nyuuzyou/wikis huggingface dataset:
Weakly Convergent Sequence in Hilbert Space with Convergent Norm is Convergent

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\HH$. 
Let $x \in X$ be such that: 
:$x_n \weakconv x$
and:
:$\norm {x_n} \to \norm x$
where $\weakconv$ denotes weak convergence.
Then: 
:$x_n \to x$
== Proof ==
Let $\norm \cdot$ be the inner product norm for $\struct {\HH, \innerprod \cdot \cdot}$.
From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we have: 
:$x_n \to x$
:
:$\norm {x_n - x} \to 0$
From Complex Sequence is Null iff Positive Integer Powers of Sequence are Null, it suffices to show that: 
:$\norm {x_n - x}^2 \to 0$
We have: 
From Weak Convergence in Hilbert Space, we have: 
:$\innerprod {x_n} x \to \innerprod x x = \norm x^2$
since $x_n$ converges weakly to $x$.
From Weak Convergence in Hilbert Space: Corollary, we also have: 
:$\innerprod x {x_n} \to \innerprod x x = \norm x^2$
From hypothesis, we have: 
:$\norm {x_n}^2 \to \norm x$ 
Then, from Sum Rule for Real Sequences, we have: 
:$\norm {x_n - x}^2 \to 0$
so:
:$x_n \to x$
== Sources ==
* $27.1$: Weak Convergence
Category:Weak Convergence (Normed Vector Spaces)
Category:Hilbert Spaces
Weakly Convergent Sequence in Normed Dual Space is Weakly-* Convergent

== Theorem ==
 
Let $\mathbb F$ be a subfield of $\C$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\mathbb F$.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $f \in X^\ast$. 
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence in $X^\ast$ converging weakly to $f$. 
Then $\sequence {f_n}_{n \mathop \in \N}$ converges weakly-$\ast$ to $f$. 
== Proof 1 ==
== Proof 2 ==
== Sources ==
* $27.4$: Weak-$\ast$ Convergence
Category:Weak Convergence (Normed Vector Spaces)
Category:Weak-* Convergence (Normed Vector Spaces)
Category:Weakly Convergent Sequence in Normed Dual Space is Weakly-* Convergent
Weakly Convergent Sequence in Normed Dual Space is Weakly-* Convergent/Proof 1

== Theorem ==
== Proof ==
Let $x \in X$.
We aim to show that: 
:$\map {f_n} x \to \map f x$
Then, since $x \in X$ was arbitrary, we will obtain that $\sequence {f_n}_{n \mathop \in \N}$ converges weakly-$\ast$ to $f$.
Since $\sequence {f_n}_{n \mathop \in \N}$ converges weakly to $f$, we have: 
:$\map F {f_n} \to \map F f$ for each $F \in \paren {X^\ast}^\ast$.
Define $x^\wedge : X^\ast \to \mathbb F$ by: 
:$\map {x^\wedge} f = \map f x$
for each $f \in X^\ast$.
From Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual, we have: 
:$x^\wedge \in \paren {X^\ast}^\ast$
So, setting $F = x^\wedge$, we obtain: 
:$\map {x^\wedge} {f_n} \to \map {x^\wedge} f$
That is: 
:$\map {f_n} x \to \map f x$
Category:Weakly Convergent Sequence in Normed Dual Space is Weakly-* Convergent
Weakly Convergent Sequence in Normed Dual Space is Weakly-* Convergent/Proof 2

== Theorem ==
== Proof ==
Let $J$ be the evaluation linear transformation on $X$.
By Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual:
:$J : X \to X^{\ast \ast}$
Thus, for each $x \in X$:
Category:Weakly Convergent Sequence in Normed Dual Space is Weakly-* Convergent
Weakly Convergent Sequence in Normed Vector Space is Bounded

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a weakly convergent sequence in $X$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is bounded.
== Proof ==
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual of $X$. 
Let $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$ be the second normed dual of $X$. 
Let $J : X \to X^{\ast \ast}$ be the evaluation linear transformation on $X$. 
Let: 
:$x^\wedge = \map J x$
for each $x \in X$. 
Consider the sequence $\sequence {x_n^\wedge}_{n \mathop \in \N}$ in $X^{\ast \ast}$. 
Then, for each $f \in X^\ast$, we have: 
:$\map {x_n^\wedge} f = \map f {x_n}$ 
Since $\sequence {x_n}_{n \mathop \in \N}$ converges weakly we have that: 
:$\sequence {\map f {x_n} }_{n \mathop \in \N}$ converges for each $f \in X^\ast$.
So, from Convergent Complex Sequence is Bounded, $\sequence {\map f {x_n} }_{n \mathop \in \N}$ is bounded for each $f \in X^\ast$. 
So for each $f \in X^\ast$ there exists a real number $M_f > 0$ such that: 
:$\cmod {\map f {x_n} } \le M_f$
for each $n \in \N$.
That is: 
:$\cmod {\map {x_n^\wedge} f} \le M_f < \infty$ 
so: 
:$\ds \sup_{n \mathop \in \N} \cmod {\map {x_n^\wedge} f}$ is finite for each $f \in X^\ast$. 
From Normed Dual Space is Banach Space, $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ is a Banach space, and so we may apply the Banach-Steinhaus Theorem.
From the Banach-Steinhaus Theorem, we then obtain: 
:$\ds \sup_{n \mathop \in \N} \norm {x_n^\wedge}_{X^{\ast \ast} }$ is finite. 
From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, we have: 
:$\norm {x_n^\wedge}_{X^{\ast \ast} } = \norm {x_n}_X$
So:
:$\ds \sup_{n \mathop \in \N} \norm {x_n}_X$ is finite.
So we may conclude that: 
:$\sequence {x_n}_{n \mathop \in \N}$ is bounded.
== Sources ==
* $27.1$: Weak Convergence
Category:Weak Convergence (Normed Vector Spaces)
Weakly Locally Compact Hausdorff Space is Strongly Locally Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Let $T$ be weakly locally compact.
Then $T$ is strongly locally compact.
== Proof==
Let $x \in S$.
As $T$ is weakly locally compact, $x$ is contained in a compact neighborhood $N_x$.
As $T$ is a $T_2$ (Hausdorff) space, we can use the result Compact Subspace of Hausdorff Space is Closed.
Thus the interior of $N_x$ has a closure which is compact.
Hence the result, from definition of strongly locally compact space.
== Also see ==
* Existence of Weakly Locally Compact Space which is not Strongly Locally Compact
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Localized Compactness Properties
Category:Hausdorff Spaces
Category:Weakly Locally Compact Spaces
Category:Strongly Locally Compact Spaces
Weakly Sigma-Locally Compact iff Weakly Locally Compact and Lindelöf

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
:$(1): \quad T$ is weakly $\sigma$-locally compact
:$(2): \quad T$ is weakly locally compact and Lindelöf
== Proof ==
=== 1 implies 2 ===
Let $T = \struct {S, \tau}$ be weakly $\sigma$-locally compact.
Then by definition:
:$T$ is $\sigma$-compact
:$T$ is weakly locally compact.
From Sigma-Compact Space is Lindelöf it follows directly that:
:$T$ is Lindelöf
:$T$ is weakly locally compact.
=== 2 implies 1 ===
Now let $T = \struct {S, \tau}$ be weakly locally compact and Lindelöf.
By definition:
:$T$ is weakly locally compact every point of $S$ is contained in a compact neighborhood.
:$T$ is Lindelöf every open cover of $S$ has a countable subcover.
Thus the interiors of the compact neighborhoods are an open cover of $S$.
As $T$ is Lindelöf, this cover has a countable subcover.
Thus $T$ is the union of countably many compact subspaces.
That is, $T$ is $\sigma$-compact, and so weakly $\sigma$-locally compact.
== Also see ==
* Sequence of Implications of Local Compactness Properties
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Localized Compactness Properties
Category:Weakly Locally Compact Spaces
Category:Weakly Sigma-Locally Compact Spaces
Category:Lindelöf Spaces
Weierstrass's Elliptic Function is Doubly Periodic

== Theorem == 
Weierstrass's elliptic function $\map \wp {z; \omega_1, \omega_2}$ is doubly periodic in $2 \omega_1$ and $2 \omega_2$.
== Proof == 
 
== Sources == 
* : $20.21$: Periodicity and other properties of $\map \wp z$
Category:Weierstrass's Elliptic Function
Weierstrass's Elliptic Function is Even in z

== Theorem == 
Let $\omega_1$ and $\omega_2$ be non-zero complex constants with $\dfrac {\omega_1} {\omega_2}$ having a positive imaginary part. 
For $z \in \C \setminus \set {2 m \omega_1 + 2 n \omega_2: \tuple {n, m} \in \Z^2}$:
:$\ds \map \wp {-z; \omega_1, \omega_2} = \map \wp {z; \omega_1, \omega_2}$
That is, Weierstrass's elliptic function is even in $z$. 
== Proof ==
== Sources == 
* : $20.21$: Periodicity and other properties of $\wp\left({z}\right)$
Category:Weierstrass's Elliptic Function
Weierstrass's Necessary Condition

== Theorem ==
Let $\mathbf y: \R \to \R^n$ be an $n$-dimensional vector-valued function such that $\map {\mathbf y} a = A$ and $\map {\mathbf y} b = B$.
Let $J$ be a functional such that:
:$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
Let $\mathbf w$ be an $n$-dimensional vector such that $\mathbf w \in \R^n$.
Let $\gamma$ be a strong minimum of $J$.
Then along $\gamma$ and for every $\mathbf w$:
:$\map E {x, \mathbf y, \mathbf y', \mathbf w} \ge 0$
where $E$ stands for the Weierstrass E-Function.
== Proof ==
== Sources ==
* : $\S 6.34$: The Weierstrass E-Function. Sufficient Conditions for a Strong Extremum
Category:Calculus of Variations
Weierstrass's Theorem/Lemma 1

== Lemma for Weierstrass's Theorem ==
Let $C \closedint 0 1$ denote the set of all real functions $f: \closedint 0 1 \to \R$ which are continuous on $\closedint 0 1$.
Let $\norm {\,\cdot \,}_\infty$ denote the supremum norm on $C$.
Let $X$ consist of the $f \in C \closedint 0 1$ such that:
:$\map f 0 = 0$
:$\map f 1 = 1$
:$\forall x \in \closedint 0 1: 0 \le \map f x \le 1$
$X$ is a complete metric space under $\norm {\,\cdot \,}_\infty$.
== Proof ==
For every $n \in \N$, let $f_n \in X$.
 that in $C \closedint 0 1$:
If we can prove that $f \in X$, we know $X$ contains all its limit points.
Hence by Closed Set iff Contains all its Limit Points, $X$ is closed.
From Topological Completeness is Weakly Hereditary, $X$ is complete.
It is now to be proved that $f \in X$.
Suppose $\map f 0 \ne 0$.
Then:
:$\forall n \in \N: \norm {f_n - f}_\infty \ge \size {\map {f_n} 0 - \map f 0} = \size {\map f 0} > 0$
This would contradict equation $(1)$.
Hence $\map f 0 = 0$.
Similarly, it is necessary that $\map f 1 = 1$.
Also, for all $n \in \N$ and $x \in \closedint 0 1$, we have that:
:$0 \le \map {f_n} x \le 1$
Suppose there is an $x \in \closedint 0 1$ such that either:
:$\map f x < 0$
or:
:$\map f x > 1$
We see that it must be that:
:$\forall n \in \N: \norm {f_n - f}_\infty \ge \norm {\map {f_n} x - \map f x} > 0$
which contradicts $(1)$.
Therefore, $f \in X$, and hence $X$ is complete.
== Sources ==
* : $8.10$ (for the core proof)
Category:Weierstrass's Theorem
Weierstrass's Theorem/Lemma 2

== Lemma for Weierstrass's Theorem ==
Let $C \closedint 0 1$ denote the set of all real functions $f: \closedint 0 1 \to \R$ which are continuous on $\closedint 0 1$.
Let $\norm {\,\cdot \,}_\infty$ denote the supremum norm on $C$.
Let $X$ consist of the $f \in C \closedint 0 1$ such that:
:$\map f 0 = 0$
:$\map f 1 = 1$
:$\forall x \in \closedint 0 1: 0 \le \map f x \le 1$
For every $f \in X$, define $\hat f: \closedint 0 1 \to \R$ as follows:
:$\map {\hat f} x = \begin {cases}
\dfrac 3 4 \map f {3 x} & : 0 \le x \le \dfrac 1 3 \\
\dfrac 1 4 + \dfrac 1 2 \map f {2 - 3 x} & : \dfrac 1 3 \le x \le \dfrac 2 3 \\
\dfrac 1 4 + \dfrac 3 4 \map f {3 x - 2} & : \dfrac 2 3 \le x \le 1
\end {cases}$
Then:
:$\hat \cdot: X \to X$ is a contraction mapping.
Furthermore, we have the following inequality:
:$\forall f, g \in X: \norm {\hat f - \hat g}_\infty \le \dfrac 3 4 \norm {f - g}_\infty$
== Proof ==
== Sources ==
* : $8.10$ (for the core proof)
Category:Weierstrass's Theorem
Weierstrass's Theorem/Lemma 3

== Lemma for Weierstrass's Theorem==
Let $C \closedint 0 1$ denote the set of all real functions $f: \closedint 0 1 \to \R$ which are continuous on $\closedint 0 1$.
Let $\norm {\,\cdot \,}_\infty$ denote the supremum norm on $C$.
Let $X$ consist of the $f \in C \closedint 0 1$ such that:
:$\map f 0 = 0$
:$\map f 1 = 1$
:$\forall x \in \closedint 0 1: 0 \le \map f x \le 1$
For every $f \in X$, define $\hat f: \closedint 0 1 \to \R$ as follows:
:$\map {\hat f} x = \begin {cases}
\dfrac 3 4 \map f {3 x} & : 0 \le x \le \dfrac 1 3 \\
\dfrac 1 4 + \dfrac 1 2 \map f {2 - 3 x} & : \dfrac 1 3 \le x \le \dfrac 2 3 \\
\dfrac 1 4 + \dfrac 3 4 \map f {3 x - 2} & : \dfrac 2 3 \le x \le 1
\end {cases}$
The Contraction Mapping Theorem assures existence of a unique $h \in X$ with $\hat h = h$.
For every $n \in \N$ and $k \in \set {1, 2, 3, 4, \ldots, 3^n}$, the following inequality holds:
:$\size {\map h {\dfrac {k - 1} {3^n} } - \map h {\dfrac k {3^n} } } \ge 2^{-n}$
== Proof ==
For all $n \in \N$ and $k \in \set {1, 2, 3, \ldots, 3^n}$:
:$1 \le k \le 3^n \implies 0 \le \dfrac{k - 1} {3^{n + 1} } < \dfrac k {3^{n + 1} } \le \dfrac 1 3$
:$3^n < k \le 2 \cdot 3^n \implies \dfrac 1 3 \le \dfrac {k - 1} {3^{n + 1} } < \dfrac k {3^{n + 1}} \le \dfrac 2 3$
:$2 \cdot 3^n < k \le 3^{n + 1} \implies \dfrac 2 3 \le \dfrac {k - 1} {3^{n + 1} } < \dfrac k {3^{n + 1} } \le 1$
== Sources ==
* : $8.10$ (for the core proof)
Category:Weierstrass's Theorem
Category:Real Analysis
Weierstrass-Casorati Theorem

== Theorem ==
Let $f$ be a holomorphic function defined on the open ball $\map B {a, r} \setminus \set a$.
Let $f$ have an essential singularity at $a$.
Then:
:$\forall s < r: f \sqbrk {\map B {a, s} \setminus \set a}$ is a dense subset of $\C$.
== Proof ==
, suppose $a = 0$ and $r = 1$.
 $\exists s < 1$ such that:
:$f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$.
Then, by definition of dense subset:
:$\exists z_0 \in \C: \exists r_0 > 0: \map B {z_0, r_0} \cap f \sqbrk {\map B {0, s} \setminus \set 0} = \O$
Hence, the function $\varphi$ defined on $\map B {z_0, r_0}$ by:
:$\map \varphi z = \dfrac 1 {\map f z - z_0}$
is analytic on $\map B {0, s} \setminus \set 0$ and bounded near to $0$, because:
:$\forall z \in \map B {0, s} \setminus \set 0: \cmod {\map f z - z_0} > r_0 \implies \cmod {\map \varphi z} < \dfrac 1 {r_0}$
Therefore, we can extend the domain of $\varphi$ (using the Analytic Continuation Principle).
Let $\map \varphi 0 \ne 0$.
Then:
:$\map f 0 = z_0 + \dfrac 1 {\map \varphi 0}$
and the singular point of $f$ was removable.
Otherwise, let the power series of $\varphi$ be written:
:$\ds \map \varphi z = \sum_{n \mathop = 1}^{+\infty} a_n z^n$
Then as $\varphi \ne 0$:
:$E = \set {k \in \N: a_k \ne 0} \ne \O$
Let $p = \min E$.
Then $0$ is a pole of order $p$ of $f$.
In each case, the assumption that:
:$\exists s < 1: f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$ contradicts the fact that $0$ is an essential singularity of $f$.
Hence the result, by Proof by Contradiction.
== Also known as ==
It is also known as the Casorati-Weierstrass Theorem.
Category:Complex Analysis
Weierstrass Approximation Theorem

== Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\Bbb I = \closedint a b$.
Then $f$ can be uniformly approximated on $\Bbb I$ by a polynomial function to any given degree of accuracy.
=== Complex Case ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also known as ==
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.33$: Weierstrass ()
* 
* 
* 
* 
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Weierstrass Approximation Theorem
Category:Real Analysis
Weierstrass Approximation Theorem/Also known as

== Weierstrass Approximation Theorem: Also known as ==
The Weierstrass approximation theorem is also seen referred to as Weierstrass's theorem, but as there are a number of results bearing 's name, it makes sense to be more specific.
== Sources ==
* 
* 
* 
* 
Category:Weierstrass Approximation Theorem
Weierstrass Approximation Theorem/Complex Case

== Theorem ==
Let $\Bbb I = \closedint a b$ be a closed real interval.
Let $f: \Bbb I \to \C$ be a continuous complex function. 
Let $\epsilon \in \R_{>0}$.
Then there exists a complex polynomial function $p : \Bbb I \to \C$ such that:
:$\norm {p - f}_\infty < \epsilon$
where $\norm f_\infty$ denotes the supremum norm of $f$ on $\Bbb I$.
== Proof ==
Let $\Re f : \Bbb \to \R$ be the real function defined by:
:$\map {\Re f} x = \map \Re {\map f x}$
where $\map \Re z$ denotes the real part of a complex number $z \in \C$.
Let $\Im f : \Bbb \to \R$ be the real function defined by:
:$\map {\Im f} x = \map \Im {\map f x}$
where $\map \Im z$ denotes the imaginary part of $z \in \C$.
From Real and Imaginary Part Projections are Continuous and Composite of Continuous Mappings is Continuous, it follows that $\Re f$ and $\Im f$ are continuous functions.
By the Weierstrass Approximation Theorem, there exist two real polynomial functions $u, v: \Bbb I \to \R$ such that:
:$\norm {u - \Re f}_\infty < \dfrac \epsilon 2$
:$\norm {v - \Im f}_\infty < \dfrac \epsilon 2$
Set $p := u + iv$, so $p$ becomes a complex polynomial function.
Then:
Category:Weierstrass Approximation Theorem
Weierstrass Approximation Theorem/Historical Note

== Historical Note on Weierstrass Approximation Theorem ==
The Weierstrass Approximation Theorem has been demonstrated to have far-reaching and important effects in every aspect of the field of analysis.
It has also been given a significant generalisation by .
== Sources ==
* : Chapter $\text {A}.33$: Weierstrass ()
Category:Historical Notes
Weierstrass Approximation Theorem/Lemma 1

== Theorem ==
Let $\map {p_{n, k} } t : \N^2 \times \closedint 0 1 \to \R$ be a real valued function defined by:
:$\map {p_{n, k} } t := \dbinom n k t^k \paren {1 - t}^{n - k}$
where:
:$n, k \in \N$
:$t \in \closedint 0 1$
:$\dbinom n k$ denotes the binomial coefficient.
Then:
:$\ds \sum_{k \mathop = 0}^n k \map {p_{n, k} } t = n t$
== Proof ==
From Binomial Theorem for Integral Index:
== Sources ==
* : Chapter $1$: Normed and Banach spaces
Category:Weierstrass Approximation Theorem
Weierstrass Approximation Theorem/Lemma 2

== Theorem ==
Let $\map {p_{n, k} } t : \N^2 \times \closedint 0 1 \to \R$ be a real valued function defined as:
:$\map {p_{n, k} } t := \dbinom n k t^k \paren {1 - t}^{n - k}$
where:
:$n, k \in \N$
:$t \in \closedint 0 1$
:$\dbinom n k$ denotes a binomial coefficient.
Then:
:$\ds \sum_{k \mathop = 0}^n \paren {k - n t}^2 \map {p_{n, k} } t = n t \paren {1 - t}$
== Proof ==
From Binomial Theorem for Integral Index:
:$\ds 1 = \sum_{k \mathop = 0}^n \binom n k t^k \paren {1 - t}^{n - k}$
From :
Then:
== Sources ==
* : Chapter $1$: Normed and Banach spaces
Category:Weierstrass Approximation Theorem
Weierstrass Approximation Theorem/Proof 1

== Theorem ==
== Proof ==
Let $\map f t: \Bbb I = \closedint a b \to \R$ be a continuous function.
Introduce $\map x t$ with a rescaled domain:
:$\map f t \mapsto \map x {a + t \paren {b - a} } : \closedint a b \to \closedint 0 1$
From now on we will work with $x: \closedint 0 1 \to \R$, which is also continuous.
Let $n \in \N$.
For $t \in \closedint 0 1$ consider the Bernstein polynomial:
:$\ds \map {B_n x} t = \sum_{k \mathop = 0}^n \map x {\frac k n} \binom n k t^k \paren {1 - t}^{n - k}$
For $t \in \closedint 0 1$, $0 \le k \le n$, let:
:$\map {p_{n, k} } t := \dbinom n k t^k \paren {1 - t}^{n - k}$
By Binomial Theorem for Integral Index:
:$\ds \sum_{k \mathop = 0}^n \map {p_{n, k} } t = 1$
=== Lemma 1 ===
=== Lemma 2 ===
Now we construct the estimates.
Here $\ds \sum_{k \mathop : \size {\frac k n \mathop - t} \ge \delta}^n$ denotes the summation over those values of $k \in \N$, $k \le n$, which satisfy the inequality $\size {\dfrac k n - t} \ge \delta$.
For some $\delta > 0$ denote:
:$\ds \map {\omega_\delta} x := \sup_{\size {t - s} < \delta} \size {\map x s - \map x t}$
Then:
where $\norm {\,\cdot \,}_\infty$ denotes the supremum norm.
Let $\epsilon > 0$.
By Continuous Function on Closed Real Interval is Uniformly Continuous, $\map x t$ is uniformly continuous.
We choose $\delta > 0$ such that $\map {\omega_\delta} x < \dfrac \epsilon 2$.
Choose $n > \dfrac {\norm x_\infty} {\epsilon \delta^2}$
Then:
:$\norm {\map {B_n x} t - \map x t}_\infty < \epsilon$.
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Weierstrass Approximation Theorem
Weierstrass Approximation Theorem/Proof 2

== Theorem ==
== Proof ==
, assume $\Bbb I = \closedint 0 1$
For each $n \in \N$, let:
:$\ds \map {P_n} x := \sum_{k \mathop = 0}^n \map f {\dfrac k n } \dbinom n k x^k \paren {1 - x}^{n - k}$
We shall show that $\lim_{n \to \infty} \norm { P_n - f}_\infty = 0$.
Let $\epsilon \in \R_{>0}$.
By Heine-Cantor Theorem, there is a $\delta \in \R_{>0}$ such that:
:$\forall x,y \in \Bbb I : \size {x - y} \le \delta \implies \size {\map f x - \map f y} \le \epsilon $
Let $p \in \Bbb I$.
Let $Z_n$ be a random variable such that:
:$\ds n Z_n \sim \Binomial n p$
where $\Binomial n p$ denotes the binomial distribution with parameters $n$ and $p$.
Observe that:
Furthermore:
On the other hand:
Therefore:
Thus for all $n \in \N_{> 2 \delta^2 / \norm f_\infty}$
we have:
:$\size {\map {P_n} p - \map f p} \le 2 \epsilon$
As the above is true for all $p \in \Bbb I$, we have:
:$\forall n \in \N_{> 2 \delta^2 / \norm f_\infty} : \norm { P_n - f}_\infty \le 2 \epsilon$
Category:Weierstrass Approximation Theorem
Weierstrass Approximation Theorem/Proof 3

== Theorem ==
== Proof ==
Let $\AA \subseteq \map C {\Bbb I, \R}$ be the set of real polynomial functions.
$\AA$ is a subalgebra of $\map C {\Bbb I, \R}$ because polynomials over $\R$ form an algebra over $\R$.
Let $I$ denote the identity mapping on $\Bbb I$, i.e.:
:$\forall x \in \Bbb I : \map I x = x$
Then $I \in \AA$.
Thus $\AA$ separates the points of $\Bbb I$, since trivially:
:$\forall x,y \in \Bbb I : x \ne y \implies \map I x \ne \map I y$
It is also clear that $1 \in \AA$.
Therefore the claim follows from Stone-Weierstrass Theorem.
Category:Weierstrass Approximation Theorem
Weierstrass Factorization Theorem

== Theorem ==
Let $f$ be an entire function.
Let $0$ be a zero of $f$ of multiplicity $m \ge 0$.
Let the sequence $\sequence {a_n}$ consist of the nonzero zeroes of $f$, repeated according to multiplicity.
=== First Form ===
Let $\sequence {p_n}$ be a sequence of non-negative integers for which the series:
:$\ds \sum_{n \mathop = 1}^\infty \size {\dfrac r {a_n} }^{1 + p_n}$
converges for every $r \in \R_{> 0}$.
Then there exists an entire function $g$ such that:
:$\ds \map f z = z^m e^{\map g z} \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$
where:
:$E_{p_n}$ are Weierstrass's elementary factors
:the continued product converges locally uniformly absolutely on $\C$.
=== Second Form ===
There exists a sequence $\sequence {p_n}$ of non-negative integers and an entire function $g$ such that:
:$\ds \map f z = z^m e^{\map g z} \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$
where:
:$E_{p_n}$ are Weierstrass's elementary factors
:the continued product converges locally uniformly absolutely on $\C$.
== Proof ==
From Weierstrass Product Theorem, the function:
:$\ds \map h z = z^m \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$
defines an entire function that has the same zeroes as $f$ counting multiplicity. 
Thus $f / h$ is both an entire function and non-vanishing.
As $f / h$ is both holomorphic and nowhere zero there exists a holomorphic function $g$ such that:
:$e^g = f / h$
Therefore:
:$f = e^g h$
as desired.
== Also known as ==
Some sources give this as the Weierstrass factor theorem or the Weierstrass infinite product theorem.
== Also see ==
* Hadamard Factorization Theorem, a refined version of this
* Weierstrass Product Theorem
== Historical Note ==
== Sources ==
* : $8.2.4$
* $\text {VII}$: Compact and Convergence in the Space of Analytic Functions: $\S5$: Weierstrass Factorization Theorem: Theorem $5.14$
Category:Complex Analysis
Category:Infinite Products
Category:Entire Functions
Weierstrass Factorization Theorem/Historical Note

== Historical Note on Weierstrass Factorization Theorem ==
The Weierstrass Factorization Theorem was proved by during his work investigating the properties of entire functions.
This theorem finally provided a satisfactory context for Euler Formula for Sine Function.
== Sources ==
* : Chapter $\text {A}.33$: Weierstrass ()
Category:Historical Notes