avewright/proofwiki-math huggingface dataset:
Theorem: Long Period Prime/Examples/97

The prime number $97$ is a long period prime:
:$\dfrac 1 {97} = 0 \cdotp \dot 01030 \, 92783 \, 50515 \, 46391 \, 75257 \, 73195 \, 87628 \, 86597 \, 93814 \, 43298 \, 96907 \, 21649 \, 48453 \, 60824 \, 74226 \, 80412 \, 37113 \, 40206 \, 18556 \, \dot 7$

Proof:
From Reciprocal of $97$:
{{:Reciprocal of 97}}
Counting the digits, it is seen that this has a period of recurrence of $96$.
Hence the result.
{{qed}}
Theorem: Long Period Prime/Examples/131

The prime number $131$ is a long period prime:
:$\dfrac 1 {131} = 0 \cdotp \dot 00763 \, 35877 \, 86259 \, 54198 \, 47328 \, 24427 \, 48091 \, 60305 \, 34351 \, 14503 \, 81679 \, 38931 \, 29770 \, 99236 \, 64122 \, 13740 \, 45801 \, 52671 \, 75572 \, 51908 \, 39694 \, 65648 \, 85496 \, 18320 \, 61068 \, 7022 \dot 9$
It also contains an equal number ($13$) of each of the digits from $0$ to $9$.

Proof:
From Reciprocal of $131$:
{{:Reciprocal of 131}}
Counting the digits, it is seen that this has a period of recurrence of $130$.
Inspecting the expansion and counting the digits, we find that each one appears exactly $13$ times.
Hence the result.
{{qed}}
Theorem: String is Substring of Itself

Let $S$ be a string.
Then $S$ is a substring of itself.

Proof:
By definition, a string $T$ is a substring of $S$ in $\AA$ {{iff}}:
:$S = S_1 T S_2$
where:
:$S_1$ and $S_2$ are strings in $\AA$ (possibly null)
:$S_1 T S_2$ is the concatenation of $S_1$, $T$ and $S_2$.
Let $S_1$ and $S_2$ both be the null string.
Then it follows that:
:$S = T$
Hence the result.
{{qed}}
Category:Substrings
78b4zb6ufidl238ias1dt3afat6kjn5
Theorem: Restriction of Mapping is Mapping/General Result

Let $f: S \to T$ be a mapping.
Let $X \subseteq S$.
Let $f \sqbrk X \subseteq Y \subseteq T$.
Let $f \restriction_{X \times Y}$ be the restriction of $f$ to $X \times Y$.
Then $f \restriction_{X \times Y}: X \to Y$ is a mapping:
: whose domain is $X$
: whose preimage is $X$
: whose codomain is $Y$.

Proof:
=== $f \restriction_{X \times Y}$ is Many-to-one ===
We have:
{{begin-eqn}}
{{eqn | q = \forall x \in X
      | l = \tuple {x, y_1}, \tuple {x, y_2} \in f \restriction_{X \times Y}
      | o = \leadsto
      | r = \tuple {x, y_1}, \tuple {x, y_2} \in f \cap \paren{X \times Y}
      | c = {{Defof|Restriction of Mapping}}
}}
{{eqn | o = \leadsto
      | r = \tuple {x, y_1}, \tuple {x, y_2} \in f
      | c = Intersection is Subset 
}}
{{eqn | o = \leadsto
      | r = y_1 = y_2
      | c = {{Defof|Mapping}} 
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Preimage of Open Sets forms Basis if Continuous Mappings Separate Points from Closed Sets

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 $\family {f_i}_{i \mathop \in I}$ separate points from closed sets.
Let $\BB = \set{f_i^{-1} \sqbrk V : i \in I, V \text{ is open in } Y_i}$.
Then:
:$\BB$ is a basis for $X$

Proof:
By definition of continuous:
:$\forall B \in \BB : B$ is open in $X$
Let $U \subseteq X$ be open in $X$.
Let $x \in U$.
By definition of closed subset:
:$X \setminus U$ is closed in $X$
By definition of mappings separating points from closed sets:
:$\exists i \in I : \map {f_i} x \notin f_i \sqbrk {X \setminus U}^-$
Hence:
:$\map {f_i} x \in Y_i \setminus f_i \sqbrk {X \setminus U}^-$
By definition of preimage of mapping:
:$x \in f_i^{-1} \sqbrk {Y_i \setminus f_i \sqbrk {X \setminus U}^-}$
From Topological Closure is Closed:
:$f_i \sqbrk {X \setminus U}^-$ is closed in $Y_i$
By definition of closed subset:
:$Y_i \setminus f_i \sqbrk {X \setminus U}^-$ is open in $Y_i$
Hence:
:$f_i^{-1} \sqbrk {Y_i \setminus f_i \sqbrk {X \setminus U}^-} \in \BB$
We have:
{{begin-eqn}}
{{eqn | l = y  
      | o = \in
      | r = f_i^{-1} \sqbrk {Y_i \setminus f_i \sqbrk {X \setminus U}^-}
}}
{{eqn | ll = \leadsto
      | l = \map {f_i} y
      | o = \in
      | r = Y_i \setminus f_i \sqbrk {X \setminus U}^-
      | c = {{Defof|Preimage of Mapping}}
}}
{{eqn | ll = \leadsto
      | l = \map {f_i} y
      | o = \in
      | r = Y_i \setminus f_i \sqbrk {X \setminus U}
      | c = Set is Subset of its Topological Closure and Set Difference with Subset is Superset of Set Difference
}}
{{eqn | ll = \leadsto
      | l = \map {f_i} y
      | o = \notin
      | r = f_i \sqbrk {X \setminus U}
      | c = {{Defof|Set Difference}}
}}
{{eqn | ll = \leadsto
      | l = y
      | o = \notin
      | r = X \setminus U
      | c = {{Defof|Image of Mapping}}
}}
{{eqn | ll = \leadsto
      | l = y
      | o = \in
      | r = U
      | c = {{Defof|Set Difference}}
}}
{{end-eqn}}
By definition of subset:
:$f_i^{-1} \sqbrk {Y_i \setminus f_i \sqbrk {X \setminus U}^-} \subseteq U$
We have shown that:
:$\forall$ open subsets $U \subseteq X, x \in U : \exists B \in \BB : x \in B \subseteq U$
Hence by definition $\BB$ is a basis for $X$.
Theorem: Group of Units in Unital Banach Algebra is Open

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra.
Let $\map G A$ be the group of units of $A$.
Then $\map G A$ is open in $A$.

Proof:
Let $x \in \map G A$.
We find an open neighborhood of $x$ contained in $\map G A$. 
Clearly $x^{-1} \ne \mathbf 0_A$, so $\norm {x^{-1} } > 0$. 
We have, for $y \in A$:
{{begin-eqn}}
{{eqn	| l = \norm {1 - x^{-1} y}
		| r = \norm {x^{-1} \paren {x - y} }
}}
{{eqn	| o = \le
		| r = \norm {x^{-1} } \norm {x - y}
}}
{{end-eqn}}
So for $y \in A$ satisfying: 
:$\ds \norm {x - y} < \frac 1 {\norm {x^{-1} } }$
we have: 
:$\norm {1 - x^{-1} y} < 1$ 
From Element of Unital Banach Algebra Close to Identity is Invertible, we have that $x^{-1} y$ is invertible.
That is, $x^{-1} y \in \map G A$. 
Since $\map G A$ is a group, from Group of Units is Group, we have: 
:$y = x \paren {x^{-1} y} \in \map G A$ 
So we have: 
:$\map {B_{\norm {x^{-1} }^{-1} } } x \subseteq \map G A$
where $\map {B_{\norm {x^{-1} }^{-1} } } x$ is the open ball centered at $x$ with radius $\norm {x^{-1} }^{-1}$.
Since $x \in \map G A$ was arbitrary, we have that $\map G A$ is open.
{{qed}}
Theorem: Necessary and Sufficient Condition for Diagonal Operator to be Invertible

Let $\mathbb F \in \set {\R, \C}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N_{> 0} }$ be a bounded sequence in $\mathbb F$.
Let $\ell^2$ be the $2$-sequence space.
Let $\map {CL} {\ell^2} := \map {CL} {\ell^2, \ell^2}$ be the continuous linear transformation space on $\ell^2$.
Let $\Lambda \in \map {CL} {\ell^2}$ be the diagonal operator such that:
:$\forall \mathbf x := \tuple {x_1, x_2, \ldots} \in \ell^2 : \map \Lambda {\mathbf x} = \tuple {\lambda_1 \cdot x_1, \lambda_2 \cdot x_2, \ldots}$
Then $\Lambda$ is invertible in $\map {CL} {\ell^2}$ {{iff}} $\ds \inf_{n \mathop \in \N_{> 0} } \sequence {\size {\lambda_n} } > 0$ where $\inf$ denotes the infimum.

Proof:
=== Necessary Condition ===
Suppose $\ds \inf_{n \mathop \in \N_{> 0} } \sequence {\size {\lambda_n} } > 0$.
Then:
{{begin-eqn}}
{{eqn | q = \forall k \in \N_{> 0}
      | l = \size {\lambda_k}
      | o = \ge
      | r = \inf_{n \mathop \in \N_{> 0} } \sequence {\size {\lambda_n} }
}}
{{eqn | o = >
      | r = 0
}}
{{eqn | ll = \leadsto
      | q = \forall k \in \N_{> 0}
      | l = \lambda_k
      | o = \ne
      | r = 0
}}
{{end-eqn}}
Moreover:
{{begin-eqn}}
{{eqn | l = \sup_{n \mathop \in \N_{> 0} } \size {\frac 1 {\lambda_n} }
      | r = \sup_{n \mathop \in \N_{> 0} } \frac 1 {\size {\lambda_n} }
}}
{{eqn | r = \frac 1 {\ds \inf_{n \mathop \in \N_{> 0} } \size {\lambda_n} }
}}
{{end-eqn}}
Let $V : \ell^2 \to \ell^2$ be a mapping such that:
:$\ds \forall \mathbf a := \sequence {a_n}_{n \mathop \in \N} = \tuple {a_1, a_2, \ldots} : \map V {\mathbf a} = \tuple {\frac {a_1} {\lambda_1}, \frac {a_2} {\lambda_2}, \ldots}$
Then $V \in \map {CL} {\ell^2}$.
Furthremore:
{{begin-eqn}}
{{eqn | q = \forall \mathbf a \in \ell^2
      | l = \map V {\map \Lambda {\mathbf a} }
      | r = \map V {\sequence {\lambda_n a_n}_{n \mathop \in \N_{> 0} } }
}}
{{eqn | r = \sequence {\frac{\lambda_n a_n}{\lambda_n} }_{n \mathop \in \N_{> 0} }
}}
{{eqn | r = \sequence {a_n}_{n \mathop \in \N_{> 0} }
}}
{{eqn | r = \map I { \sequence {a_n}_{n \mathop \in \N_{> 0} } }
}}
{{eqn | r = \sequence {\lambda_n \frac {a_n }{\lambda_n} }_{n \mathop \in \N_{> 0} }
}}
{{eqn | r = \map \Lambda {\sequence {\frac{a_n}{\lambda_n} }_{n \mathop \in \N_{> 0} } }
}}
{{eqn | r = \map \Lambda {\map V {\mathbf a} }
}}
{{end-eqn}}
Therefore:
:$\Lambda V = V \Lambda = I$
Hence, $\Lambda$ is invertible.
Finally, the unique inverse of $\Lambda$ is $\Lambda^{-1} = V$
{{qed|lemma}}
Theorem: Banach Space Valued Function is Analytic iff Weakly Analytic

Let $U$ be an open subset of $\C$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\C$.
Let $f : U \to X$ be a function.
Then $f$ is analytic {{iff}} it is weakly analytic.

Proof:
=== Necessary Condition ===
Suppose that $f$ is analytic. 
Define $f' : U \to X$ by: 
:$\ds \map {f'} z = \lim_{w \mathop \to z} \frac {\map f w - \map f z} {w - z}$
for each $z \in U$. 
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,} }$.
Let $\phi \in X^\ast$. 
Then for each $z, w \in U$ with $z \ne w$ we have: 
:$\ds \map \phi {\frac {\map f w - \map f z} {w - z} } = \frac {\map {\paren {\phi \circ f} } w - \map {\paren {\phi \circ f} } z} {w - z}$
since $\phi$ is linear. 
From  Continuity of Linear Functionals, $\phi$ is continuous. 
So, we have: 
:$\ds \lim_{w \mathop \to z} \map \phi {\frac {\map f w - \map f z} {w - z} } = \map \phi {\map {f'} z}$
So:
:$\ds \frac {\map {\paren {\phi \circ f} } w - \map {\paren {\phi \circ f} } z} {w - z}$ exists for each $z \in U$. 
So $\phi \circ f$ is analytic for each $\phi \in X^\ast$. 
So $f$ is weakly analytic.
{{qed|lemma}}
Theorem: Liouville's Theorem (Complex Analysis)/Banach Space

Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\C$.
Let $f : \C \to X$ be an analytic function that is bounded.
Then $f$ is constant.

Proof:
Take $M \ge 0$ such that: 
:$\norm {\map f x} \le M$
for each $x \in X$. 
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,} }$.
From Banach Space Valued Function is Analytic iff Weakly Analytic, $f$ is weakly analytic. 
So for each $\phi \in X^\ast$, $\phi \circ f : \C \to \C$ is analytic.
Further, for each $z \in \C$ we have: 
:$\cmod {\map \phi {\map f x} } \le \norm \phi_{X^\ast} \norm {\map f x} \le M \norm \phi_{X^\ast}$
So $\phi \circ f$ is bounded.
From Liouville's Theorem (Complex Analysis), it follows that $\phi \circ f$ is constant for each $\phi \in X^\ast$.
That is, for each $z, w \in \C$ we have: 
:$\map \phi {\map f z} = \map \phi {\map f w}$ 
for each $\phi \in X^\ast$. 
That is, since $\phi$ is linear:
:$\map \phi {\map f z - \map f w} = 0$ 
From Normed Dual Space Separates Points, we have: 
:$\map f z = \map f w$ 
for each $z, w \in \C$.
So $f$ is constant.
{{qed}}
Theorem: Null Ring is Ring with Unity

Let $R$ be the null ring.
Then $R$ is a ring with unity.

Proof:
We have that $R$ is the null ring.
That is, by definition it has a single element, which can be denoted $0_R$, such that:
:$R := \struct {\set {0_R}, +, \circ}$
where ring addition and the ring product are defined as:
{{begin-eqn}}
{{eqn | l = 0_R + 0_R
      | r = 0_R
}}
{{eqn | l = 0_R \circ 0_R
      | r = 0_R
}}
{{end-eqn}}
Hence we have that the algebraic structure $\struct {\set {0_R}, \circ}$ is a trivial group.
Thus we see that:
:$\forall a, b \in R: a \circ b = 0_R$
Thus by definition, $0_R$ is a unity.
{{qed}}
Category:Rings with Unity
Category:Null Ring
cypft4ci00pwxcdx50bxa64lurkmqvs
Theorem: Multitape Turing Machine Reduces to Turing Machine

Let $T$ be a multitape Turing machine.
Then, there exists a Turing machine $T'$ that accepts precisely the same language as $T$, and halts on exactly the inputs that $T$ does.

Proof:
{{Proofread|In particular, make sure the construction is correct. I keep spotting mistakes in it and fixing them; who knows how many more are left. The correctness proof is more or less secondary in this case, although if you want to make that more rigorous, go ahead.}}
Let $k$ be the number of tapes in $T$.
Let:
:$\Gamma_T$ be the symbols of $T$, with $B$ the blank symbol.
:$\Sigma \subset \Gamma_T \setminus \set B$ be the input symbols of $T$.
:$Q_T = \set {q_0, q_1, \dotsc, q_\beta}$ be the states of $T$, with $q_0$ the start state.
:$\delta_T$ be the transition function of $T$.
Theorem: Unary Representation of Natural Number

Let $x \in \N$ be a natural number.
Then:
:$x = \map s {\map s {\dots \map s 0} }$
where there are $x$ applications of the successor mapping to the constant $0$.

Proof:
We shall proceed by induction.
Theorem: Addition of Natural Numbers is Provable

Let $x, y \in \N$ be natural numbers.
Then there exists formal proof of:
:$\sqbrk x + \sqbrk y = \sqbrk {x + y}$
in minimal arithmetic, where $\sqbrk a$ is the unary representation of $a$.

Proof:
By Unary Representation of Natural Number, let $\sqbrk a$ denote the term $\map s {\dots \map s 0}$, where there are $a$ applications of the successor mapping to the constant $0$.
Proceed by induction on $y$.
Theorem: Uniformly Continuous Semigroup Bounded on Compact Intervals

Let $\GF \in \set {\R, \C}$.
Let $X$ be a Banach space over $\GF$. 
Let $\family {\map T t}_{t \ge 0}$ be a uniformly continuous semigroup. 
Let $\struct {\map B X, \norm {\, \cdot \,}_{\map B X} }$ be the space of bounded linear transformations equipped with the canonical norm.
Let $T > 0$ be a real number. 
Then there exists $M > 0$ such that: 
:$\norm {\map T t}_{\map B X} \le M$
for $t \in \closedint 0 T$.

Proof:
Since $\family {\map T t}_{t \ge 0}$ is a uniformly continuous semigroup, we have: 
:$\ds \lim_{t \mathop \to 0^+} \norm {\map T t - I}_{\map B X} = 0$ 
From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence and Modulus of Limit: Normed Vector Space, we have:
:$\norm {\map T t}_{\map B X} \to \norm I_{\map B X}$ as $t \to 0^+$
In particular: 
:$\norm {\map T t}_{\map B X} \to 1$
since $\norm I_{\map B X} = 1$ from Identity Mapping on Normed Vector Space is Bounded Linear Operator.
Then there exists $\delta > 0$ such that: 
:$\ds \norm {\map T t}_{\map B X} \le \frac 3 2$ for $t \in \closedint 0 \delta$.
For $t_0 \in \closedint 0 T$, write $t_0 = N \delta + r$ for $N \in \N$ and $r \in \hointr 0 \delta$.
Then we have: 
{{begin-eqn}}
{{eqn	| l = \norm {\map T t}_{\map B X}
		| r = \norm {\map T {N \delta + r} }_{\map B X}
}}
{{eqn	| r = \norm {\map T {N \delta} \map T r}_{\map B X}
		| c = {{Defof|Semigroup of Bounded Linear Operators}}
}}
{{eqn	| r = \norm {\paren {\map T \delta}^N \map T r}_{\map B X}
}}
{{eqn	| o = \le
		| r = \norm {\paren {\map T \delta}^N}_{\map B X} \norm {\map T r}_{\map B X}
		| c = Norm on Bounded Linear Transformation is Submultiplicative
}}
{{eqn	| o = \le
		| r = \norm {\map T \delta}_{\map B X}^N \norm {\map T r}_{\map B X}
		| c = Bound on Norm of Power of Element in Normed Algebra
}}
{{eqn	| r = \paren {\frac 3 2}^{N + 1}
		| c = since $r \in \closedint 0 \delta$
}}
{{end-eqn}}
Now note that if $t_0 \in \closedint 0 T$ and $t_0 = N \delta + r$ for $N \in \Z_{\ge 0}$ and $r \in \hointr 0 \delta$, we have: 
:$\ds N \le \frac T \delta$ 
So we have: 
:$\ds \norm {\map T t}_{\map B X} \le \paren {\frac 3 2}^{\frac T \delta + 1}$
for $t \in \closedint 0 T$.
$\delta$ is independent of $t$, so we have the result. 
{{qed}}
Category:Uniformly Continuous Semigroups
886i96fp9a82ox7kibi5sldx5kra83y
Theorem: Semigroup of Bounded Linear Operators Uniformly Continuous iff Continuous as Map from Non-Negative Reals to Bounded Linear Operators

Let $\GF \in \set {\R, \C}$.
Let $X$ be a Banach space over $\GF$. 
Let $\family {\map T t}_{t \ge 0}$ be a semigroup of bounded linear operators. 
Let $\struct {\map B X, \norm {\, \cdot \,}_{\map B X} }$ be the space of bounded linear transformations equipped with the canonical norm.
Then $\family {\map T t}_{t \ge 0}$ is uniformly continuous {{iff}}:
:the mapping $T : \hointr 0 \infty \to \map B X$ is continuous.

Proof:
=== Necessary Condition ===
If $T$ is continuous, then in particular it is continuous at $0$.
Since $\map T 0 = I$, we therefore have: 
:$\ds \lim_{t \mathop \to 0^+} \norm {\map T t - I}_{\map B X} = 0$
So $\family {\map T t}_{t \ge 0}$ is uniformly continuous.
{{qed|lemma}}
Theorem: Converting Decimal Expansion of Rational Number to Fraction

Let $x \in \Q$ be a rational number.
Let the decimal expansion of $x$ be:
:$0 \cdotp a_1 a_2 \ldots a_m \dot b_1 b_2 \ldots \dot b_n$
where $a_i: i \in \set {1, 2, \ldots, m}$ and $b_j: j \in \set {1, 2, \ldots, n}$ be the digits in the base $10$ expansion of $x$.
Then $x$ can be expressed as the following fraction:
:$x = \dfrac {a_1 a_2 \ldots a_m b_1 b_2 \ldots b_n - a_1 a_2 \ldots a_m} {10^m \times \paren {10^n - 1} }$

Proof:
First we note that:
{{begin-eqn}}
{{eqn | l = 0 \cdotp \dot b_1 b_2 \ldots \dot b_n \times 10^n
      | r = b_1 b_2 \ldots b_n \cdotp \dot b_1 b_2 \ldots \dot b_n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 0 \cdotp \dot b_1 b_2 \ldots \dot b_n \times 10^n - 0 \cdotp \dot b_1 b_2 \ldots \dot b_n
      | r = b_1 b_2 \ldots b_n \cdotp \dot b_1 b_2 \ldots \dot b_n - 0 \cdotp \dot b_1 b_2 \ldots \dot b_n
      | c = subtracting $0 \cdotp \dot b_1 b_2 \ldots \dot b_n$ from both sides
}}
{{eqn | ll= \leadsto
      | l = 0 \cdotp \dot b_1 b_2 \ldots \dot b_n \times \paren{10^n - 1}
      | r = b_1 b_2 \ldots b_n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 0 \cdotp \dot b_1 b_2 \ldots \dot b_n
      | r = \dfrac {b_1 b_2 \ldots b_n} {\paren {10^n - 1} }
      | c = 
}}
{{eqn | ll= \leadsto
      || l = 0 \cdotp a_1 a_2 \ldots a_m \dot b_1 b_2 \ldots \dot b_n
      | r = \dfrac {a_1 a_2 \ldots a_m} {10^m} + \dfrac 1 {10^m} \times \dfrac {b_1 b_2 \ldots b_n} {\paren {10^n - 1} }
      | c = {{Defof|Decimal Expansion}}
}}
{{eqn | r = \dfrac {a_1 a_2 \ldots a_m} {10^m} \times \dfrac {\paren {10^n - 1} } {\paren {10^n - 1} } + \dfrac 1 {10^m} \times \dfrac {b_1 b_2 \ldots b_n} {\paren {10^n - 1} }
      | c = multiplying top and bottom by $10^n - 1$
}}
{{eqn | r = \dfrac {a_1 a_2 \ldots a_m \overbrace{000 \ldots 000}^n - a_1 a_2 \ldots a_m } {10^m \paren {10^n - 1} } + \dfrac {b_1 b_2 \ldots b_n} {10^m \paren {10^n - 1} }
      | c = 
}}
{{eqn | r = \dfrac {a_1 a_2 \ldots a_m b_1 b_2 \ldots b_n - a_1 a_2 \ldots a_m }{10^m \times \paren {10^n - 1 } }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Bound on C0 Semigroup

Let $\GF \in \set {\R, \C}$.
Let $X$ be a Banach space over $\GF$. 
Let $\family {\map T t}_{t \ge 0}$ be a $C_0$ semigroup. 
Let $\struct {\map B X, \norm {\, \cdot \,}_{\map B X} }$ be the space of bounded linear transformations equipped with the canonical norm.
Then there exists $M \ge 1$ and $\omega \ge 0$ such that: 
:$\norm {\map T t}_{\map B X} \le M e^{\omega t}$ 
for each $t \in \hointr 0 \infty$.

Proof:
We first show that $\norm {\map T t}_{\map B X}$ is bounded on $\closedint 0 \delta$ for some $\delta > 0$.
{{AimForCont}} $\norm {\map T t}_{\map B X}$ is unbounded on $\closedint 0 \delta$ for each $\delta > 0$.
Then for each $n \in \N$ we can pick $t_n \in \closedint 0 {\frac 1 n}$ such that $\norm {\map T {t_n} }_{\map B X} \ge n$.
Then, we have: 
:$\ds \sup_{n \mathop \in \N} \norm {\map T {t_n} }_{\map B X} = \infty$ 
By definition, each $\map T {t_n}$ is a linear operator.
So we can apply Principle of Condensation of Singularities to obtain: 
:$\ds \sup_{n \mathop \in \N} \norm {\map T {t_n} x} = \infty$
for some $x \in X$.
But since: 
:$\ds \lim_{t \mathop \to 0^+} \map T {t_n} x = x$
and:
:$\ds \lim_{n \mathop \to \infty} t_n = 0$ by the Squeeze Theorem
we must have that:
:$\ds \sup_{n \mathop \in \N} \norm {\map T {t_n} x} < \infty$
by Convergent Sequence in Normed Vector Space is Bounded.
So we have a contradiction, and there exists $\delta > 0$ such that $\norm {\map T t}_{\map B X}$ is bounded on $\closedint 0 \delta$.
Suppose that: 
:$\norm {\map T t}_{\map B X} \le M$
for $t \in \closedint 0 \delta$.
Since $\map T 0 = I$, we have $M \ge 1$. 
Now for general $t \ge 0$, write: 
:$t = N \delta + r$ for $N \in \Z_{\ge 0}$ and $r \in \hointr 0 \delta$.
Then we have:
{{begin-eqn}}
{{eqn	| l = \norm {\map T t}_{\map B X}
		| r = \norm {\map T {N \delta + r} }_{\map B X}
}}
{{eqn	| r = \norm {\map T {N \delta} \map T r}_{\map B X}
		| c = {{Defof|Semigroup of Bounded Linear Operators}}
}}
{{eqn	| r = \norm {\paren {\map T \delta}^N \map T r}_{\map B X}
}}
{{eqn	| o = \le
		| r = \norm {\paren {\map T \delta}^N}_{\map B X} \norm {\map T r}_{\map B X}
		| c = Norm on Bounded Linear Transformation is Submultiplicative
}}
{{eqn	| o = \le
		| r = \norm {\map T \delta}_{\map B X}^N \norm {\map T r}_{\map B X}
		| c = Bound on Norm of Power of Element in Normed Algebra
}}
{{eqn	| r = M^{N + 1}
		| c = since $r \in \closedint 0 \delta$
}}
{{end-eqn}}
Since $t = N \delta + r$, we have: 
:$\ds \frac t \delta = N + \frac r \delta \ge N$
so we have:
{{begin-eqn}}
{{eqn	| l = M^{N + 1}
		| r = M M^N
}}
{{eqn	| o = \le
		| r = M M^{\frac t \delta}
}}
{{eqn	| r = M e^{\frac t \delta \log M}
}}
{{end-eqn}}
Taking: 
:$\ds \omega = \frac {\log M} \delta$ 
we have:
:$\ds \norm {\map T t}_{\map B X} \le M e^{\omega t}$ 
for each $t \in \hointr 0 \infty$.
{{qed}}
Theorem: Semigroup of Bounded Linear Operators is C0 iff Point Evaluations Continuous

Let $\GF \in \set {\R, \C}$.
Let $X$ be a Banach space over $\GF$. 
Let $\family {\map T t}_{t \ge 0}$ be a semigroup of bounded linear operators. 
For each $x \in X$, define $x^\wedge : \hointr 0 \infty \to X$ by: 
:$\map {x^\wedge} t = \map T t x$
for each $t \in \hointr 0 \infty$. 
Then $\family {\map T t}_{t \ge 0}$ is a $C_0$ semigroup {{iff}} $x^\wedge$ is continuous for each $x \in X$.

Proof:
=== Sufficient Condition ===
Suppose that $x^\wedge$ is continuous for each $x \in X$.
In particular, for each $x \in X$ we have that $x^\wedge$ is continuous at $0$. 
That is: 
:$\ds \lim_{t \mathop \to 0^+} \map T t x = \map T 0 x = x$ for each $x \in X$.
So $\family {\map T t}_{t \ge 0}$ is a $C_0$ semigroup.
{{qed|lemma}}
Theorem: Norm of Continuous Function is Continuous

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 $f : X \to Y$ be a continuous mapping.
Define $\norm f_Y : X \to \hointr 0 \infty$ by: 
:$\map {\paren {\norm f_Y} } x = \norm {\map f x}_Y$
for each $x \in X$. 
Then $\norm f_Y$ is continuous.

Proof:
Follows immediately from combining Norm on Vector Space is Continuous Function and Composite of Continuous Mappings is Continuous.
{{qed}}
Category:Normed Vector Spaces
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