Definition 9.1.1. For every $n \in \N$, let $f_n \colon S \to \R$ be a function. The sequence $\{ f_n \}_{n=1}^\infty$ converges pointwise to $f \colon S \to \R$ if for every $x \in S$, we have \begin{equation*} f(x) = \lim_{n\to\infty} f_n(x) . \end{equation*}

Consider the sequence of functions defined by $$f_n(x) \coloneqq x^{2n}.$$ It converges pointwise on $[-1,1]$ to $f \colon [-1,1] \to \R$, where \begin{equation*} f(x) = \begin{cases} 1 & \text{if } x=-1 \text{ or } x=1, \\ 0 & \text{otherwise.} \end{cases} \end{equation*}

Theorem 9.1.1. Let $f_n \colon S \to \R$ and $f \colon S \to \R$ be functions. Then $\{ f_n \}$ converges pointwise to $f$ if and only if for every $x \in S$, and every $\epsilon \gt 0$, there exists an $N \in \N$ such that for all $n \geq N$, we have \begin{equation*} \abs{\,f_n(x)-f(x)} \lt \epsilon . \end{equation*}

Definition 9.2.1. Let $f_n \colon S \to \R$ and $f \colon S \to \R$ be functions. The sequence $\{ f_n \}$ converges uniformly to $f$ if for every $\epsilon > 0$, there exists an $N \in \N$ such that for all $n \geq N$, we have \begin{equation*} \abs{\,f_n(x) - f(x)} \lt \epsilon \;\; \text{ for all } x \in S. \end{equation*}

Theorem 9.2.1. Let $\{ f_n \}$ be a sequence of functions $f_n \colon S \to \R$. If $\{ f_n \}$ converges uniformly to $f \colon S \to \R$, then converges pointwise to $f$.

Proof. It follows from Theorem 9.1.1. 📝

Let $f_n(x) := \dfrac{x}{n}$ defined on $\R$ for every $n\in \N$. Observe that $\{f_n\}$ converges pointwise to $f=0$, since

If we let $n_k:=k$ and $x_k := k ,$ then $f_{n_k}(x_k)=1$ so that

Therefore the sequence $\{f_n\}$ does not converges uniformly to $f$.

Theorem 9.2.2. Let $f_n \colon S \to \R$ and $f \colon S \to \R$ be functions.

for some $\epsilon_0\gt 0$ there is a subsequence $\{ f_{n_k} \}$ of $\{ f_n \}$ and a sequence $\{ x_k \}$ in $S$ such that \[ \abs{\,f_{n_k}\left(x_k\right)-f(x_k)}\geq \epsilon_0\, \text{ for all }\, k\in \N. \]

Definition 9.3.1. Let $f \colon S \to \R$ be a bounded function. Define \begin{equation*} \norm{\,f}_u := \sup \bigl\{ \abs{\,f(x)} : x \in S \bigr\} . \end{equation*} We call $\norm{\cdot}_u$ the uniform norm.

Theorem 9.3.1. A sequence of bounded functions $f_n \colon S \to \R$ converges uniformly to $f \colon S \to \R$, if and only if \begin{equation*} \lim_{n\to\infty} \norm{\,f_n - f}_u = 0 . \end{equation*}

Theorem 9.3.1. A sequence of bounded functions $f_n \colon S \to \R$ converges uniformly to $f \colon S \to \R$, if and only if $\lim_{n\to\infty} \norm{\,f_n - f}_u = 0 .$

Proof. $\nec$ Suppose $\{ f_n \}_{n=1}^\infty$ converges uniformly to $f$. Let $\epsilon > 0$ be given. Then find $N$ such that $\abs{f_n(x)-f(x)} \lt \epsilon$ for all $x \in S$. Taking the supremum we see that $\snorm{f_n - f}_u \leq \epsilon$. Therefore

Theorem 9.3.1. A sequence of bounded functions $f_n \colon S \to \R$ converges uniformly to $f \colon S \to \R$, if and only if $\lim_{n\to\infty} \norm{\,f_n - f}_u = 0 .$

Proof. $\suf$ Suppose $\lim_{n\to \infty} \norm{f_n - f}_u = 0$. Let $\epsilon > 0$ be given. Then there exists an $N$ such that for $n \geq N$, we have $\snorm{f_n - f}_u \lt \epsilon$. As $\norm{f_n-f}_u$ is the supremum of $\abs{f_n(x)-f(x)}$, then

Let $f_n \colon [0,1] \to \R$ be defined by \[ f_n(x) \coloneqq \frac{nx+ \sin(nx^2)}{n}. \] We claim $\{ f_n \}_{n=1}^\infty$ converges uniformly to $f(x) \coloneqq x$.

Let $f_n \colon [0,1] \to \R$ be defined by $ f_n(x) \coloneqq \frac{nx+ \sin(nx^2)}{n}. $ We claim $\{ f_n \}_{n=1}^\infty$ converges uniformly to $f(x) \coloneqq x$.

Let $f_n \colon [0,1] \to \R$ be defined by $ f_n(x) \coloneqq \frac{nx+ \sin(nx^2)}{n}. $ We claim $\{ f_n \}_{n=1}^\infty$ converges uniformly to $f(x) \coloneqq x$.

Proof. Let's use the uniform norm:

We cannot use Theorem 9.3.1 to the sequence of functions $$f_n(x):=\dfrac{x}{n}$$ defined on $\R,$ since they are not bounded.

We cannot use Theorem 9.3.1 to the sequence of functions $$f_n(x):=\dfrac{x}{n}$$ defined on $\R,$ since they are not bounded.

However, if we restrict the domain to $A:=[0,1],$ then $f_n(x):=x/n$ defined on $A$ converges uniformly. To see this note that

and therefore $\norm{f_n-f}_u\to 0.$

Definition 9.3.2. Let $f_n \colon S \to \R$ be bounded functions. The sequence is Cauchy in the uniform norm or uniformly Cauchy if for every $\epsilon > 0$, there exists an $N \in \N$ such that for all $m,k \geq N$, \begin{equation*} \norm{f_m-f_k}_u \lt \epsilon . \end{equation*}

Theorem 9.3.2. Let $f_n \colon S \to \R$ be bounded functions. Then $\{ f_n \}$ is Cauchy in the uniform norm if and only if there exists an $f \colon S \to \R$ and $\{ f_n \}$ converges uniformly to $f$.

Theorem 9.3.2. Let $f_n \colon S \to \R$ be bounded functions. Then $\{ f_n \}$ is Cauchy in the uniform norm if and only if there exists an $f \colon S \to \R$ and $\{ f_n \}$ converges uniformly to $f$.

Proof. $\nec$ First suppose $\{ f_n \}_{n=1}^\infty$ is Cauchy in the uniform norm. Fix $x$, then the sequence $\bigl\{ f_n(x) \bigr\}_{n=1}^\infty$ is Cauchy because

Thus $\bigl\{ f_n \bigr\}_{n=1}^\infty$ converges to some $f$. Define $f \colon S \to \R$ by

Theorem 9.3.2. Let $f_n \colon S \to \R$ be bounded functions. Then $\{ f_n \}$ is Cauchy in the uniform norm if and only if there exists an $f \colon S \to \R$ and $\{ f_n \}$ converges uniformly to $f$.

Proof. $\nec$ To prove that the convergence is uniform, let $\epsilon > 0$ be arbitrary. Find an $N$ such that for all $m, k \geq N$,

That is, for all $x$, we have $\ds \abs{f_m(x)-f_k(x)} \lt \dfrac{\epsilon}{2}. $

For any fixed $x$, take the limit as $k$ goes to infinity.

Theorem 9.3.2. Let $f_n \colon S \to \R$ be bounded functions. Then $\{ f_n \}$ is Cauchy in the uniform norm if and only if there exists an $f \colon S \to \R$ and $\{ f_n \}$ converges uniformly to $f$.

Proof. $\nec$ That is, or all $x$, we have $\ds \abs{f_m(x)-f_k(x)} \lt \dfrac{\epsilon}{2}. $

For any fixed $x$, take the limit as $k$ goes to infinity.

Then $\abs{f_m(x)-f_k(x)}\to \abs{f_m(x)-f(x)}.$

This implies that for all $x$, $\ds \abs{f_m(x)-f(x)}$ $\ds \leq \frac{\epsilon}{2}$ $\lt \epsilon .$

Therefore $\{ f_n \}_{n=1}^\infty$ converges uniformly. $\; \bs$

Theorem 9.3.2. Let $f_n \colon S \to \R$ be bounded functions. Then $\{ f_n \}$ is Cauchy in the uniform norm if and only if there exists an $f \colon S \to \R$ and $\{ f_n \}$ converges uniformly to $f$.

Proof. $\suf$ Now suppose $\{ f_n \}_{n=1}^\infty$ converges uniformly to $f$.

Given $\epsilon > 0$, find $N$ such that for all $n \geq N$, we have

Therefore, for all $m, k \geq N$,

Theorem 9.3.2. Let $f_n \colon S \to \R$ be bounded functions. Then $\{ f_n \}$ is Cauchy in the uniform norm if and only if there exists an $f \colon S \to \R$ and $\{ f_n \}$ converges uniformly to $f$.

Proof. $\suf$ Now suppose $\{ f_n \}_{n=1}^\infty$ converges uniformly to $f$.

Therefore, for all $m, k \geq N$,

Theorem 9.3.2. Let $f_n \colon S \to \R$ be bounded functions. Then $\{ f_n \}$ is Cauchy in the uniform norm if and only if there exists an $f \colon S \to \R$ and $\{ f_n \}$ converges uniformly to $f$.

Proof. $\suf$ Now suppose $\{ f_n \}_{n=1}^\infty$ converges uniformly to $f$.

Therefore, for all $m, k \geq N$,

Take supremum over all $x$ to obtain