7. Sequence of Functions

6.2 Uniform Convergence of a Sequence of Functions

Point-wise Convergence

For each $n \in N$ let $f_{n} : A \to R$ . ( $f_{n}$ ) converges point-wise on $A$ to $f$ if for all $x \in A$ , $(f_{n} (x)) \to f (x)$
(For all $ε > 0$ and $x \in A$ , $\exists N_{x} \in N$ so that $n > N ⟹ | f_{n} (x) - f (x) | < ε$ )
$N$ depends on $x$

Example

Not always continuous
$g_{n} (x) = x^{n}$ on $[0, 1]$
on $[0, 1), (x^{n}) \to 0$
$x = 1 ⟹ (x^{n}) \to 1$

(g_{n}) \to g = {\begin{cases} 0 & on [0, 1) \\ 1 & x = 1 \end{cases}

Example

Not always uniform
$n \cdot x \cdot e^{- n x} = f_{n}$ on $[0, 1]$
Pasted image 20250313100336.webp|236

$max (f_{n}) = f_{n} (\frac{1}{n}) = \frac{1}{e}$

$x = 0$
$f_{n} \to 0$

if $ε < 1 / ε$ , $| f_{n} (x) - f (x) | \geq \frac{1}{e} > ε$

Example

Not always differentiable
$h_{n} (x) = x^{1 + \frac{1}{2 n - 1}}$ on $[- 1, 1]$

$lim h_{n} = x lim x^{1 / (2 n - 1)} = | x |$

Uniform Convergence (function sequence)

Let $(f_{n})$ be a sequence of functions on $A \subseteq R$ . Then $(f_{n})$ converges uniformly on $A$ to $f$ if for every $ε > 0$ there exists an $N_{ε} \in N$ so that $| f_{n} (x) - f (x) | < ε$ whenever $n \geq N$ for all $x \in A$
$N$ does not depend on x

Thm Cauchy Criterion for Uniform Convergence

$(f_{n})$ converges uniformly on $A \subseteq R$ iff for all $ε > 0$ there exists an $N \in N$ so that $| f_{n} (x) - f_{m} (x) | < ε$ whenever $n, m \geq N$ and $x \in A$

Thm Dini

A monotone sequence of continuous functions converge point-wise on a compact set.
If the limit is continuous then the convergence is uniform.

Continuous Limit Theorem

Assume $f_{n}$ is continuous on $A \subseteq R$
If $f_{n} \overset{uniform}{\to} f$ and $f_{n}$ is continuous at $x \in A$ $⟹$ $f$ is continuous at $x$

Other facts

Assume $f_{n}$ is continuous
If $f_{n} \overset{uniform}{\to} f$ and $f_{n}$ is differentiable on interval $I$ , not guaranteed that $f^{'}$ exists.
Example: Weierstrass function

What if $f^{'}$ exists? Must $f_{n}^{'} \to f^{'}$ , nope
Example $f_{n} = \frac{\sin (n x)}{\sqrt{n}} \overset{u n i f}{\to} 0$
$f_{n}^{'} = \sqrt{n} \cos (n x) ↛ 0$

Thm Differentiable Limit Theorem

\begin{aligned} * & f_{n} \to f point-wise on [a, b] \\ f_{n}^{'} exists on [a, b] \\ (f_{n}^{'}) \overset{uniform}{\to} g on [a, b] \\ ⟹ f^{'} exists and f^{'} = g \end{aligned}

Proof:

Fix $c \in [a, b]$ and let $ε > 0$ .
WTS $f^{'} (c)$ exists and equals $g (c)$

f^{'} (c) = lim_{n \to c} \frac{f (x) - f (c)}{x - c}

want $δ > 0$ so that when $0 < | x - c | < δ$

\begin{array}{r} | \frac{f (x) - f (c)}{x - c} - g (c) | < ε \end{array}

\begin{aligned} | \frac{f (x) - f (c)}{x - c} - g (c) | \leq & | \frac{f (x) - f (c)}{x - c} - \frac{f_{n} (x)}{x - c} | \\ + | \frac{f_{n} (x) - f_{n} (x)}{x - c} - f_{n}^{'} (c) | + | f_{n}^{'} (c) - g (c) | \end{aligned}

ﬁnd an $f_{n}$ that forces the ﬁrst and third terms on the right-hand side to be less than /3. Once we establish which $f_{n}$ we want, we can then use the differentiability
of $f_{n}$ to produce a $δ$ that makes the middle term less than $ε$ /3 for all $x$ satisfying $0 < | x - c | < δ$

Differentiability stronger

We can actually skip proving point-wise convergence:

\begin{aligned} f_{n} \to f point-wise on [a, b] \\ f_{n}^{'} exists on [a, b] \\ (f_{n}^{'}) \overset{uniform}{\to} g on [a, b] \\ \exists x_{0} \in [a, b], f_{n} (x_{0}) \to L ⟹ f^{'} exists and f^{'} = g \end{aligned}

Moreover, the limit function $f = lim f_{n}$ is differentiable and satisﬁes $f^{'} = g$ .