(last updated: 2:23:03 PM, December 11, 2020)
Definition 8.1.1 A sequence of functions \(\{f_n\}\), where for each \(n \in\mathbb N\), \(f_n : D \rightarrow R\) with \(D\subset \mathbb R\), converges (pointwise) on \(D\) to a function \(f\) if and only if for each \(x_0\in D\) the sequence of real numbers \(\{ f_n (x_0)\}\) converges to the real number \(f (x_0)\).
Example: \(f_n(x)=x^n\) with domain \([0,1]\) converges pointwise to \[f(x)=\begin{cases} 0 & \text{ for } 0\leq x<1\\ 1 & \text{ for } x=1 \end{cases}\] Recall Theorem 2.1.13 that \(r^n\) goes to zero as \(n\rightarrow\infty\) when \(|r|<1\). Also, of course, \(1^n=1\) for all \(n\).
Example: \(f_n(x)=\frac xn\) with domain \([0,1]\) converges pointwise to \(f(x)=0\).
Definition 8.1.8 (Negation of Definition 8.1.1) Suppose that \(f_n : D \rightarrow R\) for all \(n \in \mathbb N\). Then the sequence \(\{f_n\}\) does not have a pointwise limit, that is, diverges on \(D\) if and only if there is \(x_0 \in D\) such that the sequence \(\{ f_n (x_0) \}\) diverges.
Example: \(f_n(x)=x^n\) with domain \([0,\infty)\) does not converges pointwise to a function since for all \(x_0>1\), \(x_0^n\rightarrow\infty\) as \(n\rightarrow\infty\).
Definition 8.1.11 A sequence \(\{f_n\}\) with \(f_n : [a,b] \rightarrow R\) converges in the mean to a function \(f\) if and only if there is \(\lim_{n\rightarrow\infty}\left(\int_a^b (f_n(x)-f(x))^2dx\right)^{1/2}=0\).
Definition 8.2.1 A sequence of functions \(\{f_n \}\), where for each \(n \in\mathbb N\) we have \(f_n : D \rightarrow R\) with \(D \subset\mathbb R\), converges uniformly to a function \(f\) if and only if for each \(\epsilon>0\) there exists \(n* \in\mathbb N\) such that \(|f_n(x) - f(x)| <\epsilon\) for all \(x \in D\) and \(n \geq n^*\). Such a function \(f\) is called the uniform limit of \(\{ f_n \}\) .
Example: \(f_n(x)=x^n\) converges pointwise but does not converge uniformly to \[f(x)=\begin{cases} 0 & \text{ for } 0\leq x<1\\ 1 & \text{ for } x=1 \end{cases}\]
Let \(\epsilon=\frac14\). Then suppose we have some cutoff \(n^*\). Now take \(x_0\) such that \(\sqrt[n^*]{\frac14}<x_0<1\) so that when \(n\geq n^*\) we have that \(\frac14<x_0^{n^*} \leq x_0^{n}=f_n(x_0)\), but \(f(x_0)=0\). Thus for all \(n\geq n^*\) we have that \(|f_n(x_0)-f(x_0)|=f_n(x_0)=x_0^n>\frac14=\epsilon\). This will be true no matter what we choose as our cutoff \(n^*\), there will always be such an \(x _0\) in the domain. Thus \(f_n\) cannot converge uniformly to \(f\).
Example: (see example 8.2.4 and 8.2.6 from textbook) \(f_n(x)=x^n\) on \([0,1)\)
\(f_n(x)=x^n\) on \([0,k]\) with \(k\in(0,1)\) a fixed constant.
Example: \(f_n(x)=\frac xn\) with domain \([0,1]\) converges uniformly to \(f(x)=0\).
Let \(\epsilon>0\) and pick \(n^*>\frac1\epsilon\). Then for all \(n\geq n^*\) and \(x\in[0,1]\) we have \(|f_n(x)-0|=\frac xn \leq \frac1n \leq \frac 1{n^*} <\epsilon\).
Theorem 8.3.1 If \(f_n\) is a sequence of continuous functions that converges uniformly to a function \(f\) on \(D\subset\mathbb R\), then \(f\) is continuous on \(D\).
Proof. Let \(\epsilon>0\) and \(x_0\in D\).
Since \(f_n\) converges uniformly to \(f\), we can choose and \(n^*\) such that \(n\geq n^*\) implies that \(|f_n(x)-f(x)|<\frac\epsilon3\) for all \(x\in D\).
Since each \(f_n\) is continuous, then at the \(n^*\) chosen above, we can choose a \(\delta>0\) so that \(|x-x_0|<\delta\) gives \(|f_{n^*}(x)-f_{n^*}(x_0)|<\frac\epsilon3\).
Now we can argue that for \(x_0\in D\) and \(x\in D\) with \(|x-x_0|<\delta\) we have
\[\begin{aligned} |f(x)-f(x_0)|&=|f(x)-f_{n^*}(x)+f_{n^*}(x)-f_{n^*}(x_0)+f_{n^*}(x_0)-f(x_0)|\\ &\leq|f(x)-f_{n^*}(x)|+|f_{n^*}(x)-f_{n^*}(x_0)|+|f_{n^*}(x_0)-f(x_0)|\\ &<\frac\epsilon3+\frac\epsilon3+\frac\epsilon3=\epsilon \end{aligned}\] Thus we have that \(f\) is continuous as \(x_0\). Since \(x_0\) was arbitrary, then it is continusous at all \(x\in D\). \(\blacksquare\)
Theorem 8.3.3 If \(f_n\) is a sequence of continuous functions that converges uniformly to a function \(f\) on \([a, b]\), then \[\lim_{n\rightarrow\infty}\int_a^b f_n(x) dx=\int_a^b \left[\lim_{n\rightarrow\infty} f(x)\right] dx\]
\[ \diamond \S \diamond \]