nyuuzyou/wikis huggingface dataset:
Suprema and Infima of Combined Bounded Functions/Warning

== Theorem ==
Let $f$ and $g$ be real functions.
Let $c$ be a constant.
Let both $f$ and $g$ be bounded above on $S \subseteq \R$.
We have from Suprema and Infima of Combined Bounded Functions:
:$\ds \map {\sup_{x \mathop \in S} } {\map f x + c} = c + \map {\sup_{x \mathop \in S} } {\map f x}$
:$\ds \map {\sup_{x \mathop \in S} } {\map f x + \map g x} \le \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x}$
where $\ds \map \sup {\map f x}$ is the supremum of $\map f x$.
The equalities do not apply in general.
Let us take as an example:
:$S = \closedint {-1} 1$
:$\map f x = x$
:$\map g x = -x$
where $f$ and $g$ are real functions defined on $\R$.
Then:
:$\ds \map {\sup_{x \mathop \in S} } {\map f x} = \map {\sup_{x \mathop \in S} } {\map g x} = 1$
:$\ds \map {\inf_{x \mathop \in S} } {\map f x} = \map {\inf_{x \mathop \in S} } {\map g x} = -1$
So:
:$\ds \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x} = 2$
:$\ds \map {\inf_{x \mathop \in S} } {\map f x} + \map {\inf_{x \mathop \in S} } {\map g x} = -2$
However:
:$\forall x \in S: \map f x + \map g x = x + \paren {-x} = 0$
So:
:$\ds \map {\sup_{x \mathop \in S} } {\map f x + \map g x} = \map {\inf_{x \mathop \in S} } {\map f x + \map g x} = 0$
and it is immediately clear that the equality does not hold.
== Sources ==
* : $\S 7.16 \ (6)$
Category:Suprema and Infima of Combined Bounded Functions
Suprema in Ordered Group

== Theorem ==
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group.
Let $x, y, z \in G$ be arbitrary.
Let any one of the sets $\set {x, y}$, $\set {x \circ z, y \circ z}$ or $\set {z \circ x, z \circ y}$ admit a supremum.
Then all three sets admit a supremum, and:
== Proof ==
First we recall that by definition of ordered group, $\preccurlyeq$ is compatible with $\circ$:
Let $\set {x, y}$ admit a supremum $c$.
Then by definition of supremum:
:$(1): \quad c$ is an upper bound of $\set {x, y}$ in $G$
:$(2): \quad c \preccurlyeq d$ for all upper bounds $d$ of $\set {x, y}$ in $S$.
Thus we have:
Hence $\sup \set {x, y} \circ z$ is an upper bound of $\set {x \circ z, y \circ z}$.
Let $d$ be an upper bound of $\set {x \circ z, y \circ z}$.
Then as $G$ is a group we have that:
:$d = d' \circ z$
for some $d' \in G$.
Then:
Hence $\sup \set {x, y} \circ z$ is an upper bound of $\set {x \circ z, y \circ z}$ which is smaller than an arbitrary upper bound $d$ of $\set {x \circ z, y \circ z}$.
That is, $\sup \set {x, y} \circ z$ is a supremum of $\set {x \circ z, y \circ z}$.
Let $\set {x \circ z, y \circ z}$ admit a supremum $c$.
Then by definition of supremum:
:$(1): \quad c$ is an upper bound of $\set {x \circ z, y \circ z}$ in $G$
:$(2): \quad c \preccurlyeq d$ for all upper bounds $d$ of $\set {x \circ z, y \circ z}$ in $S$.
As $G$ is a group, there exists $c' \in G$ such that $c' \circ z = c$.
Thus we have:
Hence $z \circ c'$ is an upper bound of $\set {z \circ x, z \circ y}$.
Let $d$ be an upper bound of $\set {z \circ x, z \circ y}$.
As $G$ is a group, there exists $d' \in G$ such that $z \circ d' = d$.
Then:
Hence $z \circ c'$ is an upper bound of $\set {z \circ x, z \circ y}$ which is smaller than an arbitrary upper bound $d$ of $\set {z \circ x, z \circ y}$.
That is, $z \circ c'$ is a supremum of $\set {z \circ x, z \circ y}$.
Let $\set {z \circ x, z \circ y}$ admit a supremum $c$.
Then by definition of supremum:
:$(1): \quad c$ is an upper bound of $\set {z \circ x, z \circ y}$ in $G$
:$(2): \quad c \preccurlyeq d$ for all upper bounds $d$ of $\set {z \circ x, z \circ y}$ in $S$.
As $G$ is a group, there exists $c' \in G$ such that $z \circ c' = c$.
Thus we have:
Hence $c'$ is an upper bound of $\set {x, y}$.
Let $d$ be an upper bound of $\set {x, y}$.
Then:
Hence $c'$ is an upper bound of $\set {x, y}$ which is smaller than an arbitrary upper bound $d$ of $\set {x, y}$.
That is, $c'$ is a supremum of $\set {x, y}$.
Hence by definition:
:$z \circ \sup \set {x, y} = \sup \set {z \circ x, z \circ y}$
Thus we have shown that if any of the three sets $\set {x, y}$, $\set {x \circ z, y \circ z}$ or $\set {z \circ x, z \circ y}$ admit a supremum, they all do, and:
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.10 \ \text {(a)}$
Category:Ordered Groups
Category:Suprema
Suprema of two Real Sets

== Theorem ==
Let $S$ and $T$ be real sets.
Let $S$ and $T$ admit suprema.
Then:
: $\sup S \le \sup T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$
== Proof ==
=== Necessary Condition ===
Let $\sup S \le \sup T$.
The aim is to establish that:
:$\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$
Let $\epsilon \in \R_{>0}$.
Let:
:$S' = \set {s \in S : \exists t \in T : s \le t}$
There are two cases for an $s \in S$:
:$s \in S'$, in other words: $\exists t \in T : s \le t$
and:
:$s \in S \setminus S'$, in other words: $\nexists t \in T : s \le t$
First, consider the case that $s \in S'$.
Let $s \in S'$.
Then $t \in T$ exists such that:
We have found:
: $\forall \epsilon \in \R_{>0}: \forall s \in S': \exists t \in T: s < t + \epsilon$
Next, consider the second case, that $s \in S \setminus S'$.
Let $s \in S \setminus S'$.
Then:
So, $s = \sup T$ is the only element of $S \setminus S'$.
A $t \in T$ exists such that:
We have found:
: $\forall \epsilon \in \R_{>0}: \forall s \in S \setminus S': \exists t \in T: s < t + \epsilon$
We conclude by combining the results for the two cases:
: $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$
=== Sufficient Condition ===
Let $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$.
The aim is to establish that $\sup S \le \sup T$.
Let $\epsilon \in \R_{>0}$.
An $s \in S$ exists such that:
Let $s$ be such an element of $S$.
A $t \in T$ exists such that:
Let $t$ be such an element of $T$.
Then:
== Also see ==
* Infima of two Real Sets
Category:Real Analysis
Supremum Inequality for Ordinals

== Theorem ==
Let $A \subseteq \On$ and $B \subseteq \On$ be ordinals.
Then:
:$\ds \forall x \in A: \exists y \in B: x \le y \implies \bigcup A \le \bigcup B$
== Proof ==
Therefore:
:$\ds \bigcup A \subseteq \bigcup B$
and:
:$\ds \bigcup A \le \bigcup B$
== Warning ==
The converse of this statement does not hold.
== Sources ==
* : $\S 8.6$
Category:Ordinals
Supremum Metric and L1 Metric on Closed Real Intervals are not Topologically Equivalent

== Theorem ==
Let $S$ be the set of all real functions which are continuous on the closed interval $\closedint a b$.
Let $d_1$ be the $L^1$ metric on $S$:
:$\ds \forall f, g \in S: \map {d_1} {f, g} := \int_a^b \size {\map f t - \map g t} \rd t$
Let $d_\infty$ be the supremum metric on $S$:
:$\ds \forall f, g \in S: \map {d_\infty} {f, g} := \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}$
Then $d_1$ and $d_\infty$ are not topologically equivalent.
== Proof ==
Let $f, g \in S$.
Then by definition of supremum metric:
:$\forall x \in \closedint a b: \size {\map f x - \map g x} \le \map {d_\infty} {f, g}$
Hence by ...
:$\map {d_1} {f, g} = \ds \int_a^b \size {\map f x - \map g x} \rd x \le \paren {b - a} \map {d_\infty} {f, g}$
and so:
:$\map {B_\epsilon} {f; d_\infty} \subseteq \map {b_{\paren {b - a} } } {f; d_1}$
Now let $f_0$ denote the constant mapping:
:$\forall x: \map {f_0} x = 0$
We show that $\map {B_1} {f_0; d_\infty}$ is not $d_1$-open.
 instead that $\map {B_1} {f_0; d_\infty}$ is $d_1$-open.
Then we should have:
:$\map {B_1} {f_0; d_1} \subseteq \map {B_1} {f_0; d_\infty}$
for some $\epsilon \in \R_{>0}$.
But for $\epsilon > 0$ there exists a continuous function on $\closedint a b$ such that:
:$\map {d_1} {f, f_0} < \epsilon$
yet:
:$\map {d_\infty} {f, f_0} = 1$
So:
:$f \in \map {B_1} {f_0; d_1}$
but:
:$f \notin \map {B_1} {f_0; d_\infty}$
which contradicts our deduction that $\map {B_1} {f_0; d_1} \subseteq \map {B_1} {f_0; d_\infty}$.
Hence it cannot be the case that $\map {B_1} {f_0; d_\infty}$ is $d_1$-open.
The result follows.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics: Example $2.4.6$
Category:L1 Metric
Category:Supremum Metric
Supremum Metric is Metric

== Theorem ==
Let $S$ be a set.
Let $M = \struct {A', d'}$ be a metric space.
Let $A$ be the set of all bounded mappings $f: S \to M$.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
== Proof ==
We have that the supremum metric on $A \times A$ is defined as:
:$\ds \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in S} \map {d'} {\map f x, \map g x}$
where $f$ and $g$ are bounded mappings.
First note that we have:
and so the exists.
=== Proof of ===
So holds for $d$.
=== Proof of ===
Let $f, g, h \in A$.
Let $x, y \in S$.
Thus $\map d {f, g} + \map d {g, h}$ is an upper bound for:
:$X := \set {\map {d'} {\map f c, \map g c}: c \in X}$
So:
:$\map d {f, g} + \map d {g, h} \ge \sup X = \map d {f, h}$
So holds for $d$.
=== Proof of ===
So holds for $d$.
=== Proof of ===
As $d$ is the supremum of the absolute value of the image of the pointwise sum of $f$ and $g$:
:$\forall f, g \in A: \map d {f, g} \ge 0$
Suppose $f, g \in A: \map d {f, g} = 0$.
Then:
So holds for $d$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.17$
Category:Supremum Metric
Supremum Metric on Bounded Continuous Mappings is Metric

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $A$ be the set of all continuous mappings $f: M_1 \to M_2$ which are also bounded.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
== Proof ==
The set $A$ is a subset of the set $A'$ of all bounded mappings $f: M_1 \to M_2$.
Let $d': A' \times A' \to \R$ be the supremum metric on $A'$.
From Supremum Metric is Metric, $\struct {A', d'}$ is a metric space. 
By definition, $A$ is a metric subspace of $A'$.
Hence the result.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.17$
Category:Supremum Metric
Supremum Metric on Bounded Real-Valued Functions is Metric

== Theorem ==
Let $X$ be a set.
Let $A$ be the set of all bounded real-valued functions $f: X \to \R$.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.17$
Category:Supremum Metric
Category:Supremum Metric on Bounded Real-Valued Functions is Metric
Supremum Metric on Bounded Real-Valued Functions is Metric/Proof 1

== Theorem ==
== Proof ==
We have that the supremum metric on $A \times A$ is defined as:
:$\ds \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in X} \size {\map f x - \map g x}$
where $f$ and $g$ are bounded real-valued functions.
From Real Number Line is Metric Space, the real numbers $\R$ together with the absolute value function form a metric space.
The result follows by Supremum Metric is Metric.
Category:Supremum Metric on Bounded Real-Valued Functions is Metric
Supremum Metric on Bounded Real-Valued Functions is Metric/Proof 2

== Theorem ==
== Proof ==
We have that the supremum metric on $A \times A$ is defined as:
:$\ds \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in X} \size {\map f x - \map g x}$
where $f$ and $g$ are bounded real-valued functions.
So:
:$\exists K, L \in \R: \size {\map f x} \le K, \size {\map g x} \le L$
for all $x \in X$.
First note that we have:
and so the exists.
=== Proof of ===
So holds for $d$.
=== Proof of ===
Let $f, g, h \in A$.
Let $c \in X$.
Thus $\map d {f, g} + \map d {g, h}$ is an upper bound for:
:$S := \set {\size {\map f c - \map h c}: c \in X}$
So:
:$\map d {f, g} + \map d {g, h} \ge \sup S = \map d {f, h}$
So holds for $d$.
=== Proof of ===
So holds for $d$.
=== Proof of ===
As $d$ is the supremum of the absolute value of the image of the pointwise sum of $f$ and $g$:
:$\forall f, g \in A: \map d {f, g} \ge 0$
Suppose $f, g \in A: \map d {f, g} = 0$.
Then:
So holds for $d$.
Category:Supremum Metric on Bounded Real-Valued Functions is Metric
Supremum Metric on Bounded Real Functions on Closed Interval is Metric

== Theorem ==
Let $\closedint a b \subseteq \R$ be a closed real interval.
Let $A$ be the set of all bounded real functions $f: \closedint a b \to \R$.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 2$: Metric Spaces: Exercise $5$
Category:Supremum Metric
Category:Supremum Metric on Bounded Real Functions on Closed Interval is Metric
Supremum Metric on Bounded Real Functions on Closed Interval is Metric/Proof 1

== Theorem ==
== Proof ==
The interval is an instance of a set.
Hence Supremum Metric on Bounded Real-Valued Functions is Metric can be directly applied.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.17$
Category:Supremum Metric on Bounded Real Functions on Closed Interval is Metric
Supremum Metric on Bounded Real Functions on Closed Interval is Metric/Proof 2

== Theorem ==
== Proof ==
We have that the supremum metric on $A \times A$ is defined as:
:$\ds \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}$
where $f$ and $g$ are bounded real functions.
So:
:$\exists K, L \in \R: \size {\map f x} \le K, \size {\map g x} \le L$
for all $x \in \closedint a b$.
First note that we have:
and so the exists.
=== Proof of ===
So holds for $d$.
=== Proof of ===
Let $f, g, h \in A$.
Let $c \in \closedint a b$.
Thus $\map d {f, g} + \map d {g, h}$ is an upper bound for:
:$S := \set {\size {\map f c - \map h c}: c \in \closedint a b}$
So:
:$\map d {f, g} + \map d {g, h} \ge \sup S = \map d {f, h}$
So holds for $d$.
=== Proof of ===
So holds for $d$.
=== Proof of ===
As $d$ is the supremum of the absolute value of the image of the pointwise sum of $f$ and $g$:
:$\forall f, g \in A: \map d {f, g} \ge 0$
Suppose $f, g \in A: \map d {f, g} = 0$.
Then:
So holds for $d$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.8$
Category:Supremum Metric on Bounded Real Functions on Closed Interval is Metric
Supremum Metric on Bounded Real Sequences is Metric

== Theorem ==
Let $A$ be the set of all bounded real sequences.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
== Proof 1 ==
== Proof 2 ==
Category:Supremum Metric
Category:Supremum Metric on Bounded Real Sequences is Metric
Supremum Metric on Bounded Real Sequences is Metric/Proof 1

== Theorem ==
== Proof ==
By definition, a real sequence is a mapping from the natural numbers $\N$ to the real numbers $\R$.
Thus a bounded real sequence is a bounded real-valued function.
The result follows from Supremum Metric on Bounded Real-Valued Functions is Metric.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.17$
Category:Supremum Metric on Bounded Real Sequences is Metric
Supremum Metric on Bounded Real Sequences is Metric/Proof 2

== Theorem ==
== Proof ==
We have that the supremum metric on $A \times A$ is defined as:
:$\ds \forall x, y \in A: \map d {x, y} := \sup_{n \mathop \in \N} \size {x_n - y_n}$
where $x = \sequence {x_i}$ and $y = \sequence {y_i}$ are bounded real sequences.
So:
:$\exists K, L \in \R: \size {x_n} \le K, \size {y_n} \le L$
for all $n \in \N$.
First note that we have:
and so the exists.
=== Proof of ===
So holds for $d$.
=== Proof of ===
Let $x, y, z \in A$.
Let $n \in \N$.
Thus $\map d {x, y} + \map d {y, z}$ is an upper bound for:
:$S := \set {\size {x_n - z_n}: n \in \N}$
So:
:$\map d {x, y} + \map d {y, z} \ge \sup S = \map d {x, z}$
So holds for $d$.
=== Proof of ===
So holds for $d$.
=== Proof of ===
As $d$ is the supremum of the absolute value of the difference of terms of the sequences $\sequence {x_i}$ and $\sequence {y_i}$:
:$\forall x, y \in A: \map d {x, y} \ge 0$
Suppose $x, y \in A: \map d {x, y} = 0$.
Then:
So holds for $d$.
Category:Supremum Metric on Bounded Real Sequences is Metric
Supremum Metric on Continuous Real Functions/Examples/Closure of Open 1-Ball of 0 on Unit Interval

== Example of Supremum Metric on Continuous Real Functions ==
Let $\closedint 0 1 \subseteq \R$ be the closed unit interval.
Let $\mathscr C \closedint 0 1$ be the supremum space of continuous functions $f: \closedint 0 1 \to \R$.
Then:
:$\map \cl {\map {B_1} \bszero} = \set {f \in \mathscr C \closedint 0 1: \map {d_\infty} {f, \bszero} \le 1}$
where:
:$\map {B_1} \bszero$ denotes the open $1$-ball of $\bszero$
:$\d_\infty$ denotes the Chebyshev distance
:$\bszero$ denotes the constant function $f_0$.
== Proof ==
Let $S = \set {f \in \mathscr C \closedint 0 1: \map {d_\infty} {f, \bszero} \le 1}$.
We have that:
:$\map \cl {\map {B_1} \bszero} \subseteq S$
Suppose $f \in S$.
Let $\epsilon \in \R_{>0}$ be given.
Then:
:$\paren {1 - \dfrac \epsilon 2} \in \map {B_1} \bszero \cap \map {b_\epsilon} f$
and so:
:$f \in \map \cl {\map {B_1} \bszero}$
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 36$
Category:Supremum Metric
Supremum Metric on Continuous Real Functions is Metric

== Theorem ==
Let $\closedint a b \subseteq \R$ be a closed real interval.
Let $\mathscr C \closedint a b$ be the set of all continuous functions $f: \closedint a b \to \R$.
Let $d$ be the supremum metric on $\mathscr C \closedint a b$.
Then $d$ is a metric.
== Proof 1 ==
== Proof 2 ==
Category:Supremum Metric
Category:Supremum Metric on Continuous Real Functions is Metric
Supremum Metric on Continuous Real Functions is Metric/Proof 1

== Theorem ==
== Proof ==
Let $\map {\mathscr B} {\closedint a b, \R}$ be the set of all bounded real functions $f: \closedint a b \to \R$.
From Supremum Metric on Continuous Real Functions is Subspace of Bounded, $\struct {\mathscr C \closedint a b, d_{\mathscr C} }$ is a (metric) subspace of $\struct {\map {\mathscr B} {\closedint a b, \R}, d}$.
The result follows from Subspace of Metric Space is Metric Space.
Category:Supremum Metric on Continuous Real Functions is Metric
Supremum Metric on Continuous Real Functions is Metric/Proof 2

== Theorem ==
== Proof ==
Let $A := \mathscr D^r \closedint a b$ be the set of all continuous functions $f: \closedint a b \to \R$ which are of differentiability class $r$.
Let $d_r: A \times A \to \R$ be the supremum metric on $A$, defined as:
:$\ds \forall f, g \in A: \map d {f, g} := \sup_{\substack {x \mathop \in \closedint a b \\ i \mathop \in \set {0, 1, 2, \ldots, r} } } \size {\map {f^{\paren i} } x - \map {g^{\paren i} } x}$
where:
:$f$ and $g$ are continuous functions on $\closedint a b$ which are $r$ times differentiable
:$r \in \N$ is a natural number.
From Supremum Metric on Differentiability Class is Metric, $d_r$ is a metric on $A$.
By definition, real functions which are continuous on $\closedint a b$ are of differentiability class $0$.
Thus setting $r = 0$ it follows that $\mathscr C \closedint a b$ is $\mathscr D^0 \closedint a b$.
The result follows.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.17$
Category:Supremum Metric on Continuous Real Functions is Metric
Supremum Metric on Continuous Real Functions is Subspace of Bounded

== Theorem ==
Let $\closedint a b \subseteq \R$ be a closed real interval.
Let $\mathscr C \closedint a b$ be the set of all continuous functions $f: \closedint a b \to \R$.
Let $\map {\mathscr B} {\closedint a b, \R}$ be the set of all bounded real functions $f: \closedint a b \to \R$.
Let $d$ be the supremum metric on $\map {\mathscr B} {\closedint a b, \R}$.
Then $\struct {\mathscr C \closedint a b, d_{\mathscr C} }$ is a subspace of $\struct {\map {\mathscr B} {\closedint a b, \R}, d}$.
== Proof ==
Let $f \in \mathscr C \closedint a b$.
Then by Image of Closed Real Interval is Bounded, $f$ is bounded on $\closedint a b$.
Thus $f \in \map {\mathscr B} {\closedint a b, \R}$ and the result follows.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.8$
Category:Supremum Metric
Supremum Metric on Differentiability Class/Examples/Difference between C0 and C1

== Example of Supremum Metric on Differentiability Class ==
Let $\mathscr D^1 \closedint 0 1$ be the set of all continuous functions $\phi: \closedint 0 1 \to \R$ which are of differentiability class $1$.
Let $f$ and $g$ be elements of $\mathscr D^1 \closedint 0 1$ defined as:
:$\forall x \in \closedint 0 1: \begin {cases} \map f x = 0 \\ \map g x = 10^{-6} \map \sin {10^{16} x} \end {cases}$
Let $d_0$ denote the supremum metric $C^0$ on $\mathscr D^1 \closedint 0 1$:
:$\ds \forall f, g \in A: \map {d_0} {f, g} := \sup_{x \mathop \in \closedint 0 1} \size {\map f x - \map g x}$
Let $d_1$ denote the supremum metric $C^1$ on $\mathscr D^1 \closedint 0 1$:
:$\ds \forall f, g \in A: \map {d_1} {f, g} := \sup_{x \mathop \in \closedint 0 1} \size {\map f' x - \map g' x}$
Then:
:$\map {d_0} {f, g} = 10^{-6}$
while:
:$\map {d_1} {f', g'} = 10^{10}$
=== Application ===
== Proof ==
We have that:
Then we have that:
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.17$
Category:Supremum Metric
Supremum Metric on Differentiability Class/Examples/Difference between C0 and C1/Application

== Example of Supremum Metric on Differentiability Class ==
Let $f$ and $g$ be elements of $\mathscr D^1 \closedint 0 1$ defined as:
:$\forall x \in \closedint 0 1: \begin {cases} \map f x = 0 \\ \map g x = 10^{-6} \map \sin {10^{16} x} \end {cases}$
Consider a road along a $1 \ \mathrm {km}$ route whose surface can be described as $\map g x$.
Let a journey be made along this road be charged at a rate of $1$ penny per $1000 \ \mathrm {km}$.
Suppose the distance over which this rate is measured is taken along the road surface.
Then, although the bumps are $1 \ \mathrm {mm}$ high, the cost of the journey is over $\pounds 30, 000$.
== Proof ==
We have that:
and so the bumps are $\dfrac {1 \ \mathrm {km} } {10^6} = 1 \ \mathrm {mm}$ high.
Let $I$ be the distance along the road according to $g$.
We have that:
We have that $\size {\cos 10^{16} x} \ge \dfrac 1 2$ for $\dfrac 2 3$ of the range $\closedint 0 1$.
So:
:$I \ge \dfrac {10^{10} } 3$
which costs more than $3 \times 10^6 \ \mathrm p = \pounds 30, 000$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 27$
Category:Supremum Metric
Supremum Metric on Differentiability Class is Metric

== Theorem ==
Let $\closedint a b \subseteq \R$ be a closed real interval.
Let $r \in \N$ be a natural number.
Let $A := \mathscr D^r \closedint a b$ be the set of all continuous functions $f: \closedint a b \to \R$ which are of differentiability class $r$.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.
== Proof ==
We have that the supremum metric on $A \times A$ is defined as:
:$\ds \forall f, g \in A: \map d {f, g} := \sup_{\substack {x \mathop \in \closedint a b \\ i \mathop \in \set {0, 1, 2, \ldots, r} } } \size {\map {f^{\paren i} } x - \map {g^{\paren i} } x}$
where $f$ and $g$ are continuous functions on $\closedint a b$ which are of differentiability class $r$.
=== Proof of ===
So holds for $d$.
=== Proof of ===
Let $f, g, h \in A$.
Let $c \in \closedint a b$.
Thus $\map d {f, g} + \map d {g, h}$ is an upper bound for:
:$\ds S := \set {\sup_{i \mathop \in \set {0, 1, 2, \ldots, r} } \size {\map {f^{\paren i} } c - \map {g^{\paren i} } c}: c \in \closedint a b}$
So:
:$\map d {f, g} + \map d {g, h} \ge \sup S = \map d {f, h}$
So holds for $d$.
=== Proof of ===
So holds for $d$.
=== Proof of ===
As $d$ is the supremum of the absolute value of the image of the pointwise sum of $f$ and $g$:
:$\forall f, g \in A: \map d {f, g} \ge 0$
Suppose $f, g \in A: \map d {f, g} = 0$.
Then:
So holds for $d$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.17$
Category:Supremum Metric