📂Analysis

Algebra of the Space of Continuous Functions

Definitions¹

1. A set $A$ that satisfies the following three conditions is called the algebra of $C(X)$:
   • (i): $\emptyset \ne A \subset C(X)$
   • (ii): $f,g \in A \implies (f+g) , fg \in A$
   • (iii): $f \in A , c \in \mathbb{R} \implies cf \in A$
2. Let’s say $X$ is for metric space $A \subset C(X)$. If every sequence $\left\{ f_{n} \in A : n \in \mathbb{N} \right\}$ of $A$ for some $f \in A$ satisfies $n \to \infty$, then $A$ is said to be uniformly closed.

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• $C \left( X \right)$ is a class of continuous function spaces whose domain is $X$ and codomain is $\mathbb{R}$.

Theorem

If $X$ is a compact metric space and $A$ includes constant functions and is a uniformly closed algebra of $C(X)$, then the following holds: $$ f,g \in A \implies (f \land g), ( f \lor g ) \in A $$

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• $\land$ and $\lor$ mean the following for $f,g \in C(X)$ and $x \in X$: $$ \begin{align*} (f \land g) (x) :=& \min \left\{ f(x) , g(x) \right\} \\ (f \lor g) (x) :=& \max \left\{ f(x) , g(x) \right\} \end{align*} $$

Proof

Strategy: While the above corollary might seem like a simple fact, the proof is anything but simple. $(f \land g)$ and $(f \lor g)$ can be expressed as combinations of simpler functions. Among those simpler functions, there is $|f|$, and showing that it is $f \in A \implies |f| \in A$ presents a challenge. A sequence $\left\{ M g_{n} (x) \right\}_{n \in \mathbb{N}}$ that converges to $|f|$ is directly constructed through a binomial series trick. Then, the proof concludes by the condition of being uniformly closed. This corollary can be usefully employed to prove the Stone-Weierstrass theorem.

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• Part 1. Algebra

  Since $A$ is an algebra, $$ \begin{align*} g \in A, 0 \in \mathbb{R} \implies & 0 \in A \\ g \in A, (-1) \in \mathbb{R} \implies & (-g) \in A \\ f , (-g) \in A \implies & (f - g ) \in A \end{align*} $$ . $(f \land g) (x)$ and $(f \lor g) (x)$ can respectively be expressed as combinations of: $$ \begin{align*} (f \land g) (x) = \min \left\{ f(x) , g(x) \right\} =& {{1} \over {2}} \left[ (f+g) (x) - \left| (f - g)(x) \right| \right] \\ (f \lor g) (x) = \max \left\{ f(x) , g(x) \right\} =& {{1} \over {2}} \left[ (f+g) (x) + \left| (f - g)(x) \right| \right] \end{align*} $$ and $(f + g)$ and $| f - g |$. Once again, since $A$ is an algebra, and given that $(f + g ) \in A$ and therefore $( f - g ) \in A$, it suffices to show $f \in A \implies | f | \in A$. If $| f | = 0$ is the case, obviously $| f | = 0 \in A$, so only consider the case of $M := | f | > 0$.

• Part 2. $\forall \epsilon> 0 , \exists N \in \mathbb{N} : n \ge N \implies \left| P_{n} (t) - |t| \right| < \varepsilon$

  Binomial Series : If $|x| < 1$, then for $\alpha \in \mathbb{C}$, $$ (1 + x )^{\alpha} = \sum_{k=0}^{\infty} \binom{\alpha}{k} x^{k} $$

  $$ \begin{align*} \displaystyle | t | =& \left( 1 - \left( 1- t^2 \right) \right)^{1/2} \\ =& 1 - {{1} \over {2}} \left( 1- t^2 \right) - {{1} \over {2 \cdot 4}} \left( 1- t^2 \right)^2 - \sum_{k=3}^{\infty} {{ 1 \cdot 3 \cdot 5 \cdots (2k-3 ) } \over { 2 \cdot 4 \cdot 6 \cdots (2k) }} \left( 1- t^2 \right)^{k} \end{align*} $$

  uniformly converges on $( -\sqrt{2} , \sqrt{2} )$, thus it also converges uniformly on $[-1,1]$. Let’s define $P_{n}$ according to $n \in \mathbb{N}$ as the following union.

  $$ P_{n} (t) := 1 - {{1} \over {2}} \left( 1- t^2 \right) - {{1} \over {2 \cdot 4}} \left( 1- t^2 \right)^2 - \sum_{k=3}^{n} {{ 1 \cdot 3 \cdot 5 \cdots (2k-3 ) } \over { 2 \cdot 4 \cdot 6 \cdots (2k) }} \left( 1- t^2 \right)^{k} $$

  Then, for any $\varepsilon > 0$ and $t \in [-1 , 1 ]$, there exists $N \in \mathbb{N}$ that satisfies $n \ge N \implies \left| P_{n} (t) - |t| \right| < \varepsilon$.

• Part 3. $f \in A \implies | f | \in A$

  For $x \in X$, let’s define $g_{n} (x)$ as

  $$ g_{n} (x ) : = P_{n} \left( {{ f(x) } \over {M}} \right) $$

  Since $A$ includes constant functions as an algebra, for all $c \in \mathbb{R}$, there exists a corresponding constant function $c(x) = c$. Since $\displaystyle P_{n} \left( {{ f(x) } \over {M}} \right)$ is expressed as a linear combination of constant functions and $\displaystyle \left( {{f(x)} \over {M}} \right)^{2k}$, $g_{n} \in A$ is ensured. Assuming $M = | f | > 0$ from Part 1, setting $t$ as $\displaystyle t := {{ f(x) } \over { M }}$, we have $t \in [-1,1]$, and according to Part 2,

  $$ \left| g_{n} (x) - \left| {{f(x)} \over {M}} \right| \right| = \left| P_{n} \left( {{ f(x) } \over {M}} \right) - \left| {{f(x)} \over {M}} \right|\right| = \left| P_{n} (t) - |t| \right| < \varepsilon $$

  there exists $N$ satisfying. Multiplying both ends by $M$ gives

  $$ \left| M g_{n} (x) - \left| f(x) \right| \right| < M \varepsilon $$

  thus when $x \in X$ and $n \to \infty$, $| M g_{n} - |f| | \to 0$, and since $A$ is uniformly closed,

  $$ |f| \in A $$

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