epsilon$
so that: 
:$\dfrac \epsilon {2 \norm x} x \in U$
Since $U$ is a vector subspace, we have $x \in U$. 
We also have:
:$\mathbf 0_X \in U$
so:
:$U = X$
This contradicts the hypothesis that $U \ne X$.
So by Proof by Contradition we have:
:$U^\circ = \O$
{{qed}}
Category:Normed Vector Spaces
Category:Set Interiors
124prybd8p9bn81okju3ml03a2f5ajd
Theorem: Weak Topology on Infinite Dimensional Normed Vector Space is not Metrizable

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be an infinite dimensional normed vector space over $\GF$.
Let $w$ be the weak topology on $X$.
Then $w$ is not metrizable.

Proof:
Let $X^\ast$ be the normed dual space of $X$. 
From Metric Space is First-Countable, it suffices to show that $\struct {X, w}$ is not first-countable.
{{AimForCont}} that $\struct {X, w}$ is first-countable.
Let $\sequence {U_n}_{n \mathop \in \N}$ be a local basis for $\mathbf 0_X$.
From Open Sets in Weak Topology of Topological Vector Space, for each $n \in \N$ we can pick $\epsilon_n > 0$ and a finite set $F_n \subseteq X$ such that: 
:$V_n = \set {x \in X : \cmod {\map f x} < \epsilon_n \text { for all } f \in F_n} \subseteq U_n$
Then $\sequence {V_n}_{n \mathop \in \N}$ is still a local basis for $\mathbf 0_X$.
Let $g \in X^\ast$ and consider: 
:$U = \set {x \in X : \cmod {\map g x} < 1}$ 
Then there exists $\map n g$ such that: 
:$U_{\map n g} \subseteq U$ 
Set $n = \map n g$. 
Let: 
:$\ds x \in \bigcap_{f \mathop \in F_n} \ker f$
Then $\map f x = 0$ for each $f \in F_n$. 
Then for each $\lambda \in \GF$ and $f \in F_n$ we have $\map f {\lambda x} = 0$.
So we have $\lambda x \in U$ for each $\lambda \in \GF$.
That is, $\cmod {\map g {\lambda x} } < 1$ for each $\lambda \in \GF$.
So for each $\lambda > 0$ we have: 
:$\ds \cmod {\map g x} < \frac 1 \lambda$ 
So we have $\map g x = 0$.
We conclude that: 
:$\ds \bigcap_{f \mathop \in F_n} \ker f \subseteq \ker g$
From Condition for Linear Dependence of Linear Functionals in terms of Kernel, we then have: 
:$g \in \map \span {F_n} = \map \span {F_{\map n g} }$
Since $g \in X^\ast$ was arbitrary, we have that: 
:$\ds X^\ast = \bigcup_{n \mathop = 1}^\infty \map \span {F_n}$
Then: 
:$\ds \bigcup_{n \mathop = 1}^\infty F_n$ is a generator for $X$. 
We have that $X$ has infinite dimension.
Hence, from Normed Dual Space of Infinite-Dimensional Normed Vector Space is Infinite-Dimensional, $X^\ast$ also has infinite dimension. 
Hence: 
:$\ds \bigcup_{n \mathop = 1}^\infty F_n$ is infinite.
From Countable Union of Finite Sets is Countable, it follows that: 
:$\ds \bigcup_{n \mathop = 1}^\infty F_n$ is countable.
From Generator of Vector Space Contains Basis, it follows that $X^\ast$ either has countable or finite dimension.
We have already established that it does not have finite dimension, hence it must have countable dimension.
From Normed Dual Space is Banach Space, $X^\ast$ is a Banach space of countable dimension.
This contradicts Infinite-Dimensional Banach Space has Uncountable Dimension. 
So $\struct {X, w}$ is not first-countable and hence is not metrizable.
{{qed}}
Category:Weak Topologies on Topological Vector Spaces
Category:Metrizable Spaces
l3knfsouhb080ighqcuyrithqr0l6dc
Theorem: Open Sets in Weak Topology of Topological Vector Space

Let $\GF \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\GF$ with weak topology $w$. 
Let $U \subseteq X$. 
Then $U$ is open in $\struct {X, w}$ {{iff}} for each $x \in U$ there exists $f_1, f_2, \ldots, f_n \in X^\ast$ and $\epsilon > 0$ such that: 
:$\set {y \in X : \cmod {\map {f_i} {y - x} } < \epsilon \text { for each } 1 \le i \le n} \subseteq U$

Proof:
Let $X^\ast$ be the topological dual space of $X$. 
By the definition of the weak topology, $w$ is the initial topology on $X$ with respect to $X^\ast$.
For each $f \in X^\ast$, define $p_f : X \to \hointr 0 \infty$ by: 
:$\map {p_f} x = \cmod {\map f x}$
for each $x \in X$, and set: 
:$\PP = \set {p_f : f \in X^\ast}$ 
Then from Initial Topology on Vector Space Generated by Linear Functionals is Locally Convex, $w$ is the standard topology on the locally convex space $\struct {X, \PP}$.
From Open Sets in Standard Topology of Locally Convex Space, we obtain that $U$ is open in $\struct {X, w}$ {{iff}} for each $x \in U$ there exists $f_1, f_2, \ldots, f_n \in X^\ast$ and $\epsilon > 0$ such that: 
:$\set {y \in X : \map {p_{f_i} } {y - x} < \epsilon \text { for each } 1 \le i \le n} = \set {y \in X : \cmod {\map {f_i} {y - x} } < \epsilon \text { for each } 1 \le i \le n} \subseteq U$
{{qed}}
Category:Weak Topologies on Topological Vector Spaces
6a0ggrovhzrh4gsakvjhj383ehi4iif
Theorem: 41 is Smallest Number whose Period of Reciprocal is 5

$41$ is the first positive integer the decimal expansion of whose reciprocal has a period of $5$:
:$\dfrac 1 {41} = 0 \cdotp \dot 0243 \dot 9$

Proof:
From Reciprocal of $41$:
{{:Reciprocal of 41}}
Counting the digits, it is seen that this has a period of recurrence of $5$.
It remains to be shown that $41$ is the smallest positive integer which has this property.
{{ProofWanted}}
Category:41
Category:Examples of Reciprocals
thl87a0mi1vnburqpl26wstyuqqgoy2
Theorem: Evaluation Mapping is Injective iff Mappings Separate Points

Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
Let $f : X \to Y$ be the evaluation mapping induced by $\family{f_i}_{i \mathop \in I}$.
Then:
:$f$ is an injection {{iff}} $\family {f_i : X \to Y_i}_{i \mathop \in I}$ separates points

Proof:
We have:
:$f$ is an injection
{{begin-eqn}}
{{eqn | lll = \iff
      | ll = \forall x, y \in X : x \ne y :
      | l = \map f x
      | o = \ne
      | r = \map f y
      | c = {{Defof|Injection}}
}}
{{eqn | lll = \iff
      | ll = \forall x, y \in X : x \ne y :
      | l = \family{ \map {f_i} x }_{i \in I}
      | o = \ne
      | r = \family{ \map {f_i} y }_{i \in I}
      | c = {{Defof|Evaluation Mapping (Topology)|Evaluation Mapping}}
}}
{{eqn | lll = \iff
      | ll = \forall x, y \in X : x \ne y : \exists i \in I : 
      | l = \map {f_i} x
      | o = \ne
      | r = \map {f_i} y
      | c = {{Defof|Cartesian Product of Family}}
}}
{{eqn | lll = \iff
      | ll = \family {f_i : X \to Y_i}_{i \mathop \in I} \text{ separates points}
      | o = 
      | c = {{Defof|Mappings Separating Points}}
}}
{{end-eqn}}
{{qed}}
Category:Evaluation Mappings (Topological Spaces)
rleb9qwpjd4n61o37ua2gn1f60zwxte
Theorem: Injection is Open Mapping iff Image of Sub-Basis Set is Open

Let $\struct {X_1, \tau_1}$ and $\struct {X_2, \tau_2}$ be topological spaces.
Let $\SS \subseteq \powerset {X_1}$ be a sub-basis of $\tau_1$.
Let $f: X_1 \to X_2$ be an injection. 
Then:
:$f$ is an open mapping 
{{iff}}:
:$\forall U \in \SS: f \sqbrk U \in \tau_2$

Proof:
=== Necessary Condition ===
Let $f$ be an open mapping.
By definition of open mapping:
:$\forall U \in \tau_1 : f \sqbrk U \in \tau_2$
By definition of sub-basis:
:$\SS \subseteq \tau_1$
Hence:
:$\forall U \in \SS: f \sqbrk U \in \tau_2$
{{qed|lemma}}
Theorem: Reciprocal of 11

:$\dfrac 1 {11} = 0 \cdotp \dot 0 \dot 9$

Proof:
Performing the calculation using long division:
<pre>
   0.0909...
  ------------
11)1.000000000
     99
   ----
      100
       99
      ---
      ...
</pre>
{{qed}}
Category:11
Category:Examples of Reciprocals
n9194mcn5vl7elqfpop63niwmfsm61o
Theorem: Reciprocal of 3

:$\dfrac 1 3 = 0 \cdotp \dot 3$

Proof:
Performing the calculation using long division:
<pre>
   0.33...
  --------
 3)1.00...
     9
   ---
     10
      9
     --
     ..
</pre>
{{qed}}
Category:3
Category:Examples of Reciprocals
ronv291wvqp44e48ldhjhsw2jd2zygn
Theorem: Period of Reciprocal of 43 is Odd

The decimal expansion of the reciprocal of $43$ has an odd period, that is, $21$:
:$\dfrac 1 {43} = 0 \cdotp \dot 02325 \, 58139 \, 53488 \, 37209 \, \dot 3$

Proof:
From Reciprocal of $43$:
{{:Reciprocal of 43}}
Counting the digits, it is seen that this has a period of recurrence of $21$.
Hence the result.
{{qed}}
Category:43
Category:Examples of Reciprocals
snxig1txup7n228s8gghkmlpcy7pg1l
Theorem: Long Period Prime/Examples/29

The prime number $29$ is a long period prime:
:$\dfrac 1 {29} = 0 \cdotp \dot 03448 \, 27586 \, 20689 \, 65517 \, 24137 \, 93 \dot 1$

Proof:
From Reciprocal of $29$:
{{:Reciprocal of 29}}
Counting the digits, it is seen that this has a period of recurrence of $28$.
Hence the result.
{{qed}}
Category:29
Category:Examples of Long Period Primes
cktz7jtmqcqpqhxb3a1g80a6fup7hpb
Theorem: Period of Reciprocal of 79 is One Sixth of Maximal

The decimal expansion of the reciprocal of $79$ has $\dfrac 1 6$ the maximum period, that is: $13$:
:$\dfrac 1 {79} = 0 \cdotp \dot 01265 \, 82278 \, 48 \dot 1$

Proof:
From Reciprocal of $79$:
{{:Reciprocal of 79}}
Counting the digits, it is seen that this has a period of recurrence of $13$.
From Maximum Period of Reciprocal of Prime, the maximum period of recurrence of $\dfrac 1 p$ is $p - 1$.
We have that:
:$13 = \dfrac {79 - 1} 6$
{{qed}}
Category:79
Category:Examples of Reciprocals
4k6yi12bd2ucl5gsx2ppen376soq4j1
Theorem: 101 is Smallest Number whose Period of Reciprocal is 4

$101$ is the first positive integer the decimal expansion of whose reciprocal has a period of $4$:
:$\dfrac 1 {101} = 0 \cdotp \dot 009 \dot 9$

Proof:
{{tidy}}
Let the positive integer reciprocal be $\dfrac1 k$ for some $k \in \Z_{\ge 0}$.
For it to have a period of recurrence of $4$ in base $10$, it must be able to be expressed as $\dfrac a {10^4-1}$ for some $a \in \Z_{\ge 0}$. 
{{begin-eqn}}
{{eqn | l = 10^4-1
      | r = \paren { 10^2 - 1 } \paren { 10^2 + 1 }
      | c = Difference of Two Squares
}}
{{eqn | r = 99 \cdot 101
}}
{{end-eqn}}
Since this rational number must have a period of recurrence ''minimum'' of $4$, it cannot also have a period of $2$ or $1$. It will be sufficient for this proof to only consider periods of $2$. This rational number must ''not'' be able to be expressed as $\dfrac b {10^2-1}$ for any $b \in \Z_{\ge 0}$.
{{begin-eqn}}
{{eqn | l = \dfrac1 k
      | r = \dfrac a {99 \cdot 101}
}}
{{eqn | ll= \leadsto
      | l = a k
      | r = 99 \cdot 101
}}
{{end-eqn}}
----
$\dfrac1 k \neq \dfrac b {10^2-1}$
$\leadsto \dfrac a {99 \cdot 101} \neq \dfrac b {99}$
$\leadsto a \neq 101 b$
Since $101$ is prime, $a$ is not a multiple of $101$ but $ak$ is, then $k$ must be a multiple of $101$. $k = 101$ is the first instance which satisfies these conditions and produces a reciprocal with a period of recurrence of 4.
{{qed}}
Category:101
Category:Examples of Reciprocals
ozcnrf32yjuee2ao3kc3il6g0r8zuk5
Theorem: Reciprocal of 83 has Prime Period

The decimal expansion of the reciprocal of $83$ has a prime period, that is $41$:
:$\dfrac 1 {83} = 0 \cdotp \dot 01204 \, 81927 \, 71084 \, 33734 \, 93975 \, 90361 \, 44578 \, 31325 \, \dot 3$

Proof:
From Reciprocal of $83$:
:$\dfrac 1 {83} = 0 \cdotp \dot 01204 \, 81927 \, 71084 \, 33734 \, 93975 \, 90361 \, 44578 \, 31325 \, \dot 3$
Counting the digits, it is seen that this has a period of recurrence of $41$.
Indeed, $41$ is the $13$th prime number.
{{qed}}
Category:83
Category:Examples of Reciprocals
iqso3aznit64uij86pa40cev8w4dks3
Theorem: Repunit 19 is Unique Period Prime with Period 19

The repunit prime $R_{19}$ is a unique period prime whose reciprocal has a period of $19$:
:$\dfrac 1 {1 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111} = 0 \cdotp \dot 00000 \, 00000 \, 00000 \, 000 \dot 9$

Proof:
The reciprocal of a repunit $R_n$ is of the form:
:$\dfrac 1 {R_n} = 0 \cdotp \underbrace{\dot 000 \ldots 000}_{n - 1} \dot 9$
{{TheoremWanted|Prove the above}}
Thus $\dfrac 1 {R_{19} }$ has a period of $19$.
From Period of Reciprocal of Prime, for prime numbers such that:
:$p \nmid 10$
we have that the period of such a prime is the order of $10$ modulo $p$.
That is, the smallest integer $d$ such that:
:$10^d \equiv 1 \pmod p$
The only other possible primes $p$ whose reciprocals might have a period of $19$ must also satisfy:
:$10^{19} \equiv 1 \pmod p$
that is:
:$p \divides \paren {10^{19} - 1} = 9 \times R_{19}$
Therefore the only other possible prime whose reciprocal might have a period of $19$ is $3$.
Trivially:
:$\dfrac 1 3 = 0 \cdotp \dot 3$
which has a period of $1$.
Hence the result.
{{qed}}
Category:1,111,111,111,111,111,111
Category:Examples of Reciprocals
Category:Examples of Unique Period Primes
0mezxt6zxq6kuvztlgyes2tr1mgukxi
Theorem: 73 is Smallest Number whose Period of Reciprocal is 8

$73$ is the first positive integer the decimal expansion of whose reciprocal has a period of $8$:
:$\dfrac 1 {73} = 0 \cdotp \dot 01369 \, 86 \dot 3$

Proof:
From Reciprocal of $73$:
{{:Reciprocal of 73}}
Counting the digits, it is seen that this has a period of recurrence of $8$.
It remains to be shown that $73$ is the smallest positive integer which has this property.
{{ProofWanted}}
Category:73
Category:Examples of Reciprocals
grxidb2h35auwbwuvxfidb9pl7w2q0v
Theorem: Reciprocal of 9801

:$\dfrac 1 {9801} = 0 \cdotp \dot 00010 \, 20304 \, 05060 \, 70809 \, 10111 \, 21314 \, ... \, 94959 \, 6979 \dot 9$

Proof:
{{ProofWanted}}
Category:9801
Category:Examples of Reciprocals
bpiqtpid8jh6p42qhqmr66jfo2afpjn
Theorem: Reciprocal of 909,091

:$\dfrac 1 {909 \, 091} = 0 \cdotp \dot 00000 \, 10999 \, 998 \dot 9$

Proof:
Performing the calculation using long division:
<pre>
       0.00000109999989000001...
      --------------------------------------------
909091)1.00000000000000000000...
         909091
         ------
          9090900
          8181819
          -------
           9090810
           8181819
           -------
            9089910
            8181819
            -------
             9080910
             8181819
             -------
              8990910
              8181819
              -------
               8090910
               7272728
               -------
                8181820
                8181819
                -------
                      1000000
                       909091
                      -------
                       ......
</pre>{{qed}}
Category:909,091
Category:Examples of Reciprocals
fnesz75dfmi63hncjb82u7wzf2udf1e
Theorem: Long Period Prime/Examples/47

The prime number $47$ is a long period prime:
:$\dfrac 1 {47} = 0 \cdotp \dot 02127 \, 65957 \, 44680 \, 85106 \, 38297 \, 87234 \, 04255 \, 31914 \, 89361 \, \dot 7$

Proof:
From Reciprocal of $47$:
{{:Reciprocal of 47}}
Counting the digits, it is seen that this has a period of recurrence of $46$.
Hence the result.
{{qed}}
Category:47
Category:Examples of Long Period Primes
jwxg7b83jx80qin8yzabynjlv90shp1
Theorem: Composite of Evaluation Mapping and Projection

Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
Let $f : X \to Y$ be the evaluation mapping induced by $\family{f_i}_{i \mathop \in I}$.
Let:
:$pr_i : Y \to Y_i$ denote the $i$th projection on $Y$
Then:
:$\forall i \in I: pr_i \circ f = f_i$

Proof:
By definition of projection:
{{begin-eqn}}
{{eqn | q = \forall x \in X, i \in I
      | l = \map {\paren{pr_i \circ f} } x
      | r = \map {pr_i} {\map f x}
      | c = {{Defof|Composite Mapping}}
}}
{{eqn | r = \map {pr_i} {\family{\map {f_i} x} }
      | c = {{Defof|Evaluation Mapping}}
}}
{{eqn | r = \map {f_i} x
      | c = {{Defof|Projection}}
}}
{{end-eqn}}
From Equality of Mappings:
:$\forall i \in I : pr_i \circ f = f_i$
{{qed}}
Category:Evaluation Mappings (Topological Spaces)
p3j2lkth0cjbyugcyu3jwwyeo2e9yk2
Theorem: Space of Bounded Linear Transformations forms Vector Space

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$. 
Let $\map B {X, Y}$ be the space of bounded linear transformations from $X$ to $Y$.
Let $+_B$ and $\circ_B$ be pointwise addition and pointwise scalar multiplication on $Y^X$.  
Then $\struct {\map B {X, Y}, +_B, \circ_B}_\GF$ is a vector space.

Proof:
Let $\map L {X, Y}$ be the space of linear transformations between $X$ and $Y$.
From Linear Mappings between Vector Spaces form Vector Space, $\map L {X, Y}$ is a vector space over $\GF$ with pointwise addition and pointwise scalar multiplication.
It therefore suffices to show that $\map B {X, Y}$ is a vector subspace of $\map L {X, Y}$. 
From One-Step Vector Subspace Test, it suffices to show that $\map B {X, Y} \ne \O$ and: 
:$T + \lambda S \in \map B {X, Y}$
for each $\lambda \in \GF$ and $T, S \in \map B {X, Y}$.
First, we have: 
:$\norm {\mathbf 0_{\map L {X, Y} } x}_Y = 0 \le \norm x_X$
So $\mathbf 0_{\map L {X, Y} } \in \map B {X, Y}$.
Now let $T, S \in \map B {X, Y}$.
Then there exists $M, M' > 0$ such that: 
:$\norm {T x}_Y \le M \norm x_X$ for each $x \in X$
and:
:$\norm {S x}_Y \le M' \norm x_X$ for each $x \in X$.
Then for each $\lambda \in \GF$ and $x \in X$ we have: 
{{begin-eqn}}
{{eqn	| l = \norm {\paren {T + \lambda S}x}_Y
		| r = \norm {T x + \lambda S x}_Y
}}
{{eqn	| o = \le
		| r = \norm {T x}_Y + \norm {\lambda S x}_Y
		| c = {{NormAxiomVector|3}}
}}
{{eqn	| r = \norm {T x}_Y + \cmod \lambda \norm {S x}_Y
		| c = {{NormAxiomVector|2}}
}}
{{eqn	| o = \le
		| r = M \norm x_X + \cmod \lambda M' \norm x_X
}}
{{end-eqn}}
So $T + \lambda S \in \map B {X, Y}$.
{{qed}}
Category:Space of Bounded Linear Transformations
pde1f2hu4kd8gyf81ii8s8dpvganu4r
Theorem: Norm on Space of Bounded Linear Transformations is Norm

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$. 
Let $\map B {X, Y}$ be the space of bounded linear transformations between $X$ and $Y$. 
Let $\norm {\, \cdot \,}_{\map B {X, Y} }$ be the norm on the space of bounded linear transformations.
Then $\norm {\, \cdot \,}_{\map B {X, Y} }$ is indeed a norm on $\map B {X, Y}$.

Proof:
From Norm on Bounded Linear Transformation is Finite, $\norm {\, \cdot \,}_{\map B {X, Y} }$ is real-valued.
Theorem: Reciprocal of 47

:$\dfrac 1 {47} = 0 \cdotp \dot 02127 \, 65957 \, 44680 \, 85106 \, 38297 \, 87234 \, 04255 \, 31914 \, 89361 \, \dot 7$

Proof:
Performing the calculation using long division:
<pre>
   0.021276595744680851063829787234042553191489361702...
  ------------------------------------------------------
47)1.000000000000000000000000000000000000000000000000...
     94            
   ----    
      60   270    400    140   160    150   440
      47   235    376     94   141    141   423
      --   ---    ---    ---   ---    ---   ---
      130   350    240    460   190     90   170
       94   329    235    423   188     47   141
      ---   ---    ---    ---   ---     --   ---
       360   210     50    370    200   430   290
       329   188     47    329    188   423   282
       ---   ---     --    ---    ---   ---   ---
        310   220     300   410    120    70    80
        282   188     282   376     94    47    47
        ---   ---     ---   ---    ---    --    --
         280   320     180   340    260   230   330
         235   282     141   329    235   188   329
         ---   ---     ---   ---    ---   ---   ---
          450   380     390   110    250   420    100
          423   376     376    94    235   376     94
          ---   ---     ---   ---    ---   ---    ---
           270    40     140   160    150   440   ... 
</pre>
{{qed}}
Category:47
Category:Examples of Reciprocals
cc91wxe313hyxdhnw280s1wqahb0qcr
Theorem: Normed Dual Space is Banach Space

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.
{{qed}}
Category:Normed Dual Spaces
Category:Banach Spaces
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Theorem: Bound on Norm of Power of Element in Normed Algebra

Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$. 
Let $x \in A$ and $n \in \N$. 
Then: 
:$\norm {x^n} \le \norm x^n$

Proof:
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\norm {x^n} \le \norm x^n$
Theorem: Product Rule for Sequence in Normed Algebra

Let $\GF \in \set {\R, \C}$. 
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$. 
Let $\sequence {a_n}_{n \in \N}$ and $\sequence {b_n}_{n \in \N}$ be sequences in $A$ converging to $a$ and $b$ respectively. 
Then: 
:$a_n b_n \to a b$

Proof:
From Convergent Sequence in Normed Vector Space is Bounded, there exists $M > 0$ such that: 
:$\norm {a_n} \le M$ for each $n \in \N$.
We have for $n \in \N$: 
{{begin-eqn}}
{{eqn	| l = \norm {a_n b_n - a b}
		| r = \norm {a_n b_n - a_n b + a_n b - a b}
}}
{{eqn	| r = \norm {a_n \paren {b_n - b} + b \paren {a_n - a} }
}}
{{eqn	| o = \le
		| r = \norm {a_n \paren {b_n - b} } + \norm {b \paren {a_n - a} }
		| c = {{NormAxiomVector|3}}
}}
{{eqn	| o = \le
		| r = \norm {a_n} \norm {b_n - b} + \norm b \norm {a_n - a}
}}
{{eqn	| o = \le
		| r = M \norm {b_n - b} + \norm b \norm {a_n - a}
}}
{{eqn	| o = \to
		| r = 0
}}
{{end-eqn}}
So from Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we have: 
:$a_n b_n \to a b$ 
{{qed}}
Category:Normed Algebras
Category:Convergent Sequences (Normed Vector Spaces)
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