Function sequence and function series

5.3 Function sequence and function series

Definition 5.3.1 Assuming $f, f_{n} : I \to R (C)$ , $f_{n}$ is pointwisely convergent to $f$ on $I$ if $\forall x \in I$ , $lim_{n \to + \infty} f_{n} (x) = f (x)$ , i.e. $\forall x \in I$ , $\forall ϵ > 0$ , $\exists N_{ϵ} (x)$ s.t. $\forall n \geq N_{ϵ} (x)$ , $| f_{n} (x) - f (x) | < ϵ$ .

Definition 5.3.2 Assuming $f, f_{n} : I \to R (C)$ , $f_{n}$ is uniformly convergent to $f$ on $I$ if $\forall ϵ > 0$ , $\exists N_{ϵ}$ s.t. $\forall x \in I$ , $\forall n \geq N_{ϵ}$ , $| f_{n} (x) - f (x) | < ϵ$ , marked as $f_{n} \overset{I}{⇉} f$ .

Theorem 5.3.3 Mark $∥ f ∥_{\infty} = sup {| f (x) | | x \in I}$ , then $f_{n} \overset{I}{⇉} f \Leftrightarrow lim_{n \to + \infty} ∥ f_{n} - f ∥_{\infty} = 0$ .

Note

1.: If $f_{n} \overset{I}{⇉} f$ , then $f_{n}$ is pointwisely convergent to $f$ .
2.: For uniform convergence, given $ϵ > 0$ , $\exists N > 0$ s.t. $\forall n > N$ , $\forall x \in I$ , $| f_{n} (x) - f (x) | < ϵ$ , meaning $f_{n}$ converges to $f$ synchronously at any $x$ .

Example 5.3.1 Assuming $f_{n} (x) = x^{n}$ , $I = [0, 1]$ . Then $\forall x \in [0, 1)$ , $x^{n} \to 0, n \to + \infty$ ; $x = 1$ , $x^{n} = 1 \to 1, n \to + \infty$ , meaning $f$ is pointwisely convergent, and $f (x) = {\begin{cases} 0 & x \in [0, 1) \\ 1 & x = 1 \end{cases}$ $f_{n}$ is uniformly convergent to $f$ on $[0, a]$ for any $a \in (0, 1)$ , since $| f_{n} (x) - f (x) | = | x^{n} - 0 | = x^{n} \leq a^{n} < ϵ \Rightarrow n > \frac{\ln ϵ}{\ln a}$ Yet $f_{n}$ is not uniformly convergent to $f$ on $[0, 1]$ , since $\forall n \geq 2$ , take $x_{n} = \sqrt[n]{1 - \frac{1}{n}} \in (0, 1)$ , we have $| f_{n} (x_{n}) - f (x_{n}) | = 1 - \frac{1}{n} \geq \frac{1}{2} > ϵ$ . Another way to verify it is $∥ f_{n} - f ∥_{\infty} = sup_{x \in [0, 1]} | f_{n} (x) - f (x) | \geq | f_{n} (x_{n}) - f (x_{n}) | \geq \frac{1}{2}$ whose limit is not $0$ .

Theorem 5.3.4 (Uniformly Cauchy) $f_{n} \overset{I}{⇉} f$ if and only if ${f_{n}}$ is uniformly Cauchy on $I$ , i.e. $\forall ϵ > 0$ , $\exists N_{ϵ}$ s.t. $\forall n, m \geq N_{ϵ}$ , $∥ f_{n} - f_{m} ∥ < ϵ$ or $\forall x \in I$ , $| f_{n} (x) - f_{m} (x) | < ϵ$ .

Proof $\Rightarrow$ is trivial. $\Leftarrow$ : Assuming $f_{n}$ is uniformly Cauchy on $I$ , then $\forall x \in I$ , ${f_{n} (x)}$ is a Cauchy sequence, meaning there exists the limit $f (x)$ of $f_{n} (x)$ .

Notice that $\forall ϵ > 0$ , $\exists N_{ϵ}$ s.t. $\forall n, m \geq N_{ϵ}$ , $\forall x \in I$ , $| f_{n} (x) - f_{m} (x) | < ϵ$ . Let $m \to + \infty$ , $| f_{n} (x) - f (x) | \leq ϵ$ , meaning $f_{n} \overset{I}{⇉} f$ .

Another method to prove it. $\forall ϵ > 0$ , $\exists N_{ϵ}$ s.t. $\forall n, m \geq N_{ϵ}$ , $\forall x \in I$ , $| f_{n} (x) - f_{m} (x) | < ϵ$ ; for the same $ϵ > 0$ , $\exists M_{ϵ} (x)$ s.t. $\forall m \geq M_{ϵ} (x)$ , $| f_{m} (x) - f (x) | < ϵ$ , so $\forall N_{ϵ}$ , $\forall m \geq M_{ϵ} (x) + N_{ϵ}$ , $| f_{n} (x) - f (x) | \leq | f_{n} (x) - f_{m} (x) | + | f_{m} (x) - f (x) | < 2 ϵ$ meaning $f_{n} \overset{I}{⇉} f$ . $◻$

Boundness, continuity and integrability. Mark $B (I) = {f : I \to R is bounded}$ which is a linear space, $∥ f ∥_{\infty} = sup_{x \in I} | f (x) |$ .

Theorem 5.3.5 Assuming $f_{n} \overset{I}{⇉} f$ , $f_{n} \in B (I)$ , then

1.: $f \in B (I)$ , i.e. $(B (I), ∥ \cdot ∥_{\infty})$ is complete.
2.: ${f_{n}}$ is uniformly bounded on $I$ , i.e. $\exists M > 0$ s.t. $\forall n \geq 1, \forall x \in I$ , $| f_{n} (x) | \leq M$ .

Proof

1.: $\forall ϵ > 0$ , $\exists N_{ϵ} > 0$ s.t. $\forall n \geq N_{ϵ}$ , $\forall x \in I$ , $| f_{n} (x) - f (x) | < ϵ$ , so $| f (x) | \leq | f_{N_{1}} (x) - f (x) | + | f_{N_{1}} (x) | < 1 + ∥ f_{N_{1}} ∥_{\infty}$ , meaning $f \in B (I)$ .
2.: $\forall n \geq N_{1}$ , $\forall x \in I$ , $| f_{n} (x) | \leq | f_{n} (x) - f (x) | + | f (x) | \leq 1 + ∥ f ∥_{\infty}$ meaning $∥ f_{n} ∥_{\infty} \leq 1 + ∥ f ∥_{\infty}$ . Therefore, $\forall n \geq 1$ , $∥ f_{n} ∥_{\infty} \leq ∥ f_{1} ∥_{\infty} + \dots + ∥ f_{N_{1}} ∥_{\infty} + ∥ f ∥_{\infty} + 1 = M$ meaning ${f_{n}}$ is uniformly bounded on $I$ . $◻$

Example 5.3.2 Assuming $f_{n} (x) = {\begin{cases} \frac{1}{x} & \frac{1}{n} \leq x \leq 1 \\ n & 0 < x \leq \frac{1}{n} \end{cases}$ which is bounded on $(0, 1]$ and pointwisely convergent to $f (x) = \frac{1}{x}$ , yet $f_{n}$ is not uniformly convergent since $f$ is not bounded on $(0, 1]$ .

Theorem 5.3.6 Assuming $f_{n} \in C (I)$ , $f_{n} \overset{I}{⇉} f$ , then

1.: $f \in C (I)$ , i.e. $C (I)$ is a closed set under uniform convergence.
2.: Assuming $I \subseteq R$ is a bounded closed set, then $C (I) \subseteq B (I)$ is a bounded subset under $∥ \cdot ∥_{\infty}$ .
3.: Assuming $I \subseteq R$ is a bounded closed set, then ${f_{n}}$ is uniformly equicontinuous on $I$ , i.e. $\forall ϵ > 0$ , $\exists δ (ϵ) > 0$ s.t. $\forall n \geq 1$ , $\forall x, y \in I$ , $| x - y | < δ (ϵ) \Rightarrow | f_{n} (x) - f_{n} (y) | < ϵ$ .

Proof Only 1 will be proved. $\forall x_{0} \in I$ , prove that $f$ is continuous at $x_{0}$ . Notice that $| f (x) - f (x_{0}) | \leq | f_{n} (x) - f_{n} (x_{0}) | + | f_{n} (x) - f (x) | + | f_{n} (x_{0}) - f (x_{0}) |$ $\forall ϵ > 0$ , $\exists N_{ϵ} > 0$ s.t. $\forall n \geq N_{ϵ}$ , $\forall x \in I$ , $| f_{n} (x) - f (x) | < ϵ$ . Since $f_{N_{ϵ}}$ is continuous at $x_{0}$ , so for the same $ϵ > 0$ , $\exists δ_{ϵ} (x_{0}) > 0$ s.t. $| x - x_{0} | < δ_{ϵ} (x_{0}) \Rightarrow | f_{N_{ϵ}} (x) - f_{N_{ϵ}} (x_{0}) | < ϵ$ . Therefore, $| f (x) - f (x_{0}) | \leq | f_{N_{ϵ}} (x) - f_{N_{ϵ}} (x_{0}) | + | f_{N_{ϵ}} (x) - f (x) | + | f_{N_{ϵ}} (x_{0}) - f (x_{0}) | < 3 ϵ$ meaning $f \in C (I)$ . $◻$

Theorem 5.3.7 Assuming $f_{n} \overset{I}{⇉} f$ , $f_{n} \in R (I)$ , then $f \in R (I)$ , i.e. $(R (I), ∥ \cdot ∥_{\infty})$ is complete and $\int_{I} f (x) d x = \int_{I} lim_{n \to + \infty} f_{n} (x) d x = lim_{n \to + \infty} \int_{I} f_{n} (x) d x$

Proof Since $f_{n} \in R (I)$ , so $f_{n} \in B (I)$ . Since $f_{n} \overset{I}{⇉} f$ , so $f \in B (I)$ . Assuming $D_{n}$ is the set of discontinuities of $f$ on $I$ which is a set of measure $0$ . Take $D = ⋃_{n = 1}^{+ \infty} D_{n}$ , then $\forall x \in I ∖ D$ , $\forall n \geq 1$ , $f_{n}$ is continuous at $x$ . According to the proof of theorem above (not theorem itself), $f$ is continuous at $x$ , and the set of discontinuities of $f$ on $I$ is the subset of $D$ . According to Lebesgue criterion, $D_{n}$ are all sets of measure 0, meaning $\forall ϵ > 0$ , $| D_{1} | < ϵ$ , $| D_{2} | < \frac{ϵ}{2}$ and $| D_{n} | < \frac{ϵ}{2^{n - 1}}$ , so $| D | \leq \sum_{n = 1}^{+ \infty} | D_{n} | < \sum_{n = 1}^{+ \infty} \frac{ϵ}{2^{n - 1}} = 2 ϵ$ meaning $D$ is also a set of measure 0. According to Lebesgue criterion, $f \in R (I)$ , and $| \int_{I} f_{n} (x) d x - \int_{I} f (x) d x | \leq \int_{I} | f_{n} (x) - f (x) | d x \leq \int_{I} ∥ f_{n} - f ∥_{\infty} d x = ∥ f_{n} - f ∥_{\infty} | I | \to 0$ when $n \to + \infty$ . $◻$

Theorem 5.3.8 Assuming $f_{n}, g : I \to R$ , $I = [a, b]$ satisfying

$f_{n} (a) \to A$ , $n \to + \infty$ .
$f_{n}^{'} \in C (I)$ , $f_{n}^{'} \overset{I}{⇉} g$ .

then $f_{n}$ is uniformly convergent on $I$ , mark $f = lim_{n \to + \infty} f_{n}$ , then $f \in C (I)$ and $f^{'} (x) = g (x)$ , i.e. ${(lim_{n \to + \infty} f_{n} (x))}^{'} = lim_{n \to + \infty} f_{n}^{'} (x)$

Proof Since $f_{n}^{'} \overset{I}{⇉} g$ , $f \in C (I)$ , so $g \in C (I)$ . Notice that $f (x) = A + \int_{a}^{x} g (u) d u, f_{n} (x) = f_{n} (a) + \int_{a}^{x} f_{n}^{'} (u) d u$ hence, $\begin{aligned} | f_{n} (x) - f (x) | & \leq | f_{n} (a) - A | + \int_{a}^{x} | f_{n}^{'} (u) - g (u) | d u \\ \leq | f_{n} (a) - A | + \int_{a}^{x} ∥ f_{n}^{'} - g ∥_{\infty} d u = | f_{n} (a) - A | + ∥ f_{n}^{'} - g ∥_{\infty} (x - a) \\ \leq \underset{\exists N_{1} (ϵ) s.t. \forall n \geq N_{1}}{\underset{⏟}{| f_{n} (a) - A |}} + \underset{\exists N_{2} (ϵ) s.t. \forall n \geq N_{2}}{\underset{⏟}{∥ f_{n}^{'} - g ∥_{\infty}}} (b - a) < ϵ + ϵ (b - a) \end{aligned}$

meaning $\forall ϵ > 0$ , $\exists N_{ϵ} = N_{1} (ϵ) + N_{2} (ϵ)$ s.t. $\forall n \geq N_{ϵ}$ , $| f_{n} (x) - f (x) | < ϵ$ , i.e. $f_{n} \overset{I}{⇉} f$ . $◻$

Uniform convergence of function series and properties of sum function.

Definition 5.3.9 $\sum_{n = 1}^{+ \infty} u_{n} (x) \overset{I}{⇉} S (x)$ if $S_{N} (x) = \sum_{n = 1}^{N} u_{n} (x) \overset{I}{⇉} S (x)$ , i.e. $\forall ϵ > 0$ , $\exists N_{ϵ} \geq 1$ s.t. $\forall N \geq N_{ϵ}$ , ${‖ \sum_{n = 1}^{N} u_{n} (x) - S (x) ‖}_{\infty} < ϵ$ , i.e. $\forall x \in I$ , $| \sum_{n = 1}^{N} u_{n} (x) - S (x) | < ϵ$ .

Theorem 5.3.10 Assuming $\sum_{n = 1}^{+ \infty} u_{n} \overset{I}{⇉} S$ .

1.: If $u_{n}$ is continuous at $x_{0} \in I$ , then $S$ is continuous at $x_{0}$ .
2.: If $u_{n} \in R (I)$ , then $S \in R (I)$ and $\int_{I} \sum_{n = 1}^{+ \infty} u_{n} (x) d x = \int_{I} S (x) d x = \sum_{n = 1}^{+ \infty} \int_{I} u_{n} (x) d x$

Theorem 5.3.11 If $u_{n} \in C^{1} (I)$ , $\sum_{n = 1}^{+ \infty} u_{n}^{'} \overset{I}{⇉} T$ , and $\exists x_{0} \in I$ s.t. $\sum_{n = 1}^{+ \infty} u_{n} (x_{0}) = A \in R$ , then $\sum_{n = 1}^{+ \infty} u_{n} \overset{I}{⇉} S (x) = A + \int_{x_{0}}^{x} T (t) d t$ where $S \in C (I)$ and ${(\sum_{n = 1}^{+ \infty} u_{n} (x))}^{'} = S^{'} (x) = T (x) = \sum_{n = 1}^{+ \infty} u_{n}^{'} (x)$

Criterion of uniform convergence.

Theorem 5.3.12 (Cauchy) $\sum_{n = 1}^{+ \infty} u_{n}$ is uniformly convergent on $I$ is equivalent to that ${S_{N}}$ is uniformly Cauchy on $I$ , i.e. $\forall ϵ > 0$ , $\exists N_{ϵ} \geq 1$ s.t. $\forall n \geq N_{ϵ}$ , $\forall p \geq 1$ , $∥ u_{n + 1} + \dots + u_{n + p} ∥_{\infty} < ϵ$ , i.e. $\forall x \in I$ , $| u_{n + 1} (x) + \dots + u_{n + p} (x) | < ϵ$ .

Weierstrass criterion for absolutely convergent function series.

Theorem 5.3.13 (Weierstrass) If $\forall n \geq N_{0}$ , $\forall x \in I$ , $| u_{n} (x) | \leq a_{n}$ , $\sum_{n = 1}^{+ \infty} a_{n}$ is (absolutely) convergent, then $\sum_{n = 1}^{+ \infty} u_{n} (x)$ is uniformly absolutely convergent on $I$ .

Example 5.3.3 Assuming a series $\sum_{n = 1}^{+ \infty} \frac{x^{2}}{(1 + x^{2})^{n}}$ , $x \in R$ .

When $| x | \geq δ > 0$ , we have $0 \leq \frac{x^{2}}{(1 + x^{2})^{n}} \leq \frac{1}{(1 + x^{2})^{n - 1}} \leq \frac{1}{(1 + δ^{2})^{n - 1}}$ So $\forall δ > 0$ , the original series in uniformly convergent on $(- \infty, - δ]$ and $[δ, + \infty)$ , meaning it’s internally-closed uniformly convergent.

The original series is not uniformly convergent on any set with $0$ as its cluster point. $u_{n} (x) = \frac{x^{2}}{(1 + x^{2})^{n}}$ takes the maximum at $x_{n} = \frac{\pm 1}{\sqrt{n - 1}}$ , so $u_{n} (x_{n}) = \frac{\frac{1}{n - 1}}{{(1 + \frac{1}{n - 1})}^{n}} ≃ \frac{1}{n e}$ Here $u_{n + k} (x_{n}) = \frac{\frac{1}{n - 1}}{{(1 + \frac{1}{n - 1})}^{n + k}}$ Therefore, $\sum_{k = 1}^{n} u_{n + k} (x_{n}) \geq n \cdot \frac{\frac{1}{n - 1}}{{(1 + \frac{1}{n - 1})}^{2 n}} \overset{n \to + \infty}{\to} \frac{1}{e^{2}}$ meaning $\exists N_{0}$ s.t. $\forall n > N_{0}$ , $u_{n + 1} (x_{n}) + \dots + u_{2 n} (x_{n}) \geq \frac{1}{2 e^{2}}$ , so it’s not uniformly Cauchy, meaning it’s not uniformly convergent.

Dirichlet and Abel criterion for conditionally convergent function series.

Theorem 5.3.14 $\sum_{n = 1}^{+ \infty} u_{n} (x) v_{n} (x) \overset{I}{⇉} S (x)$ if one of the following sets of condition hold,

(Dirichlet) $\sum_{n = 1}^{N} u_{n} (x)$ is uniformly bounded, i.e. $\exists M_{0} s.t. \forall x \in I, \forall N \geq 1, | \sum_{n = 1}^{N} u_{n} (x) | \leq M_{0}$ $v_{n} (x)$ is monotonously decreasing with respect to $n$ , and $v_{n} (x) \overset{I}{⇉} 0$ when $n \to + \infty$ .
(Abel) $\sum_{n = 1}^{N} u_{n} (x)$ is uniformly convergent; $v_{n} (x)$ is monotonously decreasing with respect to $n$ , and uniformly bounded.

Continuous yet never-being-derivable functions.

Example 5.3.4 Weierstrass function. $W (x) = \sum_{n = 0}^{\infty} a^{n} \cos (b^{n} π x)$ where $a \in (0, 1)$ (indicating that it’s uniformly absolutely convergent, so $W$ is continuous), $b$ is a positive odd number and $a b > 1$ . Here $W (x)$ is not derivable anywhere.

See Figure 5.3 another example of continuous yet never-being-derivable functions.

Figure 5.3: Continuous yet never-being-derivable function

Example 5.3.5 Prove that $\sum_{n = 1}^{+ \infty} \frac{\sin n x}{n} = \frac{π - x}{2}$ when $x \in (0, 2 π)$ .

Notice that $\begin{aligned} {(\sum_{n = 1}^{N} \frac{\sin n x}{n})}^{'} & = \sum_{n = 1}^{N} {(\frac{\sin n x}{n})}^{'} = \sum_{n = 1}^{N} \cos n x = \frac{1}{2 \sin \frac{x}{2}} \sum_{n = 1}^{N} 2 \sin \frac{x}{2} \cos n x \\ = \frac{1}{2 \sin \frac{x}{2}} \sum_{n = 1}^{N} [\sin (n + \frac{1}{2}) x + \sin (n - \frac{1}{2}) x] = \frac{\sin (N + \frac{1}{2}) x}{2 \sin \frac{x}{2}} - \frac{1}{2} \\ \sum_{n = 1}^{N} \frac{\sin n x}{n} & = \int_{π}^{x} \sum_{n = 1}^{N} \cos n t d t = \int_{π}^{x} [\frac{\sin (N + \frac{1}{2}) t}{2 \sin \frac{t}{2}} - \frac{1}{2}] d t \end{aligned}$

where $\begin{aligned} \int_{π}^{x} \frac{\sin (N + \frac{1}{2}) t}{2 \sin \frac{t}{2}} d t & = - {\frac{\cos (N + \frac{1}{2}) t}{(2 N + 1) \sin \frac{t}{2}} |}_{x}^{π} + \int_{π}^{x} \frac{\cos (N + \frac{1}{2}) t}{2 N + 1} d \frac{1}{\sin \frac{t}{2}} \\ \overset{δ \leq x \leq 2 π - x}{\to} \leq \frac{M_{1}}{2 N + 1} + \frac{M_{2} | x - π |}{2 N + 1} \overset{N \to + \infty}{\to} 0 \end{aligned}$

Let $N \to + \infty$ , we have $\sum_{n = 1}^{+ \infty} \frac{\sin n x}{n} = \frac{π - x}{2}$ on any $[δ, 2 π - δ]$ , meaning it holds on $(0, 2 π)$ .

Example 5.3.6 $\sum_{n = 1}^{+ \infty} \frac{\cos n x}{n^{2}}$ is uniformly absolutely convergent on $R$ , and ${(\frac{\cos n x}{n^{2}})}^{'} = - \frac{\sin n x}{n}$ which is uniformly convergent on any $[δ, 2 π - δ]$ . Therefore $\forall x \in (0, 2 π)$ , ${(\sum_{n = 1}^{+ \infty} \frac{\cos n x}{n^{2}})}^{'} = \sum_{n = 1}^{+ \infty} {(\frac{\cos n x}{n^{2}})}^{'} = - \sum_{n = 1}^{+ \infty} \frac{\sin n x}{n} = \frac{x - π}{2}$ Therefore, $C (x) = \sum_{n = 1}^{+ \infty} \frac{\cos n x}{n^{2}} = C (0) + \int_{0}^{x} \frac{t - π}{2} d t = C (0) + \frac{x^{2} - 2 π x}{4}$ Notice that $\int_{0}^{2 π} C (x) d x = \sum_{n = 1}^{+ \infty} \frac{1}{n^{2}} \int_{0}^{2 π} \cos n x d x = 0$ Hence, $C (0) = \frac{π^{2}}{6}$ , i.e. $\sum_{n = 1}^{+ \infty} \frac{1}{n^{2}} = \frac{π^{2}}{6}$ .