avewright/proofwiki-math huggingface dataset:
Theorem: Point at Infinity of Intersection of Parallel Lines

Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$ such that $\LL_1$ and $\LL_2$ are parallel.
By Condition for Straight Lines in Plane to be Parallel, $\LL_1$ and $\LL_2$ can be expressed as the general equations:
{{begin-eqn}}
{{eqn | lll = \LL_1:
      | l = l x + m y + n_1
      | r = 0
}}
{{eqn | lll = \LL_2:
      | l = l x + m y + n_2
      | r = 0
}}
{{end-eqn}}
The point at infinity of $\LL_1$ and $\LL_2$ can thence be expressed in homogeneous Cartesian coordinates as $\tuple {-m, l, 0}$.

Proof:
Let $\LL_1$ be expressed in the form:
:$l x + m y + n = 0$
Hence let $\LL_2$ be expressed in the form:
:$l x + m y + k n = 0$
where $k \ne 1$.
Let their point of intersection be expressed in homogeneous Cartesian coordinates as $\tuple {X, Y, Z}$
Then:
{{begin-eqn}}
{{eqn | l = \tuple {X, Y, Z}
      | r = \tuple {m n \paren {k - 1}, n l \paren {1 - k}, 0}
      | c = 
}}
{{eqn | r = \tuple {-m, l, 0}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Intersection of Straight Line in Homogeneous Cartesian Coordinates with Axes

Let $\LL$ be a straight line embedded in a cartesian plane $\CC$.
Let $\LL$ be given in homogeneous Cartesian coordinates by the equation:
:$l X + m Y + n Z = 0$
such that $l$ and $m$ are not both zero.
Then $\LL$ intersects:
:the $x$-axis $Y = 0$ at the point $\tuple {-n, 0, l}$
:the $y$-axis $X = 0$ at the point $\tuple {0, -n, m}$
When $l = 0$, $\LL$ is parallel to the $x$-axis with its point at infinity at $\tuple {-n, 0, 0}$
When $m = 0$, $\LL$ is parallel to the $y$-axis with its point at infinity at $\tuple {0, -n, 0}$.

Proof:
The intersection of $\LL$ with the $x$-axis is the point $\tuple {X, Y, Z}$ satisfied by:
{{begin-eqn}}
{{eqn | l = l X + m Y + n Z
      | r = 0
      | c = 
}}
{{eqn | l = Y
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = l X + n Z
      | r = 0
      | c = 
}}
{{end-eqn}}
which is satisfied by setting $X = -n$ and $Z = l$, while $Y = 0$.
The intersection of $\LL$ with the $y$-axis is the point $\tuple {X, Y, Z}$ satisfied by:
{{begin-eqn}}
{{eqn | l = l X + m Y + n Z
      | r = 0
      | c = 
}}
{{eqn | l = X
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = m Y + n Z
      | r = 0
      | c = 
}}
{{end-eqn}}
which is satisfied by setting $Y = -n$ and $Z = m$, while $X = 0$.
When $l = 0$ we have:
{{begin-eqn}}
{{eqn | l = m Y + n Z
      | r = 0
      | c = 
}}
{{eqn | l = Y
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n Z
      | r = 0
      | c = 
}}
{{end-eqn}}
When $m = 0$ we have:
{{begin-eqn}}
{{eqn | l = l X + n Z
      | r = 0
      | c = 
}}
{{eqn | l = X
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n Z
      | r = 0
      | c = 
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Equation of Straight Line through Intersection of Two Straight Lines

Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed using the general equations:
{{begin-eqn}}
{{eqn | lll = \LL_1:
      | l = l_1 x + m_1 y + n_1
      | r = 0
}}
{{eqn | lll = \LL_2:
      | l = l_2 x + m_2 y + n_2
      | r = 0
}}
{{end-eqn}}
Let $\LL_3$ be a third straight line embedded in $\CC$, passing through the point of intersection of $\LL_1$ and $\LL_2$.
$\LL_3$ can be expressed using the general equation:
:$(1): \quad \paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2} = 0$

Proof:
Let $P = \tuple {x, y}$ be the point of intersection of $\LL_1$ and $\LL_2$.
We have that:
{{begin-eqn}}
{{eqn | l = l_1 x + m_1 y + n_1
      | r = 0
}}
{{eqn | l = l_2 x + m_2 y + n_2
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = k \paren {l_2 x + m_2 y + n_2}
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2}
      | r = 0
}}
{{end-eqn}}
and so equation $(1)$:
:$\paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2} = 0$
is satisfied by the point $P$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2}
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \paren {l_1 - k l_2} x + \paren {m_1 - k m_2} y + \paren {n_1 - k n_2}
      | r = 0
}}
{{end-eqn}}
Each of $l_1 - k l_2$, $m_1 - k m_2$ and $n_1 - k n_2$ is a real number.
Hence $(1)$ is the equation of a straight line.
{{qed}}
Theorem: Pencil of Straight Lines through Intersection of Two Straight Lines

Let $u = l_1 x + m_1 y + n_1$.
Let $v = l_2 x + m_2 y + n_2$.
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed using the general equations:
{{begin-eqn}}
{{eqn | lll = \LL_1:
      | l = u
      | r = 0
}}
{{eqn | lll = \LL_2:
      | l = v
      | r = 0
}}
{{end-eqn}}
The pencil of lines through the point of intersection of $\LL_1$ and $\LL_2$ is given by:
:$\set {u + k v = 0: k \in \R} \cup \set {\LL_2}$

Proof:
Let $\LL$ denote an arbitrary straight line through the point of intersection of $\LL_1$ and $\LL_2$.
From Equation of Straight Line through Intersection of Two Straight Lines, $\LL$ can be given by an equation of the form:
:$u + k v = 0$
It remains to be seen that the complete pencil of lines through the point of intersection of $\LL_1$ and $\LL_2$ can be obtained by varying $k$ over the complete set of real numbers $\R$.
We have that $\LL$ can also be given by: 
:$\paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2} = 0$
That is:
:$\paren {l_1 - k l_2} x + \paren {m_1 - k m_2} y + \paren {n_1 - k n_2} = 0$
Let the slope of $\LL$ be $\tan \psi$ where $\psi$ is the angle $\LL$ makes with the $x$-axis.
{{finish}}
Theorem: Condition for Concurrency of Three Straight Lines

Let $3$ straight lines $\LL_1$, $\LL_2$ and $\LL_3$ be embedded in a cartesian plane $\CC$, expressed using the general equations:
{{begin-eqn}}
{{eqn | l = l_1 x + m_1 y + n_1
      | r = 0
}}
{{eqn | l = l_2 x + m_2 y + n_2
      | r = 0
}}
{{eqn | l = l_3 x + m_3 y + n_3
      | r = 0
}}
{{end-eqn}}
Then $\LL_1$, $\LL_2$ and $\LL_3$ are concurrent {{iff}}:
:$\begin {vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end {vmatrix} = 0$
where $\begin {vmatrix} \cdot \end {vmatrix}$ denotes a determinant.

Proof:
=== Necessary Condition ===
Let $\LL_1$, $\LL_2$ and $\LL_3$ be concurrent.
From Equation of Straight Line through Intersection of Two Straight Lines, $\LL_1$, $\LL_2$ and $\LL_3$ are concurrent {{iff}} $\LL_3$ has an equation of the form:
:$\paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2} = 0$
That is:
:$\paren {l_1 - k l_2} x + \paren {m_1 - k m_2} y + \paren {n_1 - k n_2} = 0$
Hence:
{{begin-eqn}}
{{eqn | l = \begin {vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end {vmatrix}
      | r = \begin {vmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \\ n_1 & n_2 & n_3 \end {vmatrix}
      | c = Determinant of Transpose
}}
{{eqn | r = \begin {vmatrix} l_1 & l_2 & \paren {l_1 - k l_2} \\ m_1 & m_2 & \paren {m_1 - k m_2} \\ n_1 & n_2 & \paren {n_1 - k n_2} \end {vmatrix}
      | c = Definition of $\LL_3$
}}
{{eqn | r = \begin {vmatrix} l_1 & l_2 & l_1 \\ m_1 & m_2 & m_1 \\ n_1 & n_2 & n_1 \end {vmatrix}
      | c = Multiple of Column Added to Column of Determinant: adding $k$ times column $2$ to column $3$
}}
{{eqn | r = 0
      | c = Square Matrix with Duplicate Columns has Zero Determinant
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Bisectors of Angles between Two Straight Lines/Normal Form

Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed in normal form as:
{{begin-eqn}}
{{eqn | lll = \LL_1:
      | l = x \cos \alpha + y \sin \alpha
      | r = p
}}
{{eqn | lll = \LL_2:
      | l = x \cos \beta + y \sin \beta
      | r = q
}}
{{end-eqn}}
The angle bisectors of the angles formed at the point of intersection of $\LL_1$ and $\LL_2$ are given by:
{{begin-eqn}}
{{eqn | l = x \paren {\cos \alpha - \cos \beta} + y \paren {\sin \alpha - \sin \beta}
      | r = p - q
}}
{{eqn | l = x \paren {\cos \alpha + \cos \beta} + y \paren {\sin \alpha + \sin \beta}
      | r = p + q
}}
{{end-eqn}}

Proof:
Let $A'SA$ and $B'SB$ be the straight lines $\LL_1$ and $\LL_2$ respectively, intersecting at the point $S$.
Let $P = \tuple {x, y}$ be an arbitrary point on either of the angle bisectors of $\angle ASB$ or $\angle BSA'$.
:500px
Drop perpendiculars $PM$ from $P$ to $SA$ and $PN$ from $P$ to $SB$.
Because:
:$\angle PSM = \angle PSN$
:$\angle PMS = \angle PNS$
:$PS$ is common
we have that:
:$\triangle PSM = \triangle PSN$
and so:
:$PM = PN$
Having regard only for the magnitude of the perpendiculars:
{{begin-eqn}}
{{eqn | l = PM
      | r = x \cos \alpha + y \sin \alpha - p
      | c = 
}}
{{eqn | l = PN
      | r = x \cos \beta + y \sin \beta - q
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x \paren {\cos \alpha - \cos \beta} + y \paren {\sin \alpha - \sin \beta}
      | r = p - q
      | c = 
}}
{{end-eqn}}
which is the equation of a straight line in normal form.
This represents either the angle bisector of $\angle ASB$ or the angle bisector of $\angle BSA'$.
Considering now the signs of the perpendiculars, it is seen that while $PM$ and $PM'$ have the same sign, $PN$ and $PN'$ have opposite signs.
Hence:
:for one bisector we have $PM = PN$
:for the other bisector we have $PM = -PN$.
So the equation for the other bisector is $x \paren {\cos \alpha + \cos \beta} + y \paren {\sin \alpha + \sin \beta} = p + q$.
{{qed}}
Theorem: Fourier Transform of 1

Let:
:$\map f x = 1$
Then:
:$\map {\hat f} s = \map \delta s$
where $\map {\hat f} s$ is the Fourier transform of $\map f x$.

Proof:
By the definition of a Fourier transform:
{{begin-eqn}}
{{eqn | l = \map {\hat f} s
      | r = \int_{-\infty}^\infty e^{-2 \pi i x s} \map f x \rd x
      | c = 
}}
{{eqn | r = \int_{-\infty}^\infty e^{-2 \pi i x s} 1 \rd x
      | c = 
}}
{{eqn | r = \int_{-\infty}^\infty \paren {\map \cos {2 \pi s x} - i \map \sin {2 \pi s x} } \rd x
      | c = Euler's Formula/Corollary
}}
{{eqn | r = \int_{-\infty}^\infty \map \cos {2 \pi s x} \rd x - i \int_{-\infty}^\infty \map \sin {2 \pi s x } \rd x
      | c = Linear Combination of Definite Integrals
}}
{{eqn | r = \dfrac 1 {2 \pi s} \paren {\paren {\lim_{\gamma \mathop \to +\infty} \bigintlimits {\map \sin {2 \pi s x } } {-\gamma} \gamma} - i \paren {\lim_{\gamma \mathop \to +\infty} \bigintlimits {-\map \cos {2 \pi s x } } {-\gamma} \gamma} }
      | c = Primitive of Sine Function and Primitive of Cosine Function
}}
{{eqn | r = \dfrac 1 {2 \pi s} \paren {\paren {\lim_{\gamma \mathop \to +\infty} \bigintlimits {\map \sin {2 \pi s x } } {-\gamma} \gamma} + i \paren {\lim_{\gamma \mathop \to +\infty} \bigintlimits {\map \cos {2 \pi s x } } {-\gamma} \gamma} }
      | c = 
}}
{{eqn | r = \dfrac 1 {2 \pi s} \paren {\paren {\lim_{\gamma \mathop \to +\infty} \bigintlimits {\map \sin {2 \pi s x } } {-\gamma} \gamma} + 0 }
      | c = Cosine of Conjugate Angle: $\map \cos {-x} = \map {\cos} x$
}}
{{eqn | r = \dfrac 1 {2 \pi s} \lim_{\gamma \mathop \to +\infty} 2 \map \sin {2 \pi s \gamma}
      | c = Sine of Conjugate Angle: $\map \sin {-x} = -\map \sin x$
}}
{{eqn | r = \dfrac 1 {\pi s} \lim_{\epsilon \mathop \to 0} \map \sin {\frac {2 \pi s} \epsilon}
      | c = Let $\epsilon = \dfrac 1 \gamma $
}}
{{eqn | r = \map \delta s
      | c = {{Defof|Dirac Delta Function}}: Limit 5
}}
{{end-eqn}}
{{qed}}
Theorem: Integral of Dirac Delta Function over Reals

Let $\map \delta x$ denote the Dirac delta function.
Then:
:$\ds \int_{-\infty}^{+\infty} \map \delta x \rd x = 1$

Proof:
We have that:
:$\map \delta x = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
where:
:$\map {F_\epsilon} x = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$
We have that:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^{+\infty} \map {F_\epsilon} x \rd x
      | r = \int_{-\infty}^{-\epsilon} 0 \rd x + \int_{-\epsilon}^\epsilon \dfrac 1 {2 \epsilon} \rd x + \int_\epsilon^\infty 0 \rd x
      | c = {{Defof|Dirac Delta Function|index = 2}}
}}
{{eqn | r = \lim_{L \mathop \to -\infty} \int_L^{-\epsilon} 0 \rd x + \int_{- \epsilon}^\epsilon \dfrac 1 {2 \epsilon} \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \rd x
      | c = {{Defof|Improper Integral}}
}}
{{eqn | r = \lim_{L \mathop \to -\infty} \paren {0 \times \paren {- \epsilon - L} } + \dfrac 1 {2 \epsilon} \paren {\epsilon - \paren {-\epsilon} } + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \epsilon} }
      | c = Definite Integral of Constant
}}
{{eqn | r = \lim_{L \mathop \to -\infty} 0 + 1 + \lim_{L \mathop \to \infty} 0
      | c = simplification
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^{+\infty} \map \delta x \rd x
      | r = \lim_{\epsilon \mathop \to 0} \int_{-\infty}^{+\infty} \map {F_\epsilon} x \rd x
      | c = {{Defof|Dirac Delta Function|index = 2}}
}}
{{eqn | r = \lim_{\epsilon \mathop \to 0} 1
      | c = from above
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Integral of Dirac Delta Function by Continuous Function over Reals

Let $\map \delta x$ denote the Dirac delta function.
Let $g$ be a continuous real function.
Then:
:$\ds \int_{- \infty}^{+ \infty} \map \delta x \map g x \rd x = \map g 0$

Proof:
We have that:
:$\map \delta x = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
where:
:$\map {F_\epsilon} x = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$
We have that:
{{begin-eqn}}
{{eqn | l = \int_{- \infty}^{+ \infty} \map {F_\epsilon} x \map g x \rd x
      | r = \int_{-\infty}^{-\epsilon} 0 \times \map g x \rd x + \int_{-\epsilon}^\epsilon \dfrac 1 {2 \epsilon} \map g x \rd x + \int_\epsilon^\infty 0 \times \map g x \rd x
      | c = {{Defof|Dirac Delta Function|index = 2}}
}}
{{eqn | r = \lim_{L \mathop \to -\infty} \int_L^{-\epsilon} 0 \times \map g x \rd x + \int_{-\epsilon}^\epsilon \dfrac 1 {2 \epsilon} \, \map g x \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \times \map g x \rd x
      | c = {{Defof|Improper Integral}}
}}
{{eqn | r = \lim_{L \mathop \to -\infty} \int_L^{-\epsilon} 0 \rd x + \dfrac 1 {2 \epsilon} \int_{-\epsilon}^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \rd x
      | c = 
}}
{{eqn | r = \lim_{L \mathop \to -\infty} \paren {0 \times \paren {-\epsilon - L} } + \dfrac 1 {2 \epsilon} \int_{-\epsilon}^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \epsilon} }
      | c = Definite Integral of Constant
}}
{{eqn | r = \lim_{L \mathop \to -\infty} 0 + \dfrac 1 {2 \epsilon} \int_{-\epsilon}^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} 0
      | c = simplification
}}
{{eqn | r = \dfrac 1 {2 \epsilon} \int_{-\epsilon}^\epsilon \map g x \rd x
      | c = 
}}
{{end-eqn}}
From Darboux's Theorem:
:$\ds m \paren {\epsilon - \paren {-\epsilon} } \le \int_{-\epsilon}^\epsilon \map g x \rd x \le M \paren {\epsilon - \paren {-\epsilon}}$
where:
:$M$ is the maximum of $\map g x$
:$m$ is the minimum of $\map g x$
on $\closedint {-\epsilon} \epsilon$.
Hence:
:$\ds 2 m \epsilon \le \int_{-\epsilon}^\epsilon \map g x \rd x \le 2 M \epsilon$
and so dividing by $2 \epsilon$:
:$\ds m \le \dfrac 1 {2 \epsilon} \int_{-\epsilon}^\epsilon \map g x \rd x \le M$
Then:
:$\ds \lim_{\epsilon \mathop \to 0} M = m = \map g 0$
and so by the Squeeze Theorem:
:$\ds \lim_{\epsilon \mathop \to 0} \int_{-\infty}^{+\infty} \map {F_\epsilon} x \map g x \rd x = \map g 0$
But by {{Defof|Dirac Delta Function|index = 2}}:
:$\ds \lim_{\epsilon \mathop \to 0} \int_{-\infty}^{+\infty} \map {F_\epsilon} x \map g x \rd x = \ds \int_{-\infty}^{+\infty} \map \delta x \map g x \rd x$
Hence the result.
{{qed}}
Category:Dirac Delta Function
s622viumgx45fgh3l6s93xrnpat8fls
Theorem: Filter Property of Dirac Delta Function

Let $\map \delta x$ denote the Dirac delta function.
Let $g$ be a continuous real function.
Then:
:$\ds \int_{-\infty}^{+ \infty} \map {\delta} {x - s} \map g x \rd x = \map g s$

Proof:
We have that:
:$\map \delta x = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
where:
:$\map {F_\epsilon} x = \begin{cases} 0 & : x < -\epsilon + s \\ \dfrac 1 {2 \epsilon} & : -\epsilon + s \le x \le \epsilon + s \\ 0 & : x > \epsilon + s \end{cases}$
We have that:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^{+ \infty} \map {F_\epsilon} x \map g x \rd x
      | r = \int_{-\infty}^{-\epsilon + s} 0 \times \map g x \rd x + \int_{-\epsilon + s}^{\epsilon + s} \dfrac 1 {2 \epsilon} \map g x \rd x + \int_{\epsilon + s}^\infty 0 \times \map g x \rd x
      | c = Definition of $F_\epsilon$
}}
{{eqn | r = \lim_{L \mathop \to -\infty} \int_L^{-\epsilon + s} 0 \times \map g x \rd x + \int_{-\epsilon + s}^{\epsilon + s} \dfrac 1 {2 \epsilon} \map g x \rd x + \lim_{L \mathop \to \infty} \int_{\epsilon + s}^L 0 \times \map g x \rd x
      | c = {{Defof|Improper Integral}}
}}
{{eqn | r = \lim_{L \mathop \to -\infty} \int_L^{-\epsilon + s} 0 \rd x + \dfrac 1 {2 \epsilon} \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x + \lim_{L \mathop \to \infty} \int_{\epsilon + s}^L 0 \rd x
      | c = 
}}
{{eqn | r = \lim_{L \mathop \to -\infty} \paren {0 \times \paren {-\epsilon + s - L} } + \dfrac 1 {2 \epsilon} \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \epsilon - s} }
      | c = Definite Integral of Constant
}}
{{eqn | r = \lim_{L \mathop \to -\infty} 0 + \dfrac 1 {2 \epsilon} \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x + \lim_{L \mathop \to \infty} 0
      | c = simplification
}}
{{eqn | r = \dfrac 1 {2 \epsilon} \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x
      | c = 
}}
{{end-eqn}}
From Darboux's Theorem:
:$\ds m \paren {\epsilon + s - \paren {-\epsilon + s} } \le \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x \le M \paren {\epsilon + s - \paren {-\epsilon + s} }$
where:
:$M$ is the maximum of $\map g x$
:$m$ is the minimum of $\map g x$
on $\closedint {-\epsilon + s} {\epsilon + s}$.
Hence:
:$\ds 2 m \epsilon \le \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x \le 2 M \epsilon$
and so dividing by $2 \epsilon$:
:$\ds m \le \dfrac 1 {2 \epsilon} \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x \le M$
Then:
:$\ds \lim_{\epsilon \mathop \to 0} M = m = \map g s$
and so by the Squeeze Theorem:
:$\ds \lim_{\epsilon \mathop \to 0} \int_{-\infty}^{+ \infty} \map {F_\epsilon} x \map g x \rd x = \map g s$
But by definition of the Dirac delta function:
:$\ds \lim_{\epsilon \mathop \to 0} \int_{-\infty}^{+ \infty} \map {F_\epsilon} x \map g x \rd x = \int_{- \infty}^{+ \infty} \map \delta x \map g x \rd x$
Hence the result.
{{qed}}
Theorem: Derivative Operator on Continuously Differentiable Function Space with Supremum Norm is not Continuous

Let $I = \closedint 0 1$ be a closed real interval.
Let $\map C I$ be the real-valued, continuous on $I$ function space.
Let $\map {C^1} I$ be the continuously differentiable function space.
Let $x \in \map {C^1} I$ be a continuously differentiable real-valued function.
Let $D : \map {C^1} I \to \map C I$ be the derivative operator such that:
:$\forall t \in \closedint 0 1 : \map {Dx} t := \map {x'} t$
Suppose $\map C I$ and $\map {C^1} I$ are equipped with the supremum norm.
Then $D$ is not continuous.

Proof:
{{AimForCont}} $D$ is continuous.
We have that the Derivative Operator is Linear Mapping.
By Continuity of Linear Transformation between Normed Vector Spaces:
:$\exists M \in \R_{> 0} : \forall x \in \map {C^1} I : \norm {\map D x}_\infty \le M \norm x_\infty$
Suppose $x = t^n$ with $n \in \N$.
Then:
:$\norm x_\infty = \norm {t^n}_\infty = 1$
:$\norm {x'}_\infty = \norm {n t^{n-1}}_\infty = n$
Hence:
{{begin-eqn}}
{{eqn | l = \norm {D x}_\infty
      | r = \norm {x'}_\infty 
}}
{{eqn | r = n 
}}
{{eqn | o = \le 
      | r = M \norm x_\infty
}}
{{eqn | r = M \cdot 1
}}
{{end-eqn}}
In other words:
:$\forall n \in \N : n \le M$
But $M$ is finite.
This is a contradiction.
Hence, $D$ is not continuous.
{{qed}}
Theorem: Derivative Operator on Continuously Differentiable Function Space with C^1 Norm is Continuous

Let $I = \closedint 0 1$ be a closed real interval.
Let $\map C I$ be the real-valued, continuous on $I$ function space.
Let $\map {C^1} I$ be the continuously differentiable function space.
Let $x \in \map {C^1} I$ be a continuously differentiable real-valued function.
Let $D : \map {C^1} I \to \map \CC I$ be the derivative operator such that:
:$\forall t \in \closedint 0 1 : \map {D x} t := \map {x'} t$
Suppose $\map C I$ and $\map {C^1} I$ are equipped with $C^0$ and $C^1$ norms respectively
Then $D$ is continuous.

Proof:
{{begin-eqn}}
{{eqn | l = \norm {D x}_\infty
      | r = \norm {x'}_\infty 
}}
{{eqn | o = \le 
      | r = \norm {x'}_\infty + \norm x_\infty 
}}
{{eqn | r = \norm x_{1, \infty}
}}
{{end-eqn}}
We have that the Derivative Operator is Linear Mapping.
By definition and Continuity of Linear Transformation between Normed Vector Spaces, $D$ is continuous.
{{qed}}
Theorem: Mean Value Theorem for Holomorphic Functions

Let $D$ be a region.
Let $f : D \to \C$ be a holomorphic function.
Let $z \in D$. 
Let $r$ be such that $\map {B_r} z \subseteq D$. 
Then:
:$\ds \map f z = \frac 1 {2 \pi} \int_0^{2 \pi} \map f {z + r e^{i \theta} } \rd \theta$

Proof:
By Cauchy's Integral Formula, we have: 
:$\ds \map f z = \frac 1 {2 \pi i} \oint_{\partial \map {B_r} z} \frac {\map f t} {t - z} \rd t$
where $\partial \map {B_r} z$ is the boundary of $\map {B_r} z$. 
That is, $\partial \map {B_r} z$ is the circle of radius $r$, centred at $z$.
Note that we can parameterise $\partial \map {B_r} z$ by the function $\gamma : \hointr 0 {2 \pi} \to \partial \map {B_r} z$ defined by: 
:$\map \gamma \theta = z + r e^{i \theta}$
for each $\theta \in \hointr 0 {2 \pi}$.
Then, by the definition of a contour integral, we have: 
{{begin-eqn}}
{{eqn | l = \frac 1 {2 \pi i} \oint_{\partial \map {B_r} z} \frac {\map f t} {t - z} \rd t
      | r = \frac 1 {2 \pi i} \int_0^{2 \pi} \paren {i r e^{i \theta} \times \frac {\map f {z + r e^{i \theta} } } {z + r e^{i \theta} - z} } \rd \theta
      | c = Derivative of Exponential Function
}}
{{eqn | r = \frac i {2 \pi i} \int_0^{2 \pi} \map f {z + r e^{i \theta} } \paren {r e^{i \theta} \times \frac 1 {r e^{i \theta} } } \rd \theta
}}
{{eqn | r = \frac 1 {2 \pi} \int_0^{2 \pi} \map f {z + r e^{i \theta} } \rd \theta
}}
{{end-eqn}}
{{qed}}
Category:Mean Value Theorem
Category:Holomorphic Functions
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