Definition: Let \(A \subseteq \mathbb R\) be non-empty, \(\mathcal F = \lbrace f: A \to \mathbb R \rbrace\) be a set of functions. A sequence of functions on A is a function
\[\begin{align*} n \in \mathbb N \mapsto f_n \in \mathcal F \end{align*}\]and we write \((f_n)_{n = 0}^\infty\)
Definition: Given a sequence of functions on \(A\), \((f_n)_n\) and a function \(f: A \to \mathbb R\), we say that \((f_n)\) converges to \(f\) pointwise when
\[\begin{align*} \forall a \in A, \forall \epsilon \gt 0, \exists N_a \in \mathbb N, \forall n \geq N_a, \vert f_n(a) - f(a) \vert \lt \epsilon \end{align*}\]We say \((f_n)\) converges uniformly on \(A\) to \(f\), when
\[\begin{align*} \forall \epsilon \gt 0, \exists N \in \mathbb N, \forall n \geq N, \forall x \in A, \vert f_n(x) - f(x) \vert \lt \epsilon \end{align*}\]and we write \(f_n \rightrightarrows_A f\).
Remark: If \(f_n \rightrightarrows_A f\), then \(f_n(x) \to_{n \to \infty} f(x), \forall x \in A\), that is, \(f_n \to_{n \to \infty} f\). That is, uniform convergence implies pointwise convergence.
Remark: If \(f_n \to_{n \to \infty} f\), then \(f_n \rightrightarrows g \implies g = f\).
Example: \(f_n(x) = x^n, x \in [0, 1]\)
Claim 1: \(f_n \to f\), where
\[\begin{align*} f(x) = \begin{cases} 0 & x \in [0, 1) \\ 1 & x = 1 \end{cases} \end{align*}\]Proof: For any \(x \in [0, 1)\), we have \(\lim_{n \to \infty} x^n = 0 = f(x)\). When \(x = 1, x^n = 1, \forall n \implies f_n(1) \to_{n \to \infty} f(1)\). \(\blacksquare\)
Claim 2: \((f_n)\) doesn’t converge uniformly to any function.
Proof: Consider \(\epsilon = 1/2\). If \(f_n \rightrightarrows f\), then \(\forall k \in \mathbb Z_+\), \(\exists N \in \mathbb N, \forall n \geq N, f_n(1 - \frac{1}{k}) \lt \frac{1}{2}\), that is, \((1 - \frac{1}{k})^n \lt \frac{1}{2}\). We have
\[\begin{align*} 1 - \frac{1}{k} \lt \frac{1}{\sqrt[n]{2}} &\iff \frac{1}{k} \gt 1 - \frac{1}{\sqrt[n]{2}} = \frac{\sqrt[n]{2} - 1}{\sqrt[n]{2}} \\ &\iff \frac{\sqrt[n]{2}}{\sqrt[n]{2} - 1} \gt k \end{align*}\]Theorem: If \(A \subseteq \mathbb R\) is a non-empty set, \((f_n)\) is a seuqence of functions continuous on \(A\), such that \(f_n \rightrightarrows_A f\), then \(f\) is continuous on \(A\).
Proof: Let \(a \in A\) and \(\epsilon \gt 0\). Choose \(N_0 \in \mathbb N\) such that \(\forall n \geq N_0, \forall x \in A, \vert f_n(x) - f(x) \vert \lt \epsilon/3\). Choose \(\delta \gt 0\) such that \(\forall x \in A, \vert x - a \vert \lt \delta \implies \vert f_{N_0}(x) - f_{N_0}(a) \vert \lt \epsilon/3\). Now, \(\forall x \in A\),
\[\begin{align*} \vert x - a \vert \lt \delta &\implies \vert f(x) - f(a) \vert \\ &= \vert f(x) - f_N(x) + f_N(x) - f_N(a) + f_N(a) - f(a) \vert \\ &\leq \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon \end{align*}\]Definition: We say that \((f_n)\) is Cauchy (on \(A\)) when
\[\begin{align*} \forall \epsilon \gt 0, \exists N_0 \in \mathbb N, \forall n, m \geq N_0, \forall x \in A \vert f_m(x) - f_n(x) \vert \lt \epsilon \end{align*}\]Theorem: A functional sequence \((f_n)\) in \(A\) is uniformly convergent on \(A\) if and only if it is Cauchy on \(A\).
Proof: (\(\rightarrow\)) Suppose \(f: A \to \mathbb R\) is such that \(f_n \rightrightarrows_A f\). Let \(\epsilon \gt 0\). Choose \(N_0 \in \mathbb N\) such that \(\forall n \geq N_0, \forall x \in A, \vert f_n(x) - f(x) \vert \lt \epsilon/2\). Then, \(\forall m, n, \geq N_0, \forall x \in A\), \(\vert f_m(x) - f_n(x) \vert = \vert (f_m(x) - f(x)) + (f(x) - f_n(x)) \vert \leq 2 \cdot \epsilon/2 = \epsilon\).
(\(\leftarrow\)) Now, suppose \((f_n)\) is Cauchy on \(A\). Then \(\forall x \in A\), \((f_n(x))_n \subseteq \mathbb R\) is Cauchy, so \(\lim_{n \to \infty} f_n(x) =: f(x)\). Hence, we have \(f: x \in A \mapsto f(x) \in \mathbb R\). We claim that \(f_n \rightrightarrows_A f\). Let \(\epsilon \gt 0\). Choose \(N_0 \in \mathbb N\) such that \(\forall m, n \geq N_0, \forall x \in A, \vert f_m(x) - f_n(x) \vert \lt \epsilon/2\). Pick an arbitrary \(n_0 \geq N_0\) and \(x_0 \in A\). Then for any \(m \geq N_0\), \(\vert f_m(x_0) - f_{n_0}(x_0) \vert \lt \epsilon/2\). So, \(f_{n_0} - \frac{\epsilon}{2} \lt f_m(x_0) \lt f_{n_0}(x_0) + \frac{\epsilon}{2}\). As \(m \to \infty\), since the left and right inequalities are constant sequences, we have
\[\begin{align*} &f_{n_0}(x_0) - \frac{\epsilon}{2} \leq f(x_0) \leq f_{n_0}(x_0) + \frac{\epsilon}{2} \\ &\implies \vert f_n(x) - f(x) \vert \leq \epsilon/2 \lt \epsilon \end{align*}\]Note that we use the fact that \(a_n \lt b_n \implies \lim a_n \leq \lim b_n\).
Definition: Given \(A \subseteq \mathbb R\) non-empty, a series of functions on \(A\) \(\sum_{n = 0}^\infty f_n\) is a pair of sequences \(((f_n), (s_n))\), where all \(f_n: A \to \mathbb R\), \((f_n)\) is the sequence of terms, \((s_n)\) is the sequence of partial sums, and \(s_n(x) = \sum_{k = 0}^n f_k(x)\) for all \(x \in A, n \in \mathbb N\).
We say \(\sum f_n\) is convergent to a function \(f: A \to \mathbb R\)
Theorem (Weierstrass M-Test): Given \(A \neq \emptyset\) and a series \(\sum_{n = 0}^\infty f_n(x)\) of functions on \(A\), suppose \(\forall n \in \mathbb N\), \(\exists M_n \geq 0\) such that \(\forall x \in A, \vert f_n(x) \vert \leq M_n\). Assume \(\sum_{n = 0}^\infty M_n\) is convergent. Then, \(\sum_{n = 0}^\infty f_n\) is absolutely and uniformly convergent on \(A\).
Proof: For every \(x \in A, 0 \leq \sum_{n = 0}^\infty \vert f_n(x) \vert \leq \sum_{n = 0}^\infty M_n\), so \(\sum \vert f_n(x) \vert\) converges by Comparison Test. So the series is absolutely convergent.
Let \((s_n)_n\) be the sequence of partial sums of \(\sum f_n\). We show that \((s_n)_n\) is Cauchy on \(A\). Let \(\epsilon \gt 0\). Choose \(N_0 \in \mathbb N\) such that \(\forall n \gt m \geq N_0\), \(\sum_{k = m + 1}^n M_k \lt \epsilon\), which is possible since \(\sum M_k\) is convergent and thus its sequence of partial sums is Cauchy. Then \(\forall n \gt m \geq N_0, \forall x \in A\),
\(\begin{align*} \vert s_n(x) - s_m(x) \vert = \left\vert \sum_{k = m + 1}^n f_k(x) \right\vert \leq \sum_{k = m + 1}^n \vert f_k(x) \vert \leq \sum_{k = m + 1}^n M_k \lt \epsilon \end{align*}\) \(\blacksquare\)
Definition: A power series centered at the centre \(c \in \mathbb R\) with coefficients \((a_n)_n \subseteq \mathbb R\) is the series \(\sum_{n = 0}^\infty a_n(x - c)^n\).
Theorem:
Definition The radius of convergence \(R\) of the power series \(S = \sum a_n(x - c)^n\) is \(\sup \lbrace r \gt 0 : S \text{ is absolutely and uniformly convergent on } [c - r, c + r] \rbrace\) if \(\exists x_0 \neq c\) such that \(S\) converges. Otherwise, \(R := 0\).
Definition: A function \(f: I \to \mathbb R\) is analytic when \(\forall c \in I, \exists (a_n)_n \subseteq \mathbb R, \exists r \gt 0\) such that \((c - r, c + r) \subseteq I\) and \(f(x) = \sum_{n = 0}^\infty a_n (x - c)^n, \forall x \in (c - r, c + r)\).
Definition: A family \(\lbrace U_\iota \rbrace_{\iota \in I}\) of open sets in \(\mathbb R\) is called an open cover of a set \(A \subseteq \mathbb R\) when \(A \subseteq \bigcup U\).
Proposition: If \(K \subseteq \mathbb R\) is a compact set, then any open cover of \(K\) admits a finite subcover.
Proof: Let \(K \subseteq \mathbb R, K \neq \emptyset\) be compact. Observe that \(\forall r \gt 0, \exists a_1, \dots, a_n \in \mathbb R\) such that \(K \subseteq \bigcup_{i = 1}^n (a_i - r, a_i + r)\). Suppose for contradiction that \(\lbrace U_\iota \rbrace_{\iota \in I}\) is an open cover of \(K\) such that no finite number of \(U_\iota\) covers \(K\). Consider \(K = \bigcup_{i = 1}^s K \cap [a_i - 1, a_i + 1]\). Pick \(K_1 := K \cap [a_1 - 1, a_1 + 1]\) such that it cannot be covered by finitely many \(U_\iota\). Consider \(K_1 = \bigcup_{j = 1}^t K_1 \cap [b_j - \frac{1}{2}, b_j + \frac{1}{2}]\). Similarily, pick \(K_2 = K_1 \cap [b_2 - \frac{1}{2}, b_2 + \frac{1}{2}]\) such that it cannot be covered by finitely many \(U_\iota\), and so on. Hence, we get a sequence of compacts \((K_n)\), \(K_1 \supseteq K_2 \supseteq \dots\) with the diameter of \(K_{n + 1}\) half of that of \(K_n\), none of which can be covered by finitely many \(U_\iota\). Pick \(\forall n \geq 1, x_n \in K_n\). Then \((x_n) \subseteq K\) is a convergent sequence. So \(p := \lim_{n \to \infty} x_n \in K\). Let \(U_{\iota_0}\) be such that \(p \in U_{\iota_0}\). Let \(\epsilon \gt 0\) such that \((p - \epsilon, p + \epsilon) \subseteq U_{\iota_0}\). For all but finitely many \(n \geq 1\), \(K_n \subseteq U_{\iota_0}\). So \(K_n\) does have a finite cover, namely \(U_{\iota_0}\); a contradiction. \(\blacksquare\)
Theorem (Stone-Weierstrass): Suppose \(K \subseteq \mathbb R\) is a non-empty compact set, \(f: K \to \mathbb R\) continuous. Then, there is a sequence \(P_n \in \mathbb R[x]\) such that \(P_n \rightrightarrows_K f\).
Proof: Let \(\mathscr A := \lbrace f \in \mathscr C(K; \mathbb R) : \forall \epsilon \gt 0, \exists P \in \mathbb R[x], \forall x \in K, \vert f(x) - P(x) \vert \lt \epsilon \rbrace\). That is, \(\mathscr A\) is the set of continuous functions \(f: K \to \mathbb R\) such that we can pick a polynomial that is arbitrarily close to \(. We want to show that\)\mathscr A = \mathscr C(K; \mathbb R)$$.
(1) If \(f \in \mathscr C(K; \mathbb R)\) and \(\forall \epsilon \gt 0\), \(\exists g \in \mathscr A\) that is arbitrarily close to \(f\), then \(f \in \mathscr A\).
Proof (1): Pick \(\epsilon \gt 0\). Choose \(g \in \mathscr A\) such that \(\forall x \in K, \vert f(x) - g(x) \vert \lt \epsilon/2\). Since \(g \in \mathscr A\), choose \(P \in \mathbb R[x]\), such that \(\forall x \in K, \vert g(x) - P(x) \vert \lt \epsilon/2\). Then, \(\forall x, \vert f(x) - P(x) \vert \leq \vert f(x) - g(x) \vert + \vert g(x) - P(x) \lt \epsilon/2 + \epsilon/2 = \epsilon\) So \(f \in \mathscr A\). \(\blacksquare\)
(2) \(f, g \in \mathscr A \implies f \cdot g \in \mathscr A\)
Proof (2): Let \(M \gt 0\) be such that \(\vert f(x) \vert, \vert g(x) \vert \lt M, \forall x \in K\). Given \(\epsilon \gt 0\), choose \(P, Q \in \mathbb R[x]\) such that \(\vert f(x) - P(x) \vert, \vert g(x) - Q(x) \vert \lt \frac{\epsilon}{2(2M + 1)}\) and \(\vert g(x) - Q(x) \vert \lt 1, \forall x \in K\). Then \(\forall x \in K\),
\[\begin{align*} \vert f(x) g(x) - P(x)Q(x) \vert &= \vert f(x)g(x) - f(x)Q(x) + f(x)Q(x) - P(x)Q(x) \vert \\ &\leq \vert f(x) \vert \cdot \vert g(x) - Q(x) \vert + \vert Q(x) \vert \cdot \vert f(x) - P(x) \vert \\ &\lt \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ &= \epsilon \end{align*}\]So \(f \cdot g \in \mathscr A\). \(\blacksquare\)
(3) \(f, g \in \mathscr A \implies f + g \in \mathscr A\)
Proof (3): TODO.
(4) \(f \in \mathscr A, c \in \mathbb R \implies c \cdot f \in \mathscr A\)
Proof (4): TODO.
(5) \(f \in \mathscr A \implies P \circ f \in \mathscr A, \forall P \in \mathbb R[x]\)
Proof (5): Follows from (2), (3), (4). For example, if \(P(x) = x^3 + 2x - 5\), then \((P \circ f)(x) = f^3(x) + 2f(x) - 5\); a finite linear combination of functions in \(\mathscr A\). \(\blacksquare\)
(6) \(f \in \mathscr A \implies \vert f \vert \in \mathscr A\)
Proof (6): By (2), \(f^2 \in \mathscr A\). Note that \(\vert f(x) \vert = \sqrt{f(x)^2}\). If \(\max_{x \in K} \vert f(x) \vert \leq 1\), by Problem Set 6, Problem 2, \(\exists P \in \mathbb R[x], \forall x \in K, \vert \sqrt{f^2(x)} - P(f(x)) \vert \lt \epsilon\). If instead \(\max_{x \in K} \vert f(x) \vert =: M \gt 1\), then \(\left\vert \frac{f}{M} \right\vert \in \mathscr A\), or \(\frac{1}{M} \cdot \vert f \vert \in \mathscr A \implies \vert f \vert = M \cdot \frac{1}{m} f \in \mathscr A\). \(\blacksquare\)
(7) \(f, g \in A \implies \max \lbrace f, g \rbrace, \min \lbrace f, g \rbrace \in \mathscr A\).
Proof (7): Observe that \(\max \lbrace \alpha, \beta \rbrace = \frac{\alpha + \beta}{2} + \frac{\vert \alpha - \beta \vert}{2}\) and \(\min \lbrace \alpha, \beta \rbrace = \frac{\alpha + \beta}{2} - \frac{\vert \alpha - \beta \vert}{2}\). So the claim follows from (2), (3), (4). \(\blacksquare\)
(8) \(\forall x_1 \neq x_2, x_{1}, x_{2} \in K, \forall \alpha, \beta \in \mathbb R, \exists h \in \mathscr A\) such that \(h(x_1) = \alpha\) and \(h(x_2) = \beta\).
Proof (8): Choose \(P \in \mathbb R[x]\) such that \(P(x_1) \neq P(x_2)\). Define
\[\begin{align*} h(x) := \alpha + (\beta - \alpha) \cdot \frac{P(x) - P(x_1)}{P(x_2) - P(x_1)} \end{align*}\]Proof of Theorem: Given \(f \in \mathscr C(K; \mathbb R), \epsilon \gt 0\).
Step 1: For any \(x, y \in K\), choose \(h_{x, y} \in \mathscr A\) such that \(h_{x, y}(x) = f(x) \land h_{x, y}(y) = f(y)\).
Step 2: Fix \(x_0 \in K\). For any \(y \in K, f(y) - h_{x_0, y}(y) = 0\). Since this is the difference of continuous functions, it is continuous. So choose an open neighborhood \(U_y\) of \(y\) such that \(\forall z \in K \cap U_y\), \(f(z) - h_{x_0, y}(z) \gt -\epsilon\). By compactness, there are finitely many such neighborhoods \(U_{y_1}, \dots, U_{y_n}\) such that their union covers \(K\). Set \(h_{x_0} := \min \lbrace h_{x_0, y_1}, \dots, h_{x_0, y_n} \rbrace\). By (7), \(h_{x_0} \in A\). Also, \(\forall z \in K, f(z) + \epsilon \gt h_x(z)\), sinec \(h_{x_0}\) is the minimum of all such functions, which themselves satisfy the inequality locally.
Step 3: Since \(\forall x \in K, h_x(x) - f(x) = 0\). Then by continuity, we can choose an open neighborhood \(V_x\) such that \(\forall z \in V_x\), \(h_x(z) - f(z) \gt -\epsilon\). Let \(x_1, \dots, x_m \in K\) be such that \(K \subseteq V_{x_1} \cup \dots \cup V_{x_m}\), which exists by the proposition about open covers. Define \(g := \max \lbrace h_{x_1}, \dots, h_{x_n} \rbrace \in A\), by (7). Then \(\forall z \in K, g(z) \gt f(z) - \epsilon\). Hence, \(\forall z \in K\), \(f(z) - \epsilon \lt g(z) \lt f(z) + \epsilon\). \(\blacksquare\).