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Extra. Stone Weierstrass Theorem

Def'n. Function Algebra

For \(\mathcal A\subset C([a,b])\), that follows

  • (linearity) \(\forall f,g\in \mathcal A. \forall a,b\in\mathbb R. af + bg\in \mathcal A\)
  • (closed under product) \(\forall f,g\in\mathcal A. f\cdot g\in \mathcal A\)

Thrm 1.

For \(\mathcal A\subset C([a,b])\) be a function algebra,
IF

  • (vanishes nowhere) \(\forall p\in [a,b]. \exists f\in \mathcal A. f(p)\neq 0\)
  • (separates points) \(\forall x\neq y \in [a,b]. \exists h\in\mathcal A. h(x)\neq h(y)\)

THEN \(\mathcal A\) is dense in \(C[a,b]\), i.e. \(\forall f\in C[a,b]. \exists \{f_n\}\in \mathcal A. \{f_n\}\rightarrow^{u.c.} f\)

Thrm 2.

\(\mathcal A_{trig}:= \{a_0 + \sum_{k=1}^N a_k\cos(kn) + \sum_{k=1}^N b_k \sin(kn) | a_k, b_k\in\mathbb R. N\in\mathbb N\}\) is dense in \(C[-\pi, \pi]\)

proof. Consider the assumptions of SWT
Let \(f\) have coefficients \(a_k, b_k, M\), \(g\) have \(a'_k, a'_k, N\), let \(c,d\in\mathbb R\). wlog, assume \(M\geq N\), extend \(a'_k, b'_k = 0. \forall k > N\)

(linearity) \(ca_0 + da'_0 + \sum (ca_k + da'_k)\cos(kn) + \sum(cb_k + db'_k)\sin(kn)\in \mathcal A_{trig}\)

(product) Using half angle formula \(\cos(a)\cos(b) = \frac{1}{2}cos(a+b) + cos(a-b)\) and other similar ones, we can break all the products of trig functions back to a sum, hence rearrange the summation coefficients.

(vanish nowhere) For any \(a_0 + \sum a_k\cos(pn) + \sum b_k\sin(pn) = 0\), take \(a_0' = a_0 + 1\) and others remain unchanged.

(separates points) Take \(a\neq b\in[-\pi,\pi]\). Using periodicity of trig functions
If \(a\neq -b. \cos(a)\neq cos(b)\), if \(a = -b. \sin(a)\neq [sin(-a) = \sin(b)]\)

Therefore, we can apply SWT