Sequences and Series of Functions

Let $(f_n)$ be a sequence of functions defined on a set $E \subseteq \mathbb{R}$.

Definition. $(f_n)$ converges pointwise to $f$ on $E$ if for every $x \in E$ and every $\varepsilon > 0$There exists $N \in \mathbb{N}$ (depending on both $x$ and $\varepsilon$) such that $|f_n(x) - f(x)| \lt \varepsilon$ for all $n \geq N$.

Example. Let $f_n(x) = x^n$ on $E = [0, 1]$. For each $x \in [0, 1)$, $f_n(x) = x^n \to 0$And $f_n(1) = 1$ for all $n$. So $f_n$ converges pointwise to

$f(x) = \begin{cases} 0 & \mathrm{if\ } 0 \leq x \lt 1 \\ 1 & \mathrm{if\ } x = 1 \end{cases}$

Note that each $f_n$ is continuous, but the pointwise limit $f$ is not continuous at $x = 1$.

Definition. $(f_n)$ converges uniformly to $f$ on $E$ if for every $\varepsilon > 0$There Exists $N \in \mathbb{N}$ (depending only on $\varepsilon$Not on $x$) such that for all $x \in E$:

$|f_n(x) - f(x)| \lt \varepsilon \quad \mathrm{for\ all\ } n \geq N$

Equivalently, $\sup_{x \in E} |f_n(x) - f(x)| \to 0$ as $n \to \infty$.

Proposition 7.1. Uniform convergence implies pointwise convergence. The converse is false.

Example (continued). $f_n(x) = x^n$ on $[0, 1]$ converges pointwise but not uniformly. We have $\sup_{x \in [0,1]} |f_n(x) - f(x)| = \sup_{x \in [0,1)} x^n = 1$ for all $n$ (since the supremum is Approached as $x \to 1^-$). This does not tend to $0$.

However, on $[0, r]$ for any $r \lt 1$: $\sup_{x \in [0,r]} |x^n| = r^n \to 0$So the convergence Is uniform on $[0, r]$.

Theorem 7.1 (Weierstrass M-Test). Let $(f_n)$ be a sequence of functions on $E$. If there exists a Sequence $(M_n)$ of non-negative real numbers such that $|f_n(x)| \leq M_n$ for all $x \in E$ and all $n$And $\sum_{n=1}^{\infty} M_n \lt \infty$Then $\sum_{n=1}^{\infty} f_n$ converges uniformly on $E$.

Proof. Let $S_n(x) = \sum_{k=1}^{n} f_k(x)$ and $T_n = \sum_{k=1}^{n} M_k$. Since $\sum M_k$ converges, $(T_n)$ is a Cauchy sequence. Given $\varepsilon > 0$There exists $N$ such that for $m > n \geq N$:

$T_m - T_n = \sum_{k=n+1}^{m} M_k \lt \varepsilon$

Then for all $x \in E$ and $m > n \geq N$:

$|S_m(x) - S_n(x)| = \left|\sum_{k=n+1}^{m} f_k(x)\right| \leq \sum_{k=n+1}^{m} |f_k(x)| \leq \sum_{k=n+1}^{m} M_k \lt \varepsilon$

So the partial sums $(S_n)$ satisfy the uniform Cauchy criterion on $E$Hence converge uniformly. $\blacksquare$

Theorem 7.2. If $(f_n)$ is a sequence of continuous functions on $E$ converging uniformly to $f$ On $E$Then $f$ is continuous on $E$.

Proof. Let $c \in E$ and $\varepsilon > 0$. Since $f_n \to f$ uniformly, choose $N$ such that $|f_N(x) - f(x)| \lt \varepsilon/3$ for all $x \in E$. Since $f_N$ is continuous at $c$Choose $\delta > 0$ such that $|x - c| \lt \delta$ implies $|f_N(x) - f_N(c)| \lt \varepsilon/3$. Then:

$|f(x) - f(c)| \leq |f(x) - f_N(x)| + |f_N(x) - f_N(c)| + |f_N(c) - f(c)| \lt \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon$

$\blacksquare$

Theorem 7.3. If $(f_n)$ is a sequence of Riemann integrable functions on $[a, b]$ converging Uniformly to $f$ on $[a, b]$Then $f$ is Riemann integrable and

$\lim_{n \to \infty} \int_a^b f_n(x)\, dx = \int_a^b f(x)\, dx$

Proof. Since $(f_n)$ converges uniformly, $f$ is the uniform limit of integrable functions. Given $\varepsilon > 0$Choose $N$ with $\sup |f_N(x) - f(x)| \lt \varepsilon/(2(b-a))$ for all $x \in [a, b]$. Then $f_N - \varepsilon/(2(b-a)) \leq f(x) \leq f_N(x) + \varepsilon/(2(b-a))$ for all $x$And by Integrability of $f_N$:

$\int_a^b f_N - \frac{\varepsilon}{2} \leq \underline{\int_a^b} f \leq \overline{\int_a^b} f \leq \int_a^b f_N + \frac{\varepsilon}{2}$

So $\overline{\int} f - \underline{\int} f \leq \varepsilon$Proving $f$ is integrable. For the limit:

$\left|\int_a^b f_n - \int_a^b f\right| \leq \int_a^b |f_n - f| \leq (b-a) \cdot \sup_{[a,b]} |f_n - f| \to 0$

$\blacksquare$

Uniform convergence of functions does not guarantee convergence of derivatives. A stronger Hypothesis is needed.

Theorem 7.4. Suppose $(f_n)$ is a sequence of differentiable functions on $[a, b]$ such that:

1. $(f_n(c))$ converges for some $c \in [a, b]$
2. $(f_n")$ converges uniformly on $[a, b]$

Then $(f_n)$ converges uniformly to a differentiable function $f$ on $[a, b]$And $f'(x) = \lim_{n \to \infty} f_n'(x)$.

Proof. Let $g = \lim f_n'$ (uniform limit). Define $f(x) = \lim_{n \to \infty} \left[f_n(c) + \int_c^x f_n'(t)\, dt\right]$. By Theorem 7.3, $\int_c^x f_n'(t)\, dt \to \int_c^x g(t)\, dt$So $f(x) = f(c) + \int_c^x g(t)\, dt$. By FTC Part 1, $f$ is differentiable and $f'(x) = g(x)$. Uniform convergence of $f_n$ to $f$ follows From the estimate $|f_n(x) - f(x)| \leq |f_n(c) - f(c)| + \int_a^b |f_n'(t) - g(t)|\, dt$. $\blacksquare$

A power series centered at $a$ is a series of the form $\sum_{n=0}^{\infty} c_n (x - a)^n$.

Theorem 7.5 (Radius of Convergence). Every power series $\sum c_n (x - a)^n$ has a radius of Convergence $R \in [0, \infty]$ such that:

• The series converges absolutely for $|x - a| \lt R$
• The series diverges for $|x - a| > R$
• The behavior at $|x - a| = R$ must be checked separately

The radius is given by $1/R = \limsup_{n \to \infty} \sqrt[n]{|c_n|}$ (Cauchy-Hadamard formula), or When the limit exists, $R = \lim_{n \to \infty} |c_n/c_{n+1}|$.

Proof. Apply the root test to $\sum |c_n (x-a)^n|$: $\limsup \sqrt[n]{|c_n|} |x-a| = |x-a|/R$ (where $1/R = \limsup \sqrt[n]{|c_n|}$). The root test gives convergence when $|x-a|/R \lt 1$ And divergence when $|x-a|/R > 1$. $\blacksquare$

Theorem 7.6. A power series converges uniformly on every compact subset of its open disk of Convergence.

Theorem 7.6a (Differentiation and Integration of Power Series). If $f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n$ Has radius of convergence $R > 0$Then:

1. $f$ is differentiable on $(a - R, a + R)$ and $f'(x) = \sum_{n=1}^{\infty} n c_n (x - a)^{n-1}$ (same $R$).
2. $f$ is infinitely differentiable on $(a - R, a + R)$And $f^{(k)}(x) = \sum_{n=k}^{\infty} \frac{n!}{(n-k)!} c_n (x - a)^{n-k}$.
3. $\int_a^x f(t)\, dt = \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x - a)^{n+1}$ for $|x - a| \lt R$.
4. $c_n = f^{(n)}(a)/n!$ (uniqueness of power series coefficients).

Proof. The differentiated series $\sum n c_n (x-a)^{n-1}$ has the same radius of convergence as The original (by the Cauchy-Hadamard formula, since $\sqrt[n]{n} \to 1$). By Theorem 7.4, the Derivative of the sum equals the sum of the derivatives. Parts (2), (3), and (4) follow by Induction and the FTC. $\blacksquare$

Theorem 7.6b (Abel’s Theorem). If $\sum_{n=0}^{\infty} c_n$ converges to $L$Then

$\lim_{x \to 1^-} \sum_{n=0}^{\infty} c_n x^n = L$

That is, the power series is continuous from the left at the endpoint $x = 1$.

Proof (sketch). Let $s_n = \sum_{k=0}^{n} c_k$ and $s_n \to L$. Write the partial sum $\sum_{k=0}^{n} c_k x^k = \sum_{k=0}^{n}(s_k - s_{k-1})x^k$ (with $s_{-1} = 0$) and use summation by Parts to express this as $s_n x^n + \sum_{k=0}^{n-1} s_k(x^k - x^{k+1})$. Letting $n \to \infty$ and using That $s_n \to L$ and $x^n \to 0$ for $|x| \lt 1$One shows the expression tends to $L$ as $x \to 1^-$. $\blacksquare$

Example. Since $\sum_{k=1}^{\infty} (-1)^{k+1}/k = \ln 2$Abel’s theorem gives $\lim_{x \to 1^-} \sum_{k=1}^{\infty} (-1)^{k+1} x^k/k = \ln 2$I.e., $\ln 2$ is the left-hand limit Of $-\ln(1 - x)$ at $x = 1$.

The Taylor series of $f$ at $a$ is $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n$.

Definition. A function $f$ is analytic at $a$ if its Taylor series at $a$ converges to $f(x)$ In some neighborhood of $a$.

Not every $C^{\infty}$ function is analytic. The standard counterexample is:

$f(x) = \begin{cases} e^{-1/x^2} & x \neq 0 \\ 0 & x = 0 \end{cases}$

$f^{(n)}(0) = 0$ for all $n$So the Taylor series at $0$ is identically zero, which Converges only to $0$Not to $f(x)$ for $x \neq 0$.

:::caution Common Pitfall Pointwise convergence does not preserve continuity, differentiability, or integrability. Uniform Convergence preserves continuity and allows interchange of limit and integral, but not limit and Derivative. For derivatives, uniform convergence of the sequence of derivatives (not the original Sequence) is required, as stated in Theorem 7.4. Also, the Weierstrass M-Test applies only to series Of functions, not sequences; for sequences, one must verify the uniform Cauchy criterion directly.

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