nyuuzyou/wikis huggingface dataset:
König's Tree Lemma

== Theorem ==
Let $T$ be a rooted tree with an infinite number of nodes, each with a finite number of children.
Then $T$ has a branch of infinite length.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Note ==
This result may not hold if there exists at least one node which has an infinite number of children.
For example, let $T$ be the rooted tree defined as follows:
* $t_0$ is the root node.
* For all $n \in \N: n > 0$: $t_n$ is a leaf node which is a child of $t_0$.
Then there is an infinite number of nodes of $T$.
However, each branch of $T$ is of length equal to $1$.
== Also known as ==
This theorem is also referred to as König Tree Lemma, König's Tree Theorem and König Tree Theorem.
== Also see ==
This is a special case of the trickier to prove König's Lemma, which is a result that applies to all connected infinite graphs whose vertices are all finite in degree.
== Linguistic Note ==
Category:Tree Theory
Category:König's Tree Lemma
König's Tree Lemma/Proof 1

== Theorem ==
== Proof ==
We will show that we can choose an infinite sequence of nodes $t_0, t_1, t_2, \ldots$ of $T$ such that:
* $t_0$ is the root node;
* $t_{n + 1}$ is a child of $t_n$;
* Each $t_n$ has infinitely many descendants.
Then the sequence $t_0, t_1, t_2, \ldots$ is such a branch of infinite length.
Take the root node $t_0$.
By definition, it has a finite number of children.
Suppose that all of these childen had a finite number of descendants.
Then that would mean that $t_0$ had a finite number of descendants, and that would mean $T$ was finite.
So $t_0$ has at least one child with infinitely many descendants.
Thus, we may pick $t_1$ as any one of those children.
Now, suppose node $t_k$ has infinitely many descendants. 
As $t_k$ has a finite number of children, by the same argument as above, $t_k$ has at least one child with infinitely many descendants.
Thus we may pick $t_{k+1}$ which has infinitely many descendants.
The assertion follows by the Axiom of Dependent Choice.
== Sources ==
* : $\S 1.11$: Compactness: Theorem $1.11.3$
Category:König's Tree Lemma
König's Tree Lemma/Proof 2

== Theorem ==
== Proof ==
We will show that we can choose an infinite sequence of nodes $t_0, t_1, t_2, \ldots$ of $T$ such that:
:$t_0$ is the root node
:$t_{n + 1}$ is a child of $t_n$
Then the sequence $t_0, t_1, t_2, \ldots$ is such a branch of infinite length.
Let $I$ be the set of all nodes in $T$ that have infinitely many descendants.
Define a relation $\RR$ on $I$ by letting $x \mathrel \RR y$ $y$ is a child of $x$.
Since $T$ has infinitely many nodes, its root node is in $I$.
Let $t \in I$.
By the premise, $t$ has a finite number of children.
 that all of these children had a finite number of descendants.
Then by Finite Union of Finite Sets is Finite, that would mean that $t$ had a finite number of descendants, contradicting the fact that $t$ is in $I$.
So $t$ has at least one child with infinitely many descendants.
That is, $\map \RR t$ is non-empty for each $t \in I$.
That is, $\RR$ is a left-total relation.
Thus by Dependent Choice for Finite Sets, there is an infinite sequence $\sequence {t_n}$ such that:
:$t_0$ is the root of $T$
:$\forall n \in \N: t_n \mathrel \RR t_{n + 1}$ (that is, $t_{n + 1}$ is a child of $t_n$).
Category:König's Tree Lemma
König's Tree Lemma/Proof 3

== Lemma ==
== Proof ==
Suppose $T$ is a labeled tree with root $r$.
Let $L_0 = \set r, L_1, L_2, \ldots$ be the levels of $T$.
We construct our path by starting with $r$.
Next we partition all vertices in levels higher then $L_0$ (that is, all vertices in levels $L_i$ where $i > 0$) into as many finite parts as is the cardinality of $L_1$.
In other words if $L_1 = \set {v_1, \ldots, v_n}$ then we partition the vertices (there are infinitely many of them) in $n$ parts:
:$P_1, P_2, \ldots, P_n$
This partitioning is done as follows:
Any vertex $w$ will be connected to $r$ via some $v_i$ owing to connectedness.
We put $w$ in $P_i$. 
Now since we have partitioned infinitely many vertices in finitely many parts, one of the parts will contain infinitely many vertices.
(This is also called the Infinite Pigeonhole Principle).
Let that part be $P_j$.
Choose $w_1 = v_j$ as the next vertex in our under-construction path.
Next we delete the vertex $r$ and consider the component containing $w_1$.
It is also an infinite tree with each level finite and having $w_1$ as its root.
By a similar piece of reasoning, we get a vertex $w_2$ which is adjacent to $w_1$ and has infinitely many vertices corresponding to it.
We continue building our path by choosing $w_2$ as a part of it.
By induction we can continue and get an infinite path $r w_1 w_2 \cdots$.
This proves the theorem.
Category:König's Tree Lemma
L'Hôpital's Rule

== Theorem ==
Let $f$ and $g$ be real functions which are differentiable on the open interval $\openint a b$.
Let:
:$\forall x \in \openint a b: \map {g'} x \ne 0$
where $g'$ denotes the derivative of $g$ $x$.
Let:
:$\ds \lim_{x \mathop \to a^+} \map f x = \lim_{x \mathop \to a^+} \map g x = 0$
Then:
:$\ds \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$
provided that the second limit exists.
=== Weak Version ===
=== Corollary 1 ===
=== Corollary 2 ===
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Also defined as ==
== Also known as ==
== Historical Note ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.4$ Limits, Maxima and Minima: Indeterminate Forms (L'Hospital's Rule): $3.4.1$
* : Chapter $\text {A}.20$: The Bernoulli Brothers
* 
* 
* 
* 
* 
Category:L'Hôpital's Rule
Category:Differential Calculus
Category:Limits of Real Functions
L'Hôpital's Rule/Also defined as

== L'Hôpital's Rule: Also defined as ==
In many elementary texts on real analysis, it appears to be commonplace for $f$ and $g$ to be specified as being continuous on $\closedint a b$.
However, this is not strictly necessary, as differentiability on $\openint a b$ is completely adequate.
If is of course noted that from Differentiable Function is Continuous, if $f$ and $g$ are differentiable on $\openint a b$, they are also continuous on $\openint a b$, just not necessarily also at $a$ or $b$.
Indeed, for , $f$ and $g$ are demonstrably not continuous either at $a$ or $b$, or possibly both.
== Sources ==
* : $\S 11.8 \ (3)$
Category:L'Hôpital's Rule
L'Hôpital's Rule/Also known as

== L'Hôpital's Rule: Also known as ==
Because of variants in the rendition of 's name, L'Hôpital's Rule is often seen as:
:L'Hospital's rule
:de L'Hôpital's rule
and so on.
== Sources ==
* : $3$: Elementary Analytic Methods: $3.4$ Limits, Maxima and Minima: Indeterminate Forms (L'Hospital's Rule)
* 
* 
* 
* 
Category:L'Hôpital's Rule
L'Hôpital's Rule/Corollary 1

== Corollary to L'Hôpital's Rule ==
Let $f$ and $g$ be real functions which are differentiable on the open interval $\openint a b$.
Suppose that $\forall x \in \openint a b: \map {g'} x \ne 0$.
Suppose that $\exists c \in \openint a b: \map f c = \map g c = 0$.
Then:
:$\ds \lim_{x \mathop \to c} \frac {\map f x} {\map g x} = \lim_{x \mathop \to c} \frac {\map {f'} x} {\map {g'} x}$
provided that the second limit exists.
== Proof ==
This follows directly from the definition of limit.
If $\ds \lim_{x \mathop \to c} \frac {\map {f'} x} {\map {g'} x}$ exists, it follows that:
:$\ds \lim_{x \mathop \to c} \frac {\map {f'} x} {\map {g'} x} = \lim_{x \mathop \to c^+} \frac {\map {f'} x} {\map {g'} x}$
That is, if there exists such a limit, it is also a limit from the right.
== Also defined as ==
== Historical Note ==
Category:L'Hôpital's Rule
L'Hôpital's Rule/Corollary 2

== Corollary to L'Hôpital's Rule ==
Let $f$ and $g$ be real functions which are differentiable on the open interval $\openint a b$.
Suppose that $\forall x \in \openint a b: \map {g'} x \ne 0$.
Suppose that $\map f x \to +\infty$ and $\map g x \to +\infty$ as $x \to a^+$.
Then:
:$\ds \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$
provided that the second limit exists.
== Proof ==
Let $\ds \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x} = L$.
Let $\sequence {x_n}$ be a sequence such that:
:$\quad x_n \in \openint a b$ for all $n \in \N$ and $\ds \lim_{n \mathop \to \infty} x_n = a$
From Intermediate Value Theorem for Derivatives and the definition of limit of real function, it follows that:
:$\ds \lim_{n \mathop \to \infty} \map g {x_n} = +\infty$
and $\sequence {\map g {x_n} }$ is strictly increasing.
Consider the range $\closedint {x_{n - 1} } {x_n} \subset \openint a b$ where $n \ge 2$.
By Cauchy Mean Value Theorem, there exists $c_n \in \openint {x_{n - 1} } {x_n}$ such that:
:$\ds \frac {\map f {x_n} - \map f {x_{n - 1} } } {\map g {x_n} - \map g {x_{n - 1} } } = \frac {\map {f'} {c_n} } {\map {g'} {c_n} }$
From the above and the Squeeze Theorem for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} c_n = a$
and:
:$\ds \lim_{n \mathop \to \infty} \frac {\map f {x_n} - \map f {x_{n - 1} } } {\map g {x_n} - \map g {x_{n - 1} } } = \lim_{n \mathop \to \infty} \frac {\map {f'} {c_n} } {\map {g'} {c_n} } = L$
So, by :
:$\ds \lim_{n \mathop \to \infty} \frac {\map f {x_n} } {\map g {x_n} } = L$
From the definition of limit of real function, we deduce that:
:$\ds \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = L = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$
== Remarks ==
* The proof does not actually use the assumption $\ds \lim_{x \mathop \to a^+} \map f x = \infty$.
* Cases $x \to b^-$, $x \to \pm \infty$ and $\map g x \to -\infty$ can be proven similarly.
== Also defined as ==
== Historical Note ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.4$ Limits, Maxima and Minima: Indeterminate Forms (L'Hospital's Rule): $3.4.1$
Category:L'Hôpital's Rule
L'Hôpital's Rule/Examples/(Root (1 + x) - 1) over x

== Example of Use of L'Hôpital's Rule ==
:$\ds \lim_{x \mathop \to 0} \dfrac {\sqrt {1 + x} - 1} x = \dfrac 1 2$
== Proof ==
Let $f: \R \to \R$ be defined as:
:$\forall x \in \R: \map f x = \sqrt {1 + x} - 1$
Let $g: \R \to \R$ be defined as:
:$\forall x \in \R: \map g x = x$
We have that:
Then:
== Sources ==
* 
* 
Category:Examples of Use of L'Hôpital's Rule
L'Hôpital's Rule/Examples/x^2 - 4 over 2 x - 4

== Example of Use of L'Hôpital's Rule ==
Let $f: \R \to \R$ and $g: \R \to \R$ be real functions defined as:
Then:
:$\ds \lim_{x \mathop \to 2} \dfrac {\map f x} {\map g x} = 2$
== Proof ==
We note that $\valueat {\dfrac {\map f x} {\map g x} } {x \mathop = 2}$ is undefined.
Hence we observe that L'Hôpital's Rule may be applicable.
We have that:
Thus both $\ds \lim_{x \mathop \to 2} \map {f'} 2$ and $\ds \lim_{x \mathop \to 2} \map {g'} 2$ exist, and we can indeed continue:
== Sources ==
* 
* 
Category:Examples of Use of L'Hôpital's Rule
L'Hôpital's Rule/Historical Note

== Historical Note on L'Hôpital's Rule ==
While attributed to , who included it in his $1696$ work , published anonymously, this result was in fact discovered by in $1694$.
After 's death, claimed that most of the content of , including L'Hôpital's Rule, was in fact his own work.
However, it was discovered in $1955$, on the publication of correspondence between and that there had been an agreement between them to allow to use 's discoveries however he wanted.
Hence was vindicated, and his name continues to be associated with this result.
== Sources ==
* : Chapter $\text {A}.20$: The Bernoulli Brothers
* 
* 
* 
* 
Category:Historical Notes
L'Hôpital's Rule/Proof 1

== Theorem ==
== Proof ==
Let $l = \ds \lim_{x \mathop \to a^+} \frac {\map {f'} x}{\map {g'} x}$.
Let $\epsilon \in \R_{>0}$.
By the definition of limit, we ought to find a $\delta \in \R_{>0}$ such that:
:$\forall x \in \R: \size {x - a} < \delta \implies \size {\dfrac {\map f x} {\map g x} - l} < \epsilon$
Fix $\delta$ such that:
:$\forall x \in \R: \size {x - a} < \delta \implies \size {\dfrac {\map {f'} x} {\map {g'} x} - l} < \epsilon$
which is possible by the definition of limit.
Define:
$\quad \map {f_0} x = \begin {cases}
\map f x & : x \in \openint a b \\
0 & : x = a
\end {cases}$
$\quad \map {g_0} x = \begin {cases}
\map f x & : x \in \openint a b \\
0 & : x = a
\end {cases}$
By definition of right-continuous, it follows that $f_0$ and $g_0$ are continuous on $\hointr a b$.
Let $x_\delta$ be such that $0 < x_\delta - a < \delta$.
We have that $f_0$ and $g_0$ are continuous on $\closedint a {x_\delta}$, and differentiable on $\openint a {x_\delta}$
Thus, by the Cauchy Mean Value Theorem with $b = x_\delta$:
:$\exists \xi \in \openint a {x_\delta}: \dfrac {\map {f'_0} \xi} {\map {g'_0} \xi} = \dfrac {\map {f_0} {x_\delta} - \map {f_0} a} {\map {g_0} {x_\delta} - \map {g_0} a}$
Since $\map {f_0} a = \map {g_0} a = 0$:
:$\exists \xi \in \openint a {x_\delta}: \dfrac {\map {f'_0} \xi} {\map {g'_0} \xi} = \dfrac {\map {f_0} {x_\delta}} {\map {g_0} {x_\delta}}$
But since $\xi, x_\delta \in \openint a b$:
:$\map {f'_0} \xi = \map {f'} \xi$
:$\map {g'_0} \xi = \map {g'} \xi$
:$\map {f_0} {x_\delta} = \map f {x_\delta}$
:$\map {g_0} {x_\delta} = \map g {x_\delta}$
Therefore:
:$\exists \xi \in \openint a {x_\delta}: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f {x_\delta}} {\map g {x_\delta}}$
Now, as $a < \xi < x_\delta$, it follows that $\size{\xi - a} < \delta$ as well.
Therefore:
:$\size {\dfrac {\map f {x_\delta}} {\map g {x_\delta}} - l} = \size {\dfrac {\map {f'} \xi} {\map {g'} \xi} - l} < \epsilon$
which leads us to the desired conclusion that:
:$\ds \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$
Category:L'Hôpital's Rule
L'Hôpital's Rule/Proof 2

== Theorem ==
== Proof ==
Take the Cauchy Mean Value Theorem with $b = x$:
:$\exists \xi \in \openint a x: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f x - \map f a} {\map g x - \map g a}$
Then if $\map f a = \map g a = 0$ we have:
:$\exists \xi \in \openint a x: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f x} {\map g x}$
Note that $\xi$ depends on $x$; that is, different values of $x$ may require different values of $\xi$ to make the above statement valid.
It follows from Limit of Function in Interval that $\xi \to a$ as $x \to a$.
Also, $\xi \ne a$ when $x > a$.
So from Hypothesis $2$ of Limit of Composite Function, it follows that:
:$\ds \lim_{x \mathop \to a^+} \dfrac {\map {f'} \xi} {\map {g'} \xi} = \lim_{x \mathop \to a^+} \dfrac {\map {f'} x} {\map {g'} x}$
Hence the result.
== Sources ==
* : $\S 11.8 \ (3)$
Category:L'Hôpital's Rule
L'Hôpital's Rule/Weak Version

== Theorem ==
Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.
Let:
:$\forall x \in \openint a b: \map {g'} x \ne 0$
where $g'$ denotes the derivative of $g$ $x$.
Let:
:$\map f a = \map g a = 0$
Then:
:$\ds \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$
provided that the second limit exists.
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Also known as ==
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.20$: The Bernoulli Brothers
* 
* 
* 
* 
* 
Category:L'Hôpital's Rule
L-2 Inner Product is Inner Product

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\mathcal L^2} {X, \Sigma, \mu}$ be the Lebesgue $2$-space of $\struct {X, \Sigma, \mu}$.
Let $\map {L^2} {X, \Sigma, \mu}$ be the $L^2$ space of $\struct {X, \Sigma, \mu}$.
Let $\innerprod \cdot \cdot$ be the $L^2$ inner product.
Then $\innerprod \cdot \cdot$ is an inner product on $\map {L^2} {X, \Sigma, \mu}$.
== Proof ==
=== Proof of Symmetry ===
Let $E, F \in \map {L^2} {X, \Sigma, \mu}$.
Let:
:$E = \eqclass f \sim$ 
and:
:$F = \eqclass g \sim$
where $\eqclass f \sim$ and $\eqclass g \sim$ are the equivalence class of $f, g \in \map {\LL^2} {X, \Sigma, \mu}$ under the $\mu$-almost everywhere equality relation. 
Then: 
=== Proof of Linearity in First Argument ===
Let $E, F, G \in \map {L^2} {X, \Sigma, \mu}$ and $\alpha \in \R$.
Let $E = \eqclass f \sim$, $F = \eqclass g \sim$ and $G = \eqclass h \sim$. 
Then: 
=== Proof of Non-Negative Definiteness and Positivity ===
Let $E \in \map {L^2} {X, \Sigma, \mu}$.
Let $E = \eqclass f \sim$.
Then, we have: 
By the definition of the $\mu$-integral of a measurable function, we then have $\innerprod E E \ge 0$. 
We have $\innerprod E E = 0$ :
:$\ds \int \size f^2 \rd \mu = 0$ 
From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we then have: 
:$\size f^2 = 0$ $\mu$-almost everywhere.
From Pointwise Exponentiation preserves A.E. Equality, we have: 
:$\size f = 0$ $\mu$-almost everywhere.
So:
:$f = 0$ $\mu$-almost everywhere.
So:
:$\eqclass f \sim = \eqclass 0 \sim = \mathbf 0_{\map {L^2} {X, \Sigma, \mu} }$
so:
:$E = \mathbf 0_{\map {L^2} {X, \Sigma, \mu} }$
as required.
Category:Lp Spaces
Category:Examples of Inner Products
L-2 Inner Product is Well-Defined

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\LL^2} {X, \Sigma, \mu}$ be the Lebesgue $2$-space of $\struct {X, \Sigma, \mu}$. 
Let $\sim$ be the $\mu$-almost-everywhere equality relation on $\map {\LL^2} {X, \Sigma, \mu}$.
Let $\map {L^2} {X, \Sigma, \mu}$ be the $L^p$ space on $\struct {X, \Sigma, \mu}$.
Let $E, F \in \map {L^2} {X, \Sigma, \mu}$. 
Then the $L^2$ inner product $\innerprod E F$ is well-defined.
== Proof ==
Let $E = \eqclass f \sim$ and $F = \eqclass g \sim$ where $\sim$ is the $\mu$-almost everywhere equality relation.
We aim to show that: 
:$\ds \int \paren {f \cdot g} \rd \mu$ 
exists as a real number and is independent of the choice of representatives $f$ and $g$ of $E$ and $F$.
From the definition of the $L^2$ space, we have $f, g \in \map {\LL^2} {X, \Sigma, \mu}$.
From Hölder's Inequality for Integrals, we have:
:$f \cdot g \in \map {\LL^1} {X, \Sigma, \mu}$
So the $\mu$-integral of $f \cdot g$ is well-defined and exists as a real number.
Now suppose that: 
:$\eqclass {f_1} \sim = \eqclass {f_2} \sim = E$
and:
:$\eqclass {g_1} \sim = \eqclass {g_2} \sim = F$
From Equivalence Class Equivalent Statements, we have: 
:$f_1 \sim f_2$
and:
:$g_1 \sim g_2$
That is: 
:$f_1 = f_2$ $\mu$-almost everywhere
and:
:$g_1 = g_2$ $\mu$-almost everywhere.
From Pointwise Multiplication preserves A.E. Equality, we have: 
:$f_1 \cdot g_1 = f_2 \cdot g_2$ $\mu$-almost everywhere.
From A.E. Equal Positive Measurable Functions have Equal Integrals, we have: 
:$\ds \int \paren {f_1 \cdot g_1} \rd \mu = \int \paren {f_2 \cdot g_2} \rd \mu$
as desired.
Category:Lp Spaces
L-2 Space forms Hilbert Space

== Theorem ==
Let $\struct{ X, \Sigma, \mu }$ be a measure space.
Let $\map {L^2} \mu$ be the $L^2$ space of $\mu$.
Let $\innerprod \cdot \cdot$ be the inner product on $\map {L^2} \mu$.
Then $\map {L^2} \mu$ endowed with $\innerprod \cdot \cdot$ is a Hilbert space.
== Proof ==
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 1.$ Elementary Properties and Examples
Category:Examples of Hilbert Spaces
Category:Lp Spaces
L-Function does not Vanish at One

== Theorem ==
Let $\psi$ be a non-trivial Dirichlet character modulo $q$.
Let $\map L {s, \chi}$ be the Dirichlet $L$-function associated to $\chi$.
Then $\map L {1, \chi} \ne 0$.
== Proof ==
Let $G^*$ be the group of characters modulo $q$.
Let $\ds \map {\zeta_q} s = \prod_{\chi \mathop \in G^*} \map L {s, \chi} R$.
Among the factors of $\zeta_q$ is the the $L$-function associated to the trivial character, which by Analytic Continuation of Dirichlet L-Function we know to have a simple pole at $s = 1$.
 $\map L {1, \psi} = 0$. 
Then the zero of this factor kills the pole of the principal $L$-function.
So by Analytic Continuation of Dirichlet L-Function $\zeta_q$ is analytic on $\map \Re s > 0$.
For $\map \Re s > 1$ we have:
For any prime $p$, let $f_p$ be the order of $p$ mod $q$.
Then:
:$\map \chi p^{f_p} = \map \chi {p^{f_p} } = \map \chi 1 = 1$
So $\map \chi p$ is an $f_p$th root of unity.
Moreover by the Orthogonality Relations for Characters each distinct such root occurs $\map \phi q / f_p$ times among the numbers $\map \chi p$ where $\chi \in G^*$.
Also, letting $\xi$ be a primitive $f_p$th root of unity we find that for any $u \in \C$:
:$\ds \prod_{i \mathop = 0}^{f_p} \paren {1 - \xi^i u} = 1 - u^\xi$
Putting these facts together:
:$\ds \prod_{\chi \mathop \in G^*} \frac 1 {1 - \map \chi p p^{-s} } = \paren {\frac 1 {1 - p^{-f_p s} } }^{\map \phi q / f_p}$
Therefore:
Also, if $\chi_0$ is the trivial character modulo $q$, by definition of Euler product we have:
:$\ds \map L {\map \phi q s, \chi_0} = \prod_{p \mathop \nmid q} \paren {1 + p^{-\map \phi q s} + p^{-2 \map \phi q s} + \cdots}$
from which we see that for $s \in \R_{\ge 0}$:
:$\map {\zeta_q} s \ge \map L {\map \phi q s, \chi_0}$
However, by Analytic Continuation of Dirichlet L-Function, $\map L {\map \phi q s, \chi_0}$ diverges for $s = \map \phi q^{-1}$, and therefore so does $\map {\zeta_q} s$.
But we showed above that $\map {\zeta_q} s$ converges for $\map \Re s > 0$, a contradiction.
Category:Analytic Number Theory
L-Infinity Norm is Norm

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {L^\infty} {X, \Sigma, \mu}$ be the $L^\infty$ vector space on $\struct {X, \Sigma, \mu}$.
Let $\norm \cdot_\infty$ be the $L^\infty$ norm.
Then $\norm \cdot_\infty$ is a norm on $\map {L^\infty} {X, \Sigma, \mu}$.
== Proof ==
Let $\map {\LL^\infty} {X, \Sigma, \mu}$ be the Lebesgue $\infty$-space. 
Let $\sim$ be the $\mu$-almost everywhere equality relation on $\map {\LL^\infty} {X, \Sigma, \mu}$.
Let $\eqclass f \sim \in \map {L^\infty} {X, \Sigma, \mu}$. 
Then, we have by the definition of the $L^\infty$ norm we have: 
:$\norm {\eqclass f \sim}_\infty = \norm f_\infty$
From P-Seminorm is Seminorm, (in the case $p = \infty$) we have: 
:$\norm f_\infty \ge 0$
so:
:$\norm {\eqclass f \sim}_\infty \ge 0$ 
So $\norm \cdot_\infty$ is a map from $\map {L^\infty} {X, \Sigma, \mu}$ to the non-negative real numbers.
It remains to verify the norm axioms.
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From P-Seminorm of Function Zero iff A.E. Zero, we have: 
:$\norm f_\infty = 0$ for $f \in \map {\mathcal L^\infty} {X, \Sigma, \mu}$ $f = 0$ $\mu$-almost everywhere.
That is: 
:$\norm f_\infty = 0$ $f \sim 0$
That is, from Equivalence Class Equivalent Statements:
:$\norm f_\infty = 0$ $\eqclass f \sim = \eqclass 0 \sim = 0_{\map {L^p} {X, \Sigma, \mu} }$.
Then, for $\eqclass f \sim \in \map {L^\infty} {X, \Sigma, \mu}$, we have: 
:$\norm {\eqclass f \sim}_\infty = 0$ $\norm f_\infty = 0$ 
from the definition of the $L^\infty$ norm.
This is in turn equivalent to $\eqclass f \sim = 0_{\map {L^\infty} {X, \Sigma, \mu} }$
So we have positive definiteness.
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Let $\eqclass f \sim \in \map {L^\infty} {X, \Sigma, \mu}$ and $\lambda \in \R$. 
Then: 
So we have positive homogeneity.
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Let $\eqclass f \sim, \eqclass g \sim \in \map {L^\infty} {X, \Sigma, \mu}$. 
Then: 
so the triangle inequality holds.
Category:L-Infinity Norm
Category:Normed Vector Spaces