nyuuzyou/wikis huggingface dataset:
Functionally Complete Logical Connectives/Negation and Disjunction

== Theorem ==
The set of logical connectives:
:$\set {\neg, \lor}$: Not and Or
is functionally complete.
== Proof ==
From Functionally Complete Logical Connectives: Negation and Conjunction, $\set {\neg, \land}$ is functionally complete.
That is: any expression can be expressed in terms of $\neg$ and $\land$.
From De Morgan's laws: Conjunction, we have that:
:$p \land q \dashv \vdash \neg \paren {\neg p \lor \neg q}$
Thus all occurrences of $\land$ can be replaced by $\lor$ and $\neg$.
Thus any expression can be expressed in terms of $\neg$ and $\lor$.
That is: $\set {\neg, \lor}$ is functionally complete.
== Sources ==
* : $\S 2.5$: Further Logical Constants
* : Chapter $2$: The Propositional Calculus $2$: $3$ Truth-Tables: Exercise $2 \ \text{(i)}$
* : $\S 1$: Some mathematical language: Connectives
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic
* : $\S 2.4.2$: Theorem $2.36$
Category:Functional Completeness
Functionally Complete Singleton Sets

== Theorem ==
The only binary logical connectives that form singleton sets which are functionally complete are NAND and NOR.
== Proof ==
Let $\uparrow$ and $\downarrow$ denote NAND and NOR respectively.
From:
: NAND is Functionally Complete
and
: NOR is Functionally Complete
the singleton sets $\left\{{\uparrow}\right\}$ and $\left\{{\downarrow}\right\}$ are functionally complete.
Suppose $\circ$ is a binary logical connective such that $\left\{{\circ}\right\}$ is a functionally complete set.
The unary logical connective $\neg$ has to be equivalent to some formula:
: $\cdots \circ \left({p \circ p}\right) \circ \cdots$
Suppose some interpretation $v$ assigns $T$ to $p$.
Then $v \left({\neg p}\right) = F$
So:
: $v \left({\cdots \circ \left({p \circ p}\right) \circ \cdots}\right) = F$
So it has to be true that $T \circ T = F$, otherwise:
: $v \left({\cdots \circ \left({T \circ T}\right) \circ \cdots}\right) = v \left({\cdots \circ \left({T}\right) \circ \cdots}\right) = T$
Similarly:
:$F \circ F = T$
Now consider $T \circ F$ and $F \circ T$.
Suppose $T \circ F = F$.
If $F \circ T = T$, then we have the function defined by this truth table:
$v \left({p}\right)$	$v \left({q}\right)$	$v \left({p \circ q}\right)$
$F$ 	$F$ 	$T$
$F$ 	$T$ 	$T$
$T$ 	$F$ 	$F$
$T$ 	$T$ 	$F$
which is $\neg p$, and hence only negation would be definable.
So if $T \circ F = F$ we need $F \circ T = F$.
This gives the truth table for the NOR function:
$v \left({p}\right)$	$v \left({q}\right)$	$v \left({p \circ q}\right)$
$F$ 	$F$ 	$T$
$F$ 	$T$ 	$F$
$T$ 	$F$ 	$F$
$T$ 	$T$ 	$F$
... which we have seen is functionally complete.
Similarly, Suppose $T \circ F = T$.
If $F \circ T = F$, then we have the function defined by this truth table:
$v \left({p}\right)$	$v \left({q}\right)$	$v \left({p \circ q}\right)$
$F$ 	$F$ 	$T$
$F$ 	$T$ 	$F$
$T$ 	$F$ 	$T$
$T$ 	$T$ 	$F$
which is $\neg q$, and hence only negation would be definable.
So if $T \circ F = T$ we need $F \circ T = T$.
This gives the truth table for the NAND function:
$v \left({p}\right)$	$v \left({q}\right)$	$v \left({p \circ q}\right)$
$F$ 	$F$ 	$T$
$F$ 	$T$ 	$T$
$T$ 	$F$ 	$T$
$T$ 	$T$ 	$F$
... which we have seen is functionally complete.
Thus it follows that there can be no functionally complete binary logical connectives apart from NAND and NOR.
== Sources ==
* : $\S 2.4.2$: Theorem $2.37$
* : $\S 2.10$: Exercise $2.10$
Category:Functional Completeness
Functionally Incomplete Logical Connectives/Conjunction and Disjunction

== Theorem ==
The set of logical connectives:
:$\set {\land, \lor}$: And and Or
is not functionally complete.
== Proof ==
Let $v_T$ be the boolean interpretation that assigns $T$ to each propositional symbol.
Then it follows by the nature of the truth functions for $\land$ and $\lor$ that:
:$\map {v_T} {\mathbf A} = T$
for each WFF $\mathbf A$ comprising only $\land$ and $\lor$.
On the other hand:
:$\map {v_T} {\neg p} = F$
Therefore, $\neg p$ cannot be expressed in terms of $\land$ and $\lor$.
Hence, $\set {\land, \lor}$ is not functionally complete.
== Sources ==
* 
* 
* : $\S 2.10$: Exercise $2.11$
Category:Functional Completeness
Functionally Incomplete Logical Connectives/Negation and Biconditional

== Theorem ==
The set of logical connectives:
:$\set {\neg, \iff}$: Not and Iff
is not functionally complete.
== Proof ==
Let $\mathbf A$ be a WFF comprising only the logical connectives $\neg$ and $\iff$.
The central claim to this proof is that:
:The number of models for $\mathbf A$ is even.
In this claim, we disregard the obvious exceptions of the form $\neg \cdots \neg p$.
Firstly, note that including any propositional symbols not occurring in $\mathbf A$ in the truth table for $\mathbf A$ simply doubles the number of boolean interpretations modelling $\mathbf A$.
Next, it is obvious that if the number of models of $\mathbf A$ is even, then this also holds for $\neg \mathbf A$.
For, by Count of Truth Functions, the total number of boolean interpretations is even.
Lastly, to consider $\mathbf A = \mathbf A_1 \iff \mathbf A_2$.
Suppose that the claim were verified for $\mathbf A_1$ and $\mathbf A_2$.
Then we have the following identities:
:$\set {v \mid \map v {\mathbf A_1} = T} = \set {v \mid \map v {\mathbf A_1} = T, \map v {\mathbf A_2} = F} \cup \set {v \mid \map v {\mathbf A_1} = T, \map v {\mathbf A_2} = T}$
:$\set {v \mid \map v {\mathbf A_1} = F} = \set {v \mid \map v {\mathbf A_1} = F, \map v {\mathbf A_2} = F} \cup \set {v \mid \map v {\mathbf A_1} = F, \map v {\mathbf A_2} = T}$
:$\set {v \mid \map v {\mathbf A_2} = T} = \set {v \mid \map v {\mathbf A_1} = T, \map v {\mathbf A_2} = T} \cup \set {v \mid \map v {\mathbf A_1} = F, \map v {\mathbf A_2} = T}$
:$\set {v \mid \map v {\mathbf A_2} = F} = \set {v \mid \map v {\mathbf A_1} = T, \map v {\mathbf A_2} = F} \cup \set {v \mid \map v {\mathbf A_1} = F, \map v {\mathbf A_2} = F}$
The sets on the are disjoint, while the sets on the have an even number of elements.
Therefore, the parity of the cardinalities of the sets on the must be equal.
Then combining the first and fourth equations, the sets:
:$\set {v \mid \map v {\mathbf A_1} = T, \map v {\mathbf A_2} = T}$
:$\set {v \mid \map v {\mathbf A_1} = F, \map v {\mathbf A_2} = F}$ 
have equal parity.
Together, these two constitute the set of models for $\mathbf A$, as can be seen from the boolean interpretation of $\iff$.
Therefore, the number of models for $\mathbf A$ is even.
This completes the proof of the claim.
On the other hand, the truth table of $p \implies q$ has three entries $F$, and one $T$.
Therefore, $p \implies q$ cannot be expressed in terms of $\neg$ and $\iff$.
Hence, $\set {\neg, \iff}$ is not functionally complete.
== Sources ==
* : $\S 2.10$: Exercise $2.12$
Category:Functional Completeness
Functions A.E. Equal iff Positive and Negative Parts A.E. Equal

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space. 
Let $f, g : X \to \overline \R$ be extended real-valued functions. 
Then $f = g$ $\mu$-almost everywhere :
:$f^+ = g^+$ $\mu$-almost everywhere and $f^- = g^-$ $\mu$-almost everywhere.
where: 
:$f^+$ and $g^+$ denote the positive parts of $f$ and $g$ respectively
:$f^-$ and $g^-$ denote the negative parts of $f$ and $g$ respectively.
== Proof ==
=== Necessary Condition ===
Suppose that $f = g$ $\mu$-almost everywhere. 
Then there exists a $\mu$-null set $N \subseteq X$ such that: 
:if $x \in X$ has $\map f x \ne \map g x$, then $x \in N$.
From the definition of the positive part, we have: 
:$\map {f^+} x = \max \set {\map f x, 0}$
and:
:$\map {g^+} x = \max \set {\map g x, 0}$
Clearly if $\map f x = \map g x$, we then have $\map {f^+} x = \map {g^+} x$. 
So from Proof by Contraposition, if $\map {f^+} x \ne \map {g^+} x$ for $x \in X$, we have that $\map f x \ne \map g x$.
That is, if $\map {f^+} x \ne \map {g^+} x$, we have $x \in N$, which is $\mu$-null set.
So:
:$f^+ = g^+$ $\mu$-almost everywhere.
From the definition of the negative part, we have:
:$\map {f^-} x = -\min \set {\map f x, 0}$
and:
:$\map {g^-} x = -\min \set {\map g x, 0}$
As before, by contraposition, if $\map {f^-} x \ne \map {g^-} x$ for $x \in X$, we have that $\map f x \ne \map g x$.
So, if $\map {f^-} x \ne \map {g^-} x$, then $x \in N$.
Again, $N$ is $\mu$-null so we also obtain: 
:$f^- = g^-$ $\mu$-almost everywhere.
=== Sufficient Condition ===
Suppose that: 
:$f^+ = g^+$ $\mu$-almost everywhere.
and:
:$f^- = g^-$ $\mu$-almost everywhere.
Then there exists a $\mu$-null set $N_1 \subseteq X$ such that: 
:if $x \in X$ is such that $\map {f^+} x \ne \map {g^+} x$, we have $x \in N_1$.
There also exists a $\mu$-null set $N_2 \subseteq X$ such that: 
:if $x \in X$ is such that $\map {f^-} x \ne \map {g^-} x$, we have $x \in N_2$.
From Difference of Positive and Negative Parts, we have:
:$f = f^+ - f^-$
and:
:$g = g^+ - g^-$
If $x \in X$ is such that $\map {f^+} x = \map {g^+} x$ and $\map {f^-} x = \map {g^-} x$, we therefore have $\map f x = \map g x$.
So by contraposition, if $\map f x \ne \map g x$, then $\map {f^+} x \ne \map {g^+} x$ or $\map {f^-} x \ne \map {g^-} x$. 
Now let $x \in X$ be such that $\map f x \ne \map g x$. 
Then either:
:$\map {f^+} x \ne \map {g^+} x$
in which case, $x \in N_1$, or: 
:$\map {f^-} x \ne \map {g^-} x$
in which case $x \in N_2$.
So if $x \in X$ is such that $\map f x \ne \map g x$, we have: 
:$x \in N_1 \cup N_2$
From Null Sets Closed under Countable Union, we have that: 
:$N_1 \cup N_2$ is $\mu$-null.
So:
:$f = g$ $\mu$-almost everywhere.
Category:Measure Theory
Category:Positive Parts
Category:Negative Parts
Functions in Vector Space of Real-Valued Functions Continuously Differentiable on Closed Interval vanish at Endpoints

== Theorem ==
Let $I := \closedint a b$ be a closed real interval.
Let $\struct {\map {C^1} I, +, \, \cdot \,}_\R$ be the continuously differentiable on closed interval real function vector space.
Let $S := \set {x \in \map {C^1} I : \map x a = y_a, \map x b = y_b}$.
Then $S$ is a vector subspace of $\struct {\map {C^1} I, +, \, \cdot \,}_\R$ $y_a = y_b = 0$.
== Proof ==
=== Necessary Condition ===
Suppose $y_a = y_b = 0$.
==== Closure under Vector Addition ====
Let $x_1, x_2 \in \map {C^1} I$.
By sum rule for derivatives, $x_1 + x_2 \in \map {C^1} I$
Evaluate the sum at both endpoints:
Hence, if $x_1, x_2 \in S$ then $x_1 + x_2 \in S$.
==== Closure under Scalar Multiplication ====
Let $x \in \map {C^1} I$ and $\alpha \in \R$.
By derivative of constant multiple, $\alpha \cdot x \in \map {C^1} I$.
Evaluation at both endpoint yields:
Hence, if $x \in S$ and $\alpha \in \R$, then $\alpha \cdot x \in S$.
==== Nonemptiness ====
Let $\map 0 x \in \map {C^1} I$ be such that:
:$\map 0 x : I \to 0$.
Then:
Hence, $S$ is a subspace of $\struct {\map {C^1} I, +, \, \cdot \,}_\R$
=== Sufficient Condition ===
Suppose $S$ is a subspace of $\struct {\map {C^1} I, +, \, \cdot \,}_\R$.
Let $x \in S$.
Then $2 \cdot x \in S$ and $\map {\paren {2 \cdot x} } a = y_a$.
However, by Pointwise Scalar Multiplication of Real-Valued Functions we have that:
:$\map {\paren {2 \cdot x} } a = 2 \map x a = 2 y_a$
Hence, $2 y_a = y_a$, or $y_a = 0$.
Analogously, $y_b = 0$.
== Sources ==
* : Chapter $1.1$: Normed and Banach spaces. Vector Spaces
Category:Vector Spaces
Category:Differentiable Real-Valued Functions
Functions of Independent Random Variables are Independent

== Theorem ==
Let $X$ and $Y$ be independent random variables on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $g$ and $h$ be real-valued functions defined on the codomains of $X$ and $Y$ respectively.
Then $\map g X$ and $\map h Y$ are independent random variables.
== Proof ==
Let $A$ and $B$ be subsets of the real numbers $\R$.
Let $g^{-1} \sqbrk A$ and $h^{-1} \sqbrk B$ denote the preimages of $A$ and $B$ under $g$ and $h$ respectively. 
Applying the definition of independent random variables:
Hence $\map g X$ and $\map h Y$ are independent random variables.
== Sources ==
* : $\S 6.4$
Category:Independent Random Variables
Functor Category is Category

== Theorem ==
Let $\mathbf C$ and $\mathbf D$ be categories.
Then the functor category $\mathrm{Fun}(\mathbf C, \mathbf D)$ is a category.
== Proof ==
We check the metacategory axioms.
=== Composition (C1) ===
By Composite Of Natural Transformations is Natural Transformation composition of morphisms in $\map {\operatorname {Fun} } {\mathbf C, \mathbf D}$ is well-defined.
=== Existence of identity morphisms (C2) ===
Let $F$ be an object of $\map {\operatorname {Fun} } {\mathbf C, \mathbf D}$, that is a functor $F:
:\mathbf C \to \mathbf D$
Define $\mathrm{id}_F : F \to F$ as the transformation, that assigns to every object $A$ of $\mathbf C$ the identity morphism $\operatorname {id}_{\map F A} : \map F A \to \map F A$.
Let $f : A \to B$ be a morphism in $\mathbf C$.
The diagram:
:$\xymatrix{
\map F A \ar[r]^{\operatorname {id}_{\map F A} } \ar[d]^{\map F f} & F(A) \ar[d]^{\map F f} \\
\map F B \ar[r]^{\operatorname {id}_{\map F B} } & \map F B
}$
is commutative, since:
:$\map F f = \operatorname {id}_{\map F B} \circ \map F f = \map F f \circ \operatorname {id}_{\map F A}$.
Hence $\operatorname {id}_F$ is a natural transformation.
Let $G: \mathbf C \to \mathbf D$ be a functor.
Let $\alpha: F \to G$ be a natural transformation.
For every object $A$ of $\mathbf C$, we have:
:$\alpha_A \circ \operatorname {id}_{\map F A} = \alpha_A$, so $\alpha \circ \operatorname {id}_F = \alpha$
Let $H : \mathbf C \to \mathbf D$ be a functor.
Let $\beta : H \to F$ be a natural transformation.
For every object $A$ of $\mathbf C$, we have:
:$\operatorname {id}_{\map F A} \circ \beta_A = \beta_A$, so $\operatorname {id}_F \circ \beta = \beta$
=== Associativity of composition (C3) ===
Let $F_1, F_2, F_3, F_4 : \mathbf C \to \mathbf D$ be functors.
Let $\alpha_1: F_1 \to F_2$, $\alpha_2: F_2 \to F_3$, $\alpha_3: F_3 \to F_4$ be natural transformations.
For every object $A$ of $\mathbf C$, we have morphisms:
:$\alpha_{1, A}: \map {F_1} A \to \map {F_2} A$
:$\alpha_{2, A}: \map {F_2} A \to \map {F_3} A$
:$\alpha_{3, A}: \map {F_3} A \to \map {F_4} A$
By associativity of composition in $\mathbf D$:
:$\paren {\alpha_{3, A} \circ \alpha_{2, A} } \circ \alpha_{1, A} = \alpha_{3, A} \circ \paren {\alpha_{2, A} \circ \alpha_{1, A} }$
The associativity:
:$\paren {\alpha_3 \circ \alpha_2} \circ \alpha_1 = \alpha_3 \circ \paren {\alpha_2 \circ \alpha_1}$
follows.
Functor between Monoid Categories

== Theorem ==
Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be monoids.
Let $\mathbf S$ and $\mathbf T$ be the associated monoid categories.
Let $F: \mathbf S \to \mathbf T$ be a functor.
Then the morphism functor $F_1$ of $F$ is a monoid homomorphism.
== Proof ==
By definition of monoid category, we have:
:$\mathbf S_1 = S, \mathbf T_1 = T$
which in particular are sets.
Hence the morphism functor $F_1: \mathbf S_1 \to \mathbf T_1$ is a mapping $F_1: S \to T$.
For $F_1$ to be a monoid homomorphism, it suffices to show that the following hold:
:$F_1 \left({e_S}\right) = e_T$
:$\forall s_1, s_2 \in S: F_1 \left({s_1 \circ s_2}\right) = F_1 \left({s_1}\right) * F_1 \left({s_2}\right)$
Since all $s_1, s_2 \in S$ are composable morphisms of $\mathbf S$, these conditions are exactly equivalent to $F$ being a functor.
Hence the result.
== Sources ==
* : $\S 1.4.13$
Category:Monoid Categories
Category:Monoid Homomorphisms
Functor between Order Categories

== Theorem ==
Let $\struct {S, \preceq}$ and $\struct {T, \preceq'}$ be ordered sets.
Let $\mathbf S$ and $\mathbf T$ be their associated order categories, respectively.
Let $F: \mathbf S \to \mathbf T$ be a functor.
Then its object functor $F: S \to T$ is a monotone mapping.
== Proof ==
Suppose that for some $a, b \in S$, we have:
:$a \preceq b$
Then there is a morphism $a \to b$ in $\mathbf S$.
As $F$ is a functor, it follows that there is a morphism:
:$F a \to F b$
in $\mathbf T$ as well, that is:
:$F a \preceq' F b$
Hence the result.
== Sources ==
* : $\S 1.4.8$
Category:Order Categories
Fundamental Group is Independent of Base Point for Path-Connected Space

== Theorem ==
Let $X$ be a path-connected space.
For $x \in X$, let $\map {\pi_1}{X, x}$ denote the fundamental group.
For $x, y \in X$, there is an isomorphism:
:$\phi: \map {\pi_1} {X, x} \to \map {\pi_1} {X, y}$
== Proof ==
Since $X$ is path-connected there exists a path $f$ connecting $y$ and $x$.
We define $\phi_f: \map {\pi_1} {X, x} \to \map {\pi_1} {X, y}$ by $\map {\phi_f} {\sqbrk g} =\sqbrk {f^{-1} g f}$.
$\phi$ is a homomorphism of groups, as is seen from:
:$\ds \map {\phi_f} {\sqbrk {g h} } = \sqbrk {f^{-1} g h f} = \sqbrk {f^{-1} g f f^{-1} h f} = \sqbrk {f^{-1} g f} \sqbrk {f^{-1} h f} = \map {\phi_f} {\sqbrk g} \map {\phi_f} {\sqbrk h}$ 
Also, $f^{-1}$ is a path from $x$ to $y$. Then by the same argument as before, $\phi_{f^{-1} }: \map {\pi_1} {X, y} \to \map {\pi_1} {X, x}$ where $\map {\phi_{f^{-1} } } {\sqbrk g} = \sqbrk {f g f^{-1} }$, is a homomorphism of groups.
Trivially, $\phi_f \circ \phi_{f^{-1} } = I_{\map {\pi_1} {X, y} }$ and $\phi_{f^{-1} } \circ \phi_f = I_{\map {\pi_1} {X, x} }$.
Then $\phi_f^{-1} = \phi_{f^{-1} }$, so $\phi_f$ is bijective and thus an isomorphism.
Category:Algebraic Topology
Fundamental Law of Universal Class

== Theorem ==
:$\forall x: x \in \Bbb U$
where:
:$\Bbb U$ denotes the universal class
:$x$ denotes a set.
== Proof ==
From the definition of the universal class:
:$\Bbb U = \set {x: x = x}$
From this, it follows immediately that:
:$\forall x: \paren {x \in \Bbb U \iff x = x}$
From Equality is Reflexive, $x = x$ is a tautology.
Thus the asserted statement is also tautologous.
== Sources ==
* : $\S 6.8$
Category:Universal Class
Fundamental Principle of Counting

== Theorem ==
Let $A$ be a finite set.
Let $\sequence {B_n}$ be a sequence of distinct finite subsets of $A$ which form a partition of $A$.
Let $p_k = \size {B_k}$ for each $k \in \closedint 1 n$.
Then:
:$\ds \size A = \sum_{k \mathop = 1}^n p_k$
That is, the sum of the numbers of elements in the subsets of a partition of a set is equal to the total number of elements in the set.
== Proof ==
Let $r_0 = 0$, and let:
:$\ds \forall k \in \closedint 1 n: r_k = \sum_{j \mathop = 1}^k {p_j}$
Then:
:$r_{k - 1} + p_k = r_k$
so:
:$r_{k - 1} < r_k$.
Thus by Isomorphism to Closed Interval, $\closedint {r_{k - 1} } {r_k}$ has $r_k - r_{k - 1} = p_k$ elements.
As a consequence, there exists a bijection $\sigma_k: B_k \to \closedint {r_{k - 1} } {r_k}$ for each $k \in \closedint 1 n$.
Let $\sigma: A \to \N$ be the mapping that satisfies:
:$\forall x \in B_k: \forall k \in \N: \map \sigma x = \map {\sigma_k} x$
By Strictly Increasing Sequence on Ordered Set, $\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ is a strictly increasing sequence of natural numbers.
Thus by Strictly Increasing Sequence induces Partition, $\sigma: A \to \closedint 1 {r_n}$ is a bijection.
By Isomorphism to Closed Interval, $\closedint 1 {r_n}$ has $r_n$ elements.
Hence the result.
== Also known as ==
== Also see ==
* Product Rule for Counting
== Sources ==
* : $\S 2.3$. Partitions
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.4$
* 
* : Chapter $2$: 'And you do addition?': $\S 2.4$: Counting and mathematical induction: Theorem $2.4.1$
Category:Fundamental Principle of Counting
Category:Sequences
Category:Fundamental Theorems
Category:Set Theory
Category:Combinatorics
Category:Counting Arguments
Fundamental Property of Norm on Bounded Linear Functional

== Theorem ==
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 $\struct {X, \norm \cdot_X}$. 
Then for all $x \in X$ and $f \in X^\ast$:
:$\cmod {\map f x} \le \norm f_{X^\ast} \norm x_X$
== Proof ==
Let $x \in X$ and $f \in X^\ast$.
If $x = 0$, the claim is trivial, since by the linearity of $f$:
:$\cmod {\map f x} = \cmod {\map f 0} = 0$
Let $x \ne 0$.
Recall by :
:$\norm f_{X^\ast} = \sup \set {\dfrac {\cmod {\map f x} } {\norm x_X} : \forall x \in X, x \ne 0}$
Thus, by $\paren 1$ of :
:$\cmod {\map f x} = \dfrac {\cmod {\map f x} }{\norm x_X} {\norm x_X} \le \norm f_{X^\ast} \norm x_X$
Category:Bounded Linear Functionals
Fundamental Property of Norm on Bounded Linear Transformation

== Theorem ==
Let $\HH, \KK$ be Hilbert spaces.
Let $A: \HH \to \KK$ be a bounded linear transformation.
Let $\norm A$ denote the norm of $A$ defined by:
:$\norm A = \inf \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$
Then:
:$\forall h \in \HH: \norm {A h}_\KK \le \norm A \norm h_\HH$
== Proof ==
From Norm on Bounded Linear Transformation is Finite:
:$\norm A = \inf \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$ exists
and
:$\norm A < \infty$
Let $x \in \HH \setminus \set{0_\HH}$
Let $\lambda \in \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$.
Then:
As $c$ was arbitrary, then:
:$\forall \lambda \in \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}: \dfrac {\norm {A x}_\KK} {\norm x_\HH} \le \lambda$
By the definition of the infimum:
:$\dfrac {\norm {A x}_\KK} {\norm x_\HH} \le \norm A$
Hence:
:$\norm {A x}_\KK \le \norm A \norm x_\HH$
Since $x$ was arbitrary:
:$\forall h \in \HH \setminus \set {0_\HH}: \norm {A h}_\KK \le \norm A \norm h_\HH$
Lastly, we have:
It follows that:
:$\forall h \in \HH: \norm {A h}_\KK \le \norm A \norm h_\HH$
== Sources ==
* $\S \text {II}.1$
Category:Hilbert Spaces
Fundamental Solution to 1D Laplace's Equation

== Theorem ==
Let $\ds \map f x = \frac {\size x} 2$ where $\size x$ is the absolute value function.
Let $\delta \in \map {\DD'} \R$ be the Dirac delta distribution.
Let $T_f \in \map {\DD'} \R$ be the distribution associated with $f$.
Then $T_f$ is the fundamental solution to the $1$-dimensional Laplace's equation.
That is:
:$T_f = \delta$
or in the distributional sense:
:$\dfrac {\d^2} {\d x^2} f = \delta$
== Proof ==
$\ds x \stackrel f {\longrightarrow} \frac {\size x} 2$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.
By Differentiable Function as Distribution we have that $T'_f = T_{f'}$.
Moreover:
:$\ds x < 0 \implies \paren {\frac {\size x} 2}' = -\frac 1 2$
:$\ds x > 0 \implies \paren {\frac {\size x} 2}' = \frac 1 2$
Altogether:
:$\ds \forall x \in \R \setminus \set 0 : \paren {\frac {\size x} 2}' = \map g x$
where:
:$\map g x = \begin{cases}
\ds \frac 1 2 & : x > 0 \\
\ds - \frac 1 2 & : x < 0
\end{cases}$
Furthermore:
:$\ds \map f {0^+} - \map f {0^-} = 0$
By the Jump Rule:
:$T'_f = T_g$
$x \stackrel g {\longrightarrow} \map g x$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.
By similar arguments:
:$\ds \forall x \in \R \setminus \set 0 : \map {g'} x = 0$
Furthermore:
:$\ds \map g {0^+} - \map g {0^-} = 1$
By the Jump Rule:
:$T'_g = \delta$
By the definition of distributional derivative:
Hence, if $\ds \map f x = \frac {\size x} 2$, then $T_f = \delta$.
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Laplace's Equation
Category:Examples of Fundamental Solutions