Let $S\subseteq \R.$ 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 $f \colon D\subseteq\R \to \R$ if for every $x
\in D,$ then
\begin{equation*}
f(x) =
\lim_{n\to\infty} f_n(x) .
\end{equation*}
Limits of sequences of numbers are unique. That is, if a sequence
$\{ f_n \}_{n=1}^\infty$
converges pointwise, the limit function $f$ is unique.
We also say that $f_n \colon D \to \R$
converges pointwise to $f$ on $E \subset D$
for some $f \colon E \to \R.$ In this context we mean
that
$f(x) = \ds \lim_{n\to\infty} f_n(x)$ for every $x \in E.$
In other words, the restrictions of $f_n$ to $E$ converge pointwise to $f.$
The sequence of functions defined on $[-1,1]$ by $f_n(x) \coloneqq x^{2n}$
converges pointwise to $f \colon [-1,1] \to \R,$ where
Below, you can find a visual represenation of this sequence of functions.
Drag the slider below to explore the sequence for different values of $n.$
The sequence of functions defined on $[-1,1]$ by $f_n(x) \coloneqq x^{2n}.$
To prove that the sequence converges, consider $x \in (-1,1).$ Then
$0 \leq x^2 \lt 1.$ This implies that
Hence, $\ds \lim_{n\to\infty}f_n(x) = 0.$
Now for $x = 1$ or $x=-1,$ then we have that $x^{2n} = 1$ for all $n$ and therefore
$\ds \lim_{n\to\infty}f_n(x) = 1.$
Finally, for $x\in \R\setminus[-1,1],$ the sequence
$\bigl\{ f_n(x) \bigr\}_{n=1}^\infty$ does not converge.
Let $f_n:S\subseteq\R \to \R$ defined by $f_n(x) \coloneqq \sin(nx).$
Is this sequence pointwise convergent? If so, over which interval $I$ does it
converge pointwise?
In the applet below you can appreciate the behaviour of $f_n$ defined on $\R$ for different
value of $n.$
The sequence of functions defined on $\R$ by $f_n(x) \coloneqq \sin(nx).$
In this case $f_n$ does not converge pointwise
to any function on any interval. However, it
may converge at certain points, such
as when $x=0$ or $x=\pi.$
Try to prove as an exercise that in any interval
$[a,b],$ there exists an $x$ such that $\sin(xn)$ does not have a limit
as $n$ goes to infinity.
We can state pointwise
convergence in a different way. It is a simple application of the
definition of convergence of a sequence of real numbers.
Let $f_n \colon D \to \R$ and $f \colon D \to \R$ be functions.
Then $\{ f_n \}_{n=1}^\infty$ converges pointwise to $f$ if and only if
for every $x \in D,$ and every $\epsilon > 0,$ there exists
an $N \in \N$ such that for all
$n \geq N,$ we have
Let $f_n \colon D \to \R$
and $f \colon D \to \R$
be functions. The sequence $\{ f_n \}_{n=1}^\infty$
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 \quad \text{for all } \quad x \in D.
\end{equation*}
Uniform convergence.
Let $\{ f_n \}_{n=1}^\infty$ be a sequence of functions $f_n \colon D \to \R.$
If $\{ f_n \}_{n=1}^\infty$ converges
uniformly to $f \colon S \to \R,$ then
$\{ f_n \}_{n=1}^\infty$ converges pointwise to $f.$
The converse does not hold. For instance,
the functions $f_n(x) \coloneqq x^{2n}$ do not converge uniformly on $[-1,1],$
even though they converge pointwise. We can prove this by contradiction.
That is, assume that the convergence is uniform.
Let's choose $\epsilon \coloneqq \dfrac{1}{2}.$
Then there exists an $N\in\N$ such that
since $f_n(x)$ converges to 0 on $(-1,1).$
This means that for every sequence
$\{ x_k \}_{k=1}^\infty$ in $(-1,1)$
such that $\ds \lim_{k\to\infty} x_k=1,$
we have $x_k^{2N} \lt \dfrac{1}{2}$ for all $k.$
On the other hand, notice that
$x^{2N}$ is a continuous function of $x$ (it is a polynomial).
Therefore, we obtain
However, if we restrict our domain to $[-a,a]$ where $0 \lt a \lt 1,$ then
$\{ f_n \}_{n=1}^\infty$ converges uniformly to 0 on $[-a,a].$
Also note that
$a^{2n} \to 0$ as $n \to \infty.$ For every $\epsilon > 0,$
pick $N \in \N$ such that
$a^{2n} \lt \epsilon$ for all $n \geq N.$
When $x \in [-a,a],$ we have
$\abs{x} \leq a.$ Thus
for all $n \geq N$ and all $x \in [-a,a],$
\begin{equation*}
\abs{x^{2n}} = \abs{x}^{2n} \leq a^{2n} \lt \epsilon .
\end{equation*}
Let $f_n(x) := \dfrac{x}{n}$ defined on $\R$ for every $n\in \N.$
Observe that $\{f_n\}$ converges to $f=0,$ since
\begin{equation*}
\lim f_n(x) = \lim \frac{x}{n} = x\lim\frac{1}{n}=0 \;\text{ for all }\,x\in \R.
\end{equation*}
If we let $n_k:=k$ and $x_k := k .$ Then $f_{n_k}(x_k)=1$
so that
\begin{equation*}
\abs{\,f_{n_k}(x_k)-f(x_k)}= \abs{1-0}=1.
\end{equation*}
Therefore the sequence $\{f_n\}$ does not converges uniformly to $f.$
Let $f_n \colon S \to \R$
and $f \colon S \to \R$
be functions. The sequence $\{ f_n \}$ does not converge uniformly to $f$ if and only if
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
\begin{equation*}
\abs{\,f_{n_k}\left(x_k\right)-f(x_k)}\geq \epsilon_0\; \text{ for all } \;k\in \N.
\end{equation*}
