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Functional Analysis 3 -- Continuous Functions

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Functional Analysis 3 -- Continuous Functions

This is course note from Siran Li-Shanghai Jiao Tong University.

Proposition 3.1:

Let \(E\) be a metric space. Then \(C_b(E) = \{f: E \rightarrow \mathbb{K} : f \text{ is continuous and bounded}\}\) is a Banach space under supremum norm \(||\cdot||.\)

Observation: \(C_b(E)\) is a Banach algebra with \((f\cdot g)(x) = f(x)g(x).\)

Theorem 3.2:

Let \(E \subset \mathbb{R}^d\) be compact. Then \(A = \{p \mid p \text{ is a polynomial over } \mathbb{R}^d \}\) is dense in \(C(E).\) By definition, dense means that \(\bar{A} = C(E) \Leftrightarrow \forall f \in C(E), \exists \{p_n\} \subset A, p_n \rightarrow f\) uniformly.

Lemma 3.3:

If \(A \subset C_b(E)\) is a subalgebra, then \(\bar{A}\) is a subalgebra.

proof:

Let \(f,g \in A\), then \(\exists \{f_n\}, \{g_n\} \subset A, s.t. f_n \rightarrow f, g_n \rightarrow g\) uniformly. Then \(||f_ng_n - fg|| \le ||f_n - f|| ||g_n|| + ||g_n - g|| ||f|| \rightarrow 0 \sim n \rightarrow \infty.\) \(\Box\)

Theorem 3.4:(Stone-Weierstrass)

Let \(E\) be compact metric space. Let \(A \subset C(E; \mathbb{R})\) be a subalgebra. If \(A\) satisfies

  1. separate points, i.e. \(\forall x\neq y \in E, \exists p \in A, s.t. p(x)\neq p(y).\)
  2. contains constant functions, i.e. \(\underset{c \in \mathbb{R}}{\cup}\{f \equiv c\} \subset A.\)

then \(\bar{A} = C(E).\)

Example: \(A = \{p \mid p \text{ is a polynomial over } \mathbb{R}^d \}.\) Theorem 3.2 can be derived directly from Theorem 3.4.

proof:[[TODO]]

The proof requires lattice property of functions in \(A.\)

However, Stone-Weierstrass theorem does not apply for complex case.

A counter example: Let \(K \subset \mathbb{C}\) be a compact subset. Consider \(K = \{z \in \mathbb{C}\mid \frac{1}{2} \le |z| \le 1 \}.\) Then \[ \int_K \frac{1}{z} dz = 2\pi i \] but \(\int_K p dz = 0, \forall p\) as polynomial. The problem here is that area specified by \(K\) is not a simply connected space.

With simply connected condition added, we get the following theorem from Theorem 3.2.

Theorem 3.5: Runge's Approximation Theorem

Let \(K \subset \mathbb{C}\) be compact and simply-connected. Then \(\forall f \in C^0(K, \mathbb{C}), \exists \{p_n\} s.t. p_n \rightarrow f\) uniformly on \(K.\)

Similarly, Stone-Weierstrass theorem can also be expanded to the following theorem.

Theorem 3.6

Let \(E\) be a compact metric space. Let \(A\) be a subalgebra. \(A \subset C(E, \mathbb{C})\) which separates points and contains constant functions. Assume that \(A\) is closed under complex conjugation. Then \(A\) is dense in \(C(E,\mathbb{C}).\)

Observation: here complex conjugation is not a holomorphic function.

Example: Let \(E = \mathbb{S}^1 = \{z \in \mathbb{C} \mid |z| = 1\}, z = e^{i\theta}, \theta \in [0, 2\pi].\) Let \(A\) be the algebra of trigonometric polynomials. \(A = \{\sum_{n=-N}^N c_ne^{in\theta} \mid c_n \in \mathbb{C}, N \in \mathbb{N}\} = span\{e_n: n\in \mathbb{Z}\}.\) Then by Theorem 3.6, we have that \(A\) is dense in \(C(E, \mathbb{C}).\)

Corollary 3.7

Any \(2\pi\)-periodic function from \(\mathbb{R} \rightarrow \mathbb{R}\) can be uniformly approximated by real trigonometric polynomials \(p(x), s.t. p(x) = \sum_{k=0}^N \alpha_k \cos(kx)+\beta_k\sin(kx).\)

The above theorem is commonly seen in "Harmonic Analysis" field.

As we know that in finite dimension space, bounded + closed = compact. But this is not true when we go to infinite dimensional space. One problem is that without the limit of dimension, in infinite dimensional space, the function change may be tooooo rapid!

Definition: Equicontinuous

Let \(E\) be a metric space. \(\tilde{f} \subset C(E)\) is equicontinuous iff \(\forall \epsilon > 0, \forall x \in E, \exists \delta > 0, s.t. \underset{f \in \tilde{f}}{\sup} |f(x) - f(y)| \le \epsilon\) whenever \(d(x, t) \le \delta.\)

Remark: here the choice of \(\delta\) is independent of \(f.\)

Lemma: If \(E\) is compact, we can choose \(\delta\) independent of \(x.\)

proof:[[TODO]]

Theorem: Arzela–Ascoli

Let \(E\) be a compact metric space, \(\tilde{f} \subset C(E)\). TFAE:

  1. \(\tilde{f}\) is equicontinuous and \(\underset{f \in \tilde{f}}{\sup} ||f||_{\infty} < \infty.\)

  2. \(\tilde{f}\) is precompact(\(\bar{\tilde{f}}\) is compact).

  3. \(\forall \{f_n\} \subset \tilde{f}, \exists \text{subsequences} \{f_{n_j}\}\) that converges (uniformly) in \(C(E).\)

proof:[[TODO]]

A related result from Arzela-Ascoli theorem is as below.

Lemma

Let \(X, Y, Z\) be Banach spaces, \(X \subset Y \subset Z\) and \(X\) is compact in \(Y\). Suppose that \(\{u_n\} \subset C([0, T], X)\) and is equicontinuous in \(C([0, T], Z).\) Then \(\{u_n\}\) is precompact in \(C([0, T], Y).\)