nyuuzyou/wikis huggingface dataset:
Maximum Likelihood Estimator for Variance of Normal Distribution

== Theorem ==
Let $X$ be a continuous random variable with a normal distribution.
Let $\sigma^2$ be the variance of $X$.
Let $S = \set {x_1, x_2, \ldots, x_n}$ be a sample from $X$.
Let $\overline {\sigma^2}$ be the variance of $S$.
$\overline {\sigma^2}$ itself is a biased estimator.
Hence the maximum likelihood estimator for $\sigma^2$ $X$ needs to be $\dfrac n {n - 1} \overline {\sigma^2}$.
== Proof ==
== Sources ==
* 
* 
Category:Maximum Likelihood Estimators
Maximum Likelihood Estimator from Likelihood Function

== Theorem ==
Let $\FF$ be a one-parameter family of probability distributions whose parameter is $\theta$.
Let $X$ be a continuous random variable belonging to a member of $\FF$.
Let $\map {\mathrm L} \theta$ be the likelihood function of $\theta$ $X$.
Let $\EE$ be a maximum likelihood estimator for $\theta$ $X$.
Then $\EE$ can be found by:
:calculating $\theta$ for which $\map {\dfrac \d {\d \theta} } {\map \ln {\map {\mathrm L} \theta} } = 0$
:determine which of these solutions corresponds to a maximum.
It is often the case that an iterative solution is needed.
== Proof ==
== Sources ==
* 
* 
Category:Maximum Likelihood Estimators
Maximum Likelihood Estimator is Function of Sufficient Statistic

== Theorem ==
Let $P$ be a population.
Let $\theta$ be a parameter describing $P$.
Let there exist a sufficient statistic $s$ for the estimation of $\theta$.
Then the maximum likelihood estimator for $\theta$ is a function of $s$.
== Proof ==
== Sources ==
* 
* 
Category:Sufficient Statistics
Category:Maximum Likelihood Estimators
Maximum Modulus Principle

== Theorem ==
Let $D$ be an open region of the complex plane $\C$.
Let $f: D \to \C$ be a non-constant holomorphic function.
Then $\cmod f$ does not have any maximum points in the interior of $D$.
That is, for each $z \in D$ and $\delta > 0$, there exists some $\omega \in \map {B_\delta} z \cap D$, such that:
:$\cmod {\map f \omega} > \cmod {\map f z}$
== Proof ==
Pick some $r > 0$ such that $\map {B_r} z \subset D$.
By the Mean Value Theorem for Holomorphic Functions:
:$\ds \map f z = \dfrac 1 {2 \pi} \int_0^{2 \pi} \map f {z + r e^{i \theta} } \rd \theta$
Then:
So it must be that there exists $\omega \in \map {C_r} z$ such that:
:$\cmod {\map f z} \le \cmod {\map f \omega}$
where $\map {C_r} z$ is the circle of radius $r$ centered at $z$.
Note that equality is only obtained when $\cmod f$ is constant on $\map {C_r} z$.
However, since this holds for all sufficiently small $r > 0$, $\cmod f$ would be constant in $\map {B_r} z$.
Then $f$ must be constant in $D$, contradicting our assumption.
It follows that there exists $\omega \in \map {C_r} z$ such that: 
:$\cmod {\map f z} < \cmod {\map f \omega}$
== Also known as ==
== Sources ==
* 
Category:Maximum Modulus Principle
Category:Complex Analysis
Category:Named Theorems
Maximum Number of Arcs in Digraph

== Theorem ==
Let $D_n$ be a digraph of order $n$ such that $n \ge 1$.
Let $D_n$ have the greatest number of arcs of all digraphs of order $n$.
The number of arcs in $D$ is given by:
:$\size {D_n} = n \paren {n - 1}$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\size {D_n} = n \paren {n - 1}$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\size {D_k} = k \paren {k - 1}$
from which it is to be shown that:
:$\size {D_{k + 1} } = \paren {k + 1} k$
=== Induction Step ===
This is the induction step:
Let $D_{k + 1}$ be constructed by adding a new vertex $v_{k + 1}$ to $D_k$.
To do so, it is necessary to add $2$ arcs to join $v_{k + 1}$ to every vertex of $D_k$.
Thus there are a total of $2 k$ arcs more in $D_{k + 1}$ than there are in $D_k$.
So:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \size {D_n} = n \paren {n - 1}$
== Examples ==
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.5$: Directed Graphs as Mathematical Models: Problem $33 \ \text {(d)}$
Category:Digraphs
Category:Maximum Number of Arcs in Digraph
Category:Proofs by Induction
Maximum Number of Arcs in Digraph/Examples/Order 3

== Examples of Use of Maximum Number of Arcs in Digraph ==
Let $D$ be the digraph of order $3$ whose edge set $E$ is as large as possible.
Then the number of arcs of $G$ is given by:
:$\size E = 6$
== Proof ==
By Maximum Number of Arcs in Digraph:
:$\size E = 3 \times \paren {3 - 1} = 6$
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.5$: Directed Graphs as Mathematical Models: Problem $33 \ \text {(a)}$
Category:Maximum Number of Arcs in Digraph
Maximum Number of Arcs in Digraph/Examples/Order 4

== Examples of Use of Maximum Number of Arcs in Digraph ==
Let $D$ be the digraph of order $4$ whose edge set $E$ is as large as possible.
Then the number of arcs of $G$ is given by:
:$\size E = 12$
== Proof ==
By Maximum Number of Arcs in Digraph:
:$\size E = 4 \times \paren {4 - 1} = 12$
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.5$: Directed Graphs as Mathematical Models: Problem $33 \ \text {(b)}$
Category:Maximum Number of Arcs in Digraph
Maximum Number of Arcs in Digraph/Examples/Order 5

== Examples of Use of Maximum Number of Arcs in Digraph ==
Let $D$ be the digraph of order $5$ whose edge set $E$ is as large as possible.
Then the number of arcs of $G$ is given by:
:$\size E = 20$
== Proof ==
By Maximum Number of Arcs in Digraph:
:$\size E = 5 \times \paren {5 - 1} = 20$
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.5$: Directed Graphs as Mathematical Models: Problem $33 \ \text {(c)}$
Category:Maximum Number of Arcs in Digraph
Maximum Period of Reciprocal of Prime

== Theorem ==
Let $p$ be a prime number such that $p$ is not a divisor of $10$.
The maximum period of recurrence of the reciprocal of $p$ when expressed in decimal notation is $p - 1$.
== Proof ==
When $p \divides 10$, $\dfrac 1 p$ expressed in decimal notation is a terminating fraction:
So let $p$ be such that $p \nmid 10$.
From Period of Reciprocal of Prime, the period of recurrence is the order of $10$ modulo $p$.
That is, it is the smallest integer $d$ such that:
:$10^d \equiv 1 \pmod p$
From Fermat's Little Theorem:
:$10^{p - 1} \equiv 1 \pmod p$
Hence the maximum period of recurrence occurs when the order of $10$ modulo $p$ is $p - 1$.
== Also see ==
* $7$ is Long Period Prime: demonstrating that the maximum period of recurrence occurs for at least one prime number
Category:Reciprocals
Category:Prime Numbers
Maximum Rate of Change of X Coordinate of Cycloid

== Theorem ==
Let a circle $C$ of radius $a$ roll without slipping along the x-axis of a cartesian plane at a constant speed such that the center moves with a velocity $\mathbf v_0$ in the direction of increasing $x$.
Consider a point $P$ on the circumference of this circle.
Let $\tuple {x, y}$ be the coordinates of $P$ as it travels over the plane.
The maximum rate of change of $x$ is $2 \mathbf v_0$, which happens when $P$ is at the top of the circle $C$.
== Proof ==
From Rate of Change of Cartesian Coordinates of Cycloid, the rate of change of $x$ is given by:
:$\dfrac {\d x} {\d t} = \mathbf v_0 \paren {1 - \cos \theta}$
This is a maximum when $1 - \cos \theta$ is a maximum.
That happens when $\cos \theta$ is at a minimum.
That happens when $\cos \theta = -1$.
That happens when $\theta = \pi, 3 \pi, \ldots$
That is, when $\theta = \paren {2 n + 1} \pi$ where $n \in \Z$.
That is, when $P$ is at the top of the circle $C$.
When $\cos \theta = -1$ we have:
Hence the result.
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $4 \ \text{(b)}$
Category:Cycloids
Maximum Rate of Change of Y Coordinate of Astroid

== Theorem ==
Let $C_1$ and $C_2$ be the epicycle and deferent respectively of an astroid $H$.
Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.
Let the center $C$ of $C_1$ move at a constant angular velocity $\omega$ around the center of $C_2$.
Let $P$ be the point on the circumference of $C_1$ whose locus is $H$.
Let $C_1$ be positioned at time $t = 0$ so that $P$ its point of tangency to $C_2$, located on the $x$-axis.
Let $\theta$ be the angle made by $OC$ to the $x$-axis at time $t$.
Then the maximum rate of change of the $y$ coordinate of $P$ in the first quadrant occurs when $P$ is at the point where:
:$x = a \paren {\dfrac 1 3}^{3/2}$
:$y = a \paren {\dfrac 2 3}^{3/2}$
== Proof ==
:
The rate of change of $\theta$ is given by:
:$\omega = \dfrac {\d \theta} {\d t}$
From Equation of Astroid: Parametric Form, the point $P = \tuple {x, y}$ is described by the parametric equation:
:$\begin {cases}
x & = a \cos^3 \theta \\
y & = a \sin^3 \theta
\end{cases}$
The rate of change of $y$ is given by:
By Derivative at Maximum or Minimum, when $\dfrac {\d y} {\d t}$ is at a maximum:
:$\dfrac {\d^2 y} {\d t^2} = 0$
Thus:
Hence:
We can assume $\sin \theta \ne 0$ because in that case $\theta = 0$ and so $\dfrac {\d y} {\d t} = 0$.
Thus when $\sin \theta = 0$, $y$ is not a maximum.
So we can divide by $\sin \theta$ to give:
We have:
From Equation of Astroid: Cartesian Form:
:$x^{2/3} + y^{2/3} = a^{2/3}$
Hence:
Similarly:
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $10$
Category:Astroids
Maximum Rate of Change of Y Coordinate of Cycloid

== Theorem ==
Let a circle $C$ of radius $a$ roll without slipping along the x-axis of a cartesian plane at a constant speed such that the center moves with a velocity $\mathbf v_0$ in the direction of increasing $x$.
Consider a point $P$ on the circumference of this circle.
Let $\tuple {x, y}$ be the coordinates of $P$ as it travels over the plane.
The maximum rate of change of $y$ is $\mathbf v_0$, which happens when $\theta = \dfrac \pi 2 + 2 n \pi$ where $n \in \Z$.
== Proof ==
From Rate of Change of Cartesian Coordinates of Cycloid, the rate of change of $y$ is given by:
:$\dfrac {\d y} {\d t} = \mathbf v_0 \sin \theta$.
This is a maximum when $\sin \theta$ is a maximum.
That happens when $\sin \theta = 1$.
That happens when $\theta = \dfrac \pi 2 + 2 n \pi$ where $n \in \Z$.
When $\sin \theta = 1$ we have:
:$\dfrac {\d y} {\d t} = \mathbf v_0$
Hence the result.
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $4 \ \text{(c)}$
Category:Cycloids
Maximum Rule for Continuous Functions

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $f, g: S \to \R$ be continuous real-valued functions.
Let $\max \set {f, g}: S \to \R$ denote the pointwise maximum of $f$ and $g$.
Then:
:$\max \set {f, g}$ is continuous.
== Proof ==
From Sum Less Minimum is Maximum:
:$\forall x \in S : \max \set {\map f x, \map g x} = \map f x + \map g x - \min \set {\map f x, \map g x}$
Thus: 
:$\max \set {f, g} = f + g - \min \set{f, g}$
From Minimum Rule for Continuous Functions:
:$\min \set {f, g}$ is continuous
From Multiple Rule for Continuous Mappings into Topological Ring:
:$-\min \set {f, g}$ is continuous 
From Sum Rule for Continuous Mappings into Topological Ring:
:$f + g - \min \set {f, g}$ is continuous 
Thus:
:$\max \set {f, g}$ is continuous 
== Also see ==
* Minimum Rule for Continuous Functions
Category:Max Operation
Category:Continuous Mappings
Maximum Rule for Real Sequences

== Theorem == 
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
:$\ds \lim_{n \mathop \to \infty} \max \set {x_n, y_n} = \max \set {l, m}$
== Proof ==
=== Case $1$: $l = m$ ===
Let $l = m$.
Then:
:$\max \set {l, m} = l = m$
Let $\epsilon > 0$ be given.
By definition of the limit of a real sequence, we can find $N_1$ such that:
:$\forall n > N_1: \size {x_n - l} < \epsilon$
Similarly we can find $N_2$ such that:
:$\forall n > N_2: \size {y_n - m} < \epsilon$
Let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true:
:$n > N_1$
:$n > N_2$
Thus $\forall n > N$:
:if $y_n < x_n$ then:
::$\size {\max \set {x_n, y_n} - \max \set {l, m} } = \size {x_n - l} < \epsilon$
:if $x_n \le y_n$ then:
::$\size {\max \set {x_n, y_n} - \max \set {l, m} } = \size {y_n - m} < \epsilon$
In either case:
: $\size {\max \set {x_n, y_n} - \max \set {l, m} } < \epsilon$
The result follows.
=== Case 2: $l > m$ ===
Let $l > m$.
Then:
:$l = \max \set {l, m}$
Let $\delta = \dfrac {l - m} 2$.
Then:
:$\delta > 0$
Let $\epsilon > 0$ be given.
Let $\epsilon' = \min \set {\delta, \epsilon}$.
Then:
:$\epsilon' > 0$
By definition of the limit of a real sequence, we can find $N_1$ such that:
:$\forall n > N_1: \size {x_n - l} < \epsilon'$
Similarly we can find $N_2$ such that:
:$\forall n > N_2: \size {y_n - m} < \epsilon'$
Let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true:
:$n > N_1$
:$n > N_2$
From Corollary 3 of Negative of Absolute Value, we have that $\forall n > N$:
:$\size {x_n - l} < \epsilon' \implies x_n > l - \epsilon'$
:$\size {y_n - m} < \epsilon' \implies y_n < m + \epsilon'$
As $\epsilon' \le \delta$:
and:
So:
:$x_n > l - \epsilon' \ge \dfrac {l + m} 2 \ge m + \epsilon' > y_n$
Hence:
:$x_n = \max \set{x_n, y_n}$
So $\forall n > N$:
:$\size {\max \set {x_n, y_n} - \max \set {l, m} } = \size {x_n - l} < \epsilon' \le \epsilon$
The result follows.
=== Case 3: $m > l$ ===
Similar to Case 2 by interchanging $l$ with $m$ and $x_n$ with $y_n$.
== Also see ==
* Minimum Rule for Real Sequences
Category:Real Sequences
Maximum Speed of Rotation of Plane of Oscillation of Foucault's Pendulum

== Theorem ==
Let $P$ be a Foucault pendulum.
The maximum angular speed of the plane of oscillation of $P$ occurs at Earth's poles.
== Proof ==
From Angular Speed of Rotation of Plane of Oscillation of Foucault's Pendulum, the angular speed $\alpha$ of the plane of oscillation of $P$ is given by:
:$\alpha = \omega \sin \lambda$ where:
:$\lambda$ denotes the latitude on Earth at which $P$ is located
:$\omega$ denotes the angular speed of rotation of Earth.
So $\size \alpha$ reaches its maximum when $\sin \lambda = \pm 1$.
That is, when $\lambda = \pm 90 \degrees$.
This happens at Earth's poles.
== Sources ==
* 
* 
Category:Foucault's Pendulum
Maximum Volume of Unit Radius Sphere in Fractional Dimensions

== Theorem ==
The volume of a unit sphere in $x$-dimensional Euclidean space for real $x$ occurs when $x$ is given as:
:$x = 5 \cdotp 25694 \, 64048 \, 60 \ldots$
The corresponding volume at that dimension is given by:
:$V = 5 \cdotp 27776 \, 80211 \, 13400 \, 997 \ldots$
== Proof ==
== Historical Note ==
== Sources ==
* : $5 \cdotp 256 \, 946 \, 404 \, 860 \ldots$
* : $5 \cdotp 25694 \, 64048 \, 60 \ldots$
Category:Geometry
Maximum Weight Edge in all Minimum Spanning Trees is Bridge

== Theorem ==
Let $G$ be an undirected network.
Let every edge of $G$ have a unique weight.
Let $e$ be an edge of $G$ that belongs to every minimum spanning tree of $G$.
Let $e$ have maximum weight in $G$.
Then $e$ is a bridge in $G$.
 
== Proof ==
Category:Network Theory
Maximum is Greater than or Equal to Geometric Mean

== Theorem ==
Let $x_1, x_2, \ldots, x_n \in \R_{\ge 0}$ be positive real numbers.
Let $p \in \R$ be a real number.
Let $\map G {x_1, x_2, \ldots, x_n}$ denote the geometric mean of $x_1, x_2, \ldots, x_n$.
Then:
:$\max \set {x_1, x_2, \ldots, x_n} \ge \map G {x_1, x_2, \ldots, x_n}$
Equality holds :
:$x_1 = x_2 = \cdots x_n$
or $x_k = 0$ for some $k \in \set {1, 2, \ldots, n}$.
== Proof ==
From Limit of Hölder Mean as Exponent tends to Zero is Geometric Mean:
:$\map G {x_1, x_2, \ldots, x_n} = \ds \lim_{p \mathop \to 0} \map {M_p} {x_1, x_2, \ldots, x_n}$
where $\map {M_p} {x_1, x_2, \ldots, x_n}$ denotes the Hölder mean with exponent $p$ of $x_1, x_2, \ldots, x_n$.
Then from Maximum is Greater than or Equal to Hölder Mean:
$\max \set {x_1, x_2, \ldots, x_n} \ge \map {M_p} {x_1, x_2, \ldots, x_n}$
where equality holds :
:$x_1 = x_2 = \cdots x_n$
or:
:$p < 0$ and $x_k = 0$ for some $k \in \set {1, 2, \ldots, n}$.
In this context, if such an $x_k = 0$ then:
:$\ds \prod_{k \mathop = 1}^n x_k = 0$
and so by definition:
$\map G {x_1, x_2, \ldots, x_n} = 0$
Thus we have:
:$\max \set {x_1, x_2, \ldots, x_n} \ge \map G {x_1, x_2, \ldots, x_n}$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.2$ Inequalities: Relation Between Arithmetic, Geometric, Harmonic and Generalized Means: $3.2.3$
Category:Geometric Mean
Maximum is Greater than or Equal to Hölder Mean

== Theorem ==
Let $x_1, x_2, \ldots, x_n \in \R_{\ge 0}$ be positive real numbers.
Let $p \in \R$ be a real number.
Let $\map {M_p} {x_1, x_2, \ldots, x_n}$ denote the Hölder mean with exponent $p$ of $x_1, x_2, \ldots, x_n$.
Then:
:$\max \set {x_1, x_2, \ldots, x_n} \ge \map {M_p} {x_1, x_2, \ldots, x_n}$
Equality holds :
:$x_1 = x_2 = \cdots x_n$
or:
:$p < 0$ and $x_k = 0$ for some $k \in \set {1, 2, \ldots, n}$.
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.2$ Inequalities: Relation Between Arithmetic, Geometric, Harmonic and Generalized Means: $3.2.2$
Category:Hölder Mean
Maximum of Seven Colors Needed for Proper Vertex Coloring on Torus

== Theorem ==
Let $G$ be a graph embedded on the surface of a torus.
$G$ can be assigned a proper vertex $k$-coloring such that $k \le 7$.
== Proof ==
== Historical Note ==
== Sources ==
* : $7$
* : $7$
* 
* 
Category:Graph Theory
Category:7
Maximum of Three Mutually Perpendicular Lines in Ordinary Space

== Theorem ==
In ordinary space, there can be no more than $3$ straight lines which are pairwise perpendicular.
Thus, in a configuration of $4$ straight lines in space, at least one pair will not be perpendicular to each other.
== Proof ==
By assumption and popular belief, ordinary space (on the measurable local level) is an instance of a $3$-dimensional Euclidean space.
Hence the results of can be applied.
Let there be $4$ straight lines $a$, $b$, $c$ and $d$ in space.
Let $a$ be perpendicular to each of $b$, $c$ and $d$.
From Three Intersecting Lines Perpendicular to Another Line are in One Plane, $b$, $c$ and $d$ are all in the same plane.
Let $b$ be perpendicular to $c$ and $d$.
Either $c$ and $d$ are the same line or not.
If $c = d$ then $c$ is not perpendicular to $d$.
Otherwise, from Equal Corresponding Angles implies Parallel Lines, $c$ is parallel to $d$.
In either case $c$ is not perpendicular to $d$.
Hence the result.
== Sources ==
* : $3$
* : $3$
Category:Solid Geometry
Maxwell-Faraday Equation

== Physical Law ==
:$\nabla \times \mathbf E = -\dfrac {\partial \mathbf B} {\partial t}$
where:
:$\nabla \times \mathbf E$ denotes the curl of the electric field $\mathbf E$
:$\dfrac {\partial \mathbf B} {\partial t}$ denotes the partial derivative of the magnetic flux density $\mathbf B$ time.
== Sources ==
* : Introduction: Electromagnetic Theory
* 
* 
Category:Maxwell's Equations
Mazur's Theorem

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} } $ be a normed vector space over $\GF$ with weak topology $w$. 
Let $C \subseteq X$ be a convex subset of $X$. 
Then: 
:$\map {\cl_w} C = \map \cl C$
where $\cl_w$ denotes the weak closure. 
=== Corollary ===
== Proof ==
From Topological Closure in Coarser Topology is Larger: 
:$\map \cl C \subseteq \map {\cl_w} C$
Now let $x \not \in \map \cl C$. 
From Finite Topological Space is Compact, $\set x$ is compact. 
Applying: 
:Hahn-Banach Separation Theorem: Compact Convex Set and Closed Convex Set (Real Case) if $\GF = \R$
:Hahn-Banach Separation Theorem: Compact Convex Set and Closed Convex Set (Complex Case) if $\GF = \C$
for $A = \set x$ and $B = \map \cl C$, there exists a bounded linear functional $f : X \to \GF$ such that: 
:$\ds \map \Re {\map f x} < \inf_{y \in \map \cl C} \map \Re {\map f y}$
Now let $\alpha \in \R$ be such that: 
:$\ds \map \Re {\map f x} < \alpha < \inf_{y \in \map \cl C} \map \Re {\map f y}$
Consider: 
:$E = \set {x \in X : \map \Re {\map f x} \ge \alpha}$ 
We know that $C \subseteq E$. 
From Characterization of Continuity of Linear Functional in Weak Topology, $f : \struct {X, w} \to \GF$ is continuous.
So, by Real and Imaginary Part Projections are Continuous and Composite of Continuous Mappings is Continuous, $\map \Re f : \struct {X, w} \to \R$ is continuous.
From Continuity Defined from Closed Sets, $E$ is then weakly closed.
From Topological Closure of Subset is Subset of Topological Closure, $\map {\cl_w} C \subseteq \map {\cl_w} E$.
From Set is Closed iff Equals Topological Closure, we have $\map {\cl_w} E = E$.
So: 
:$\map {\cl_w} C \subseteq E = \set {x \in X : \map \Re {\map f x} \ge \alpha}$ 
Since $\map \Re {\map f x} < \alpha$, we have $x \in X \setminus \map {\cl_w} C$. 
So we obtain: 
:$\map {\cl_w} C \subseteq \map \cl C$
and so: 
:$\map {\cl_w} C = \map \cl C$
Category:Mazur's Theorem
Category:Weak Topologies on Topological Vector Spaces
Category:Convex Sets (Vector Spaces)
Category:Weakly Closed Sets
Mazur's Theorem/Corollary

== Corollary to Mazur's Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} } $ be a normed vector space over $\GF$ with weak topology $w$. 
Let $C \subseteq X$ be a convex subset of $X$. 
Then:
:$C$ is weakly closed it is $\norm {\, \cdot \,}$-closed.
== Proof ==
From Mazur's Theorem, we have: 
:$\map {\cl_w} C = \map \cl C$
From Set is Closed iff Equals Topological Closure, we have that $C$ is weakly closed :
:$\map {\cl_w} C = C$
That is, :
:$\map \cl C = C$
From Set is Closed iff Equals Topological Closure, this is equivalent to $C$ being $\norm {\, \cdot \,}$-closed.
So $C$ is weakly closed it is $\norm {\, \cdot \,}$-closed.
Category:Mazur's Theorem