Chapter 15: Sequences of Functions
- Page ID
- 211985
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)
Suppose that \(f_n: \mathbb{T} \to\mathbb{R} \), \(n\in \mathbb{N} \), \(S\subset \mathbb{T} \).
We say that the sequence \(\{f_n\}_{n\in \mathbb{N}} \) converges pointwise, i.e., at each point, to a function \(f \) defined on \(S \) if
\[\lim_{n \to\infty}f_n(t)=f(t) \quad\mbox{for all}\quad t\in S. \notag\]
We often write \(\lim_{n \to\infty}f_n=f \) pointwise on \(S \) or \(f_n \to f \) pointwise on \(S \).
Let \( \mathbb{T}=\left\{\frac{1}{n}:n\in \mathbb{N}\right\}\cup\{0\} \). Consider
\[f_n(t)=t^2,\quad t\in[0,1]. \notag\]
We have
\[\lim_{n \to\infty}f_n(t)=\lim_{n \to\infty}t^n
=\begin{cases}
0 & \quad\mbox{for}\quad t\in[0,1) \\ \\
1 & \quad\mbox{for}\quad t=1.
\end{cases} \notag\]
If
\[f(t)=\begin{cases}
0 & \quad\mbox{for}\quad t\in[0,1) \\ \\
1 & \quad\mbox{for}\quad t=1,
\end{cases} \notag\]
then \(f_n \to f \) pointwise on \([0,1] \).
Let \( \mathbb{T}=2^{ \mathbb{N}_0} \). If
\[f_n(t)=t\left(1+e^{-nt}\right), \notag\]
then
\[\lim_{n \to\infty}f_n(t)=\lim_{n \to\infty}t\left(1+e^{-nt}\right)=t. \notag\]
If we set \(f(t)=t \), then \(f_n \to f \) pointwise on \( \mathbb{T} \).
If \(f_n(t)=\frac{n+1}{n+t^2} \), then
\[\lim_{n \to\infty}f_n(t)=\lim_{n \to\infty}\frac{n+1}{n+t^2}=1. \notag\]
If \(f(t)=1 \), \(t\in \mathbb{T} \), then \(f_n \to f \) pointwise on \( \mathbb{T} \).
Let
\[f_n(t)=\frac{nt^2}{n+t},\quad f(t)=t^2. \notag\]
Prove that \(f_n \to f \) pointwise on \( \mathbb{T} \).
We say that the sequence \(\{f_n\}_{n\in \mathbb{N}} \) converges uniformly on \(S \) to a function \(f \) defined on \(S \) if for every \(\varepsilon>0 \), there exists \(N\in \mathbb{N} \) such that
\[|f_n(t)-f(t)|<\varepsilon \quad\mbox{for all}\quad n>N \quad\mbox{and all}\quad t\in S. \notag\]
Let \( \mathbb{T}=\left\{\frac{1}{n}:n\in \mathbb{N}\right\}\cup\{0\} \). Consider
\[f_n(t)=t^n,\quad 0\leq t\leq a,\quad 0<a<1. \notag\]
Note that
\[\lim_{n \to\infty}f_n(t)=\lim_{n \to\infty}t^n=0 \quad\mbox{for all}\quad t\in \mathbb{T}. \notag\]
Let \(\varepsilon>0 \) be arbitrarily chosen. We choose \(N\in \mathbb{N} \) such that
\[N>\frac{\log\varepsilon}{\log a}. \notag\]
Hence, \(a^N<\varepsilon \).
Then, for every \(n>N \), we have
\[a^n\leq a^N<\varepsilon \notag\]
and
\[|t^n-0|=t^n\leq a^n<\varepsilon. \notag\]
Therefore, \(f_n \to 0 \) uniformly on \([0,a] \).
Let \( \mathbb{T}=2^{ \mathbb{N}_0} \). Consider
\[f_n(t)=\sqrt{t^2+\frac{1}{n^2}}. \notag\]
If we take \(f(t)=t \) and let \(\varepsilon>0 \) be arbitrarily chosen, then
\[ \begin{aligned}
|f_n(t)-f(t)|
&=& \left|\sqrt{t^2+\frac{1}{n^2}}-t\right| \\ \\
&=& \frac{\left|\left(\sqrt{t^2+\frac{1}{n^2}}-t\right)
\left(\sqrt{t^2+\frac{1}{n^2}}+t\right)\right|}
{\sqrt{t^2+\frac{1}{n^2}}+t} \\ \\
&=& \frac{t^2+\frac{1}{n^2}-t^2}{\sqrt{t^2+\frac{1}{n^2}}+t} \\ \\
&=& \frac{1}{n^2\left(\sqrt{t^2+\frac{1}{n^2}}+t\right)} \\ \\
&\leq& \frac{1}{n^2\frac{1}{n}} \\ \\
&=& \frac{1}{n}.
\end{aligned} \notag\]
If we take \(N=\frac{1}{\varepsilon} \), then, for every \(n>N \), we have
\[frac{1}{n}<\frac{1}{N}=\varepsilon \quad\mbox{and}\quad
|f_n(t)-f(t)|<\varepsilon \quad\mbox{for any}\quad t\in \mathbb{T}. \notag \]
Hence, \(\{f_n\}_{n\in \mathbb{N}} \) is uniformly convergent to \(f \) on \( \mathbb{T} \).
Let \( \mathbb{T}=\left\{\frac{1}{n}:n\in \mathbb{N}\right\}\cup\{0\} \). Consider
\[f_n(t)=\frac{nt}{2+n^3t^3},\quad t\in \mathbb{T}. \notag\]
If \(f(t)=0 \), then \(f_n \to f \) pointwise on \( \mathbb{T} \). Assume that \(\{f_n\}_{n\in \mathbb{N}} \) is uniformly convergent to \(f \) on \( \mathbb{T} \). We take \(0<\varepsilon<\frac{1}{3} \). Then there exists \(N=N(\varepsilon) \) such that for every \(n>N \), we have
\[|f_n(t)|<\varepsilon \quad \mbox{for any}\quad t\in \mathbb{T}. \notag\]
In particular, when \(n>N \) and \(t=\frac{1}{n} \), we get
\[\left|f_n\left(\frac{1}{n}\right)\right|<\varepsilon, \notag\]
which is a contradiction because \(f_n\left(\frac{1}{n}\right)=\frac{1}{3} \). Therefore, \(\left\{f_n\right\}_{n\in \mathbb{N}} \) is not uniformly convergent to \(f \) on \( \mathbb{T} \).
Let \( \mathbb{T}=\left\{\frac{1}{n}: n\in \mathbb{N}\right\}\cup\{0\} \). Check if the following sequences are uniformly convergent on \( \mathbb{T} \).
1. \(f_n(t)=e^{-(t-3n)^2} \),
2. \(f_n(t)=\frac{t}{4+2n^2t^2} \),
3. \(f_n(t)=\frac{1}{2+3nt} \).
Answer
3. uniformly convergent to \(0 \) on \( \mathbb{T} \),
2. uniformly convergent to \(0 \) on \( \mathbb{T} \),
3. not uniformly convergent to \(0 \) on \( \mathbb{T} \).
If \(\left\{f_n\right\}_{n\in \mathbb{N}} \) converges pointwise to \(f \) on \(D\subset \mathbb{T} \), then \(\left\{f_n\right\}_{n\in \mathbb{N}} \) converges uniformly to \(f \) on \(D \) if and only if
\begin{equation}\label{191}
\lim_{n \to\infty}\sup_{t\in D}|f_n(t)-f(t)|=0.
\end{equation}
Proof
1. Supose that \(\left\{f_n\right\}_{n\in \mathbb{N}} \) converges uniformly to \(f \) on \(D \). Then, for every \(\varepsilon>0 \), there exists \(N=N(\varepsilon) \) so that \(n>N \) implies
\[|f_n(t)-f(t)|<\varepsilon \quad \mbox{for any}\quad t\in D. \notag\]
Hence,
\[\sup_{t\in D}|f_n(t)-f(t)|<\varepsilon \quad\mbox{for any}\quad n>N. \notag\]
2. Suppose that \(\left\{f_n\right\}_{n\in \mathbb{N}} \) converges pointwise to \(f \) on \(D \) and \eqref{191} holds. Then, for every \(\varepsilon>0 \), there exists \(N=N(\varepsilon) \) so that \(n>N \) implies
\[\sup_{t\in D}|f_n(t)-f(t)|<\varepsilon. \notag\]
Hence, for any \(n>N \), we have
\[|f_n(t)-f(t)|<\varepsilon \quad\mbox{for any}\quad t\in D. \notag\]
The proof is complete.
Let \( \mathbb{T}=2^{ \mathbb{N}_0} \), \(f_n(t)=\frac{1}{n+t^2} \) and \(f(t)=0 \), \(t\in \mathbb{T} \). We have that \(f_n \to f \) pointwise on \( \mathbb{T} \). Also,
\[\sup_{t\in \mathbb{T}}|f_n(t)-f(t)|
=\sup_{t\in \mathbb{T}}\frac{1}{n+t^2}=\frac{1}{n+1} \to 0 \quad\mbox{as}\quad n \to\infty. \notag\]
Hence, it follows that \(\left\{f_n\right\}_{n\in \mathbb{N}} \) converges uniformly to \(f \) on \( \mathbb{T} \).
Let \( \mathbb{T}= \mathbb{N} \), \(f_n(t)=\frac{nt}{nt+1} \), \(f(t)=1 \), \(t\in \mathbb{T} \). We have that \(f_n \to 1 \) pointwise on \( \mathbb{T} \). Also,
\[ \begin{aligned}
|f_n(t)-f(t)|
&=& \left|\frac{nt}{nt+1}-1\right| \\ \\
&=& \left|-\frac{1}{nt+1}\right| \\ \\
&=& \frac{1}{nt+1}, \\ \\
\sup_{t\in \mathbb{T}}|f_n(t)-f(t)|
&=& \sup_{t\in \mathbb{T}}\frac{1}{nt+1} \\ \\
&=& \frac{1}{n+1} \to 0 \quad \mbox{as}\quad n \to\infty.
\end{aligned} \notag\]
Hence, it follows that \(\{f_n\}_{n\in \mathbb{N}} \) converges uniformly to \(f \) on \( \mathbb{T} \).
Let \( \mathbb{T}= \mathbb{Z} \). We will investigate for uniform convergence of the sequence \(\left\{f_n(t)=\frac{n+1}{n+t^2}\right\}_{n\in \mathbb{N}} \) on \(D_2=[-1,1] \) and \(D_2=[1,\infty) \). Let \(f(t)=1 \). Note that \(f_n \to f \) pointwise on \( \mathbb{Z} \). Moreover,
\[|f_n(t)-f(t)|
=\left|\frac{n+1}{n+t^2}-1\right|=\left|\frac{1-t^2}{n+t^2}\right|. \notag\]
1. If \(t\in D_1 \), then
\[ \begin{aligned}
|f_n(t)-f(t)|
&=& \frac{1-t^2}{n+t^2}, \\ \\
\left(\frac{1-t^2}{n+t^2}\right) ^\Delta
&=& \frac{(1-t^2) ^\Delta(n+t^2)-(1-t^2)(n+t^2) ^\Delta}
{(n+(\sigma(t))^2)(n+t^2)} \\ \\
&=& \frac{-(\sigma(t)+t)(n+t^2)-(1-t^2)(\sigma(t)+t)}
{\left(n+(t+1)^2\right)\left(n+t^2\right)} \\ \\
&=& -\frac{(n+1)(2t+1)}{\left(n+(t+1)^2\right)(n+t^2)} \\ \\
&\leq& 0 \quad\mbox{if}\quad t\geq-\frac{1}{2}, \\ \\
\left(\frac{1-t^2}{n+t^2}\right) ^\nabla
&=& \frac{(1-t^2) ^\nabla(n+t^2)-(1-t^2)(n+t^2) ^\nabla}{(n+(\mathbb{R}ho(t))^2)(n+t^2)} \\ \\
&=& \frac{-(\mathbb{R}ho(t)+t)(n+t^2)-(1-t^2)(\mathbb{R}ho(t)+t)}
{\left(n+(t+1)^2\right)\left(n+t^2\right)} \\ \\
&=& -\frac{(n+1)(2t-1)}{\left(n+(t-1)^2\right)(n+t^2)} \\ \\
&\geq& 0 \quad \mbox{if}\quad t\geq\frac{1}{2}.
\end{aligned} \notag\]
Therefore, the function \(\frac{1-t^2}{n+t^2} \) has a maximum at \(t=0 \). Note that
\[\frac{1-t^2}{n+t^2}\Bigl|_{t=0}=\frac{1}{n} \to 0 \quad\mbox{as}\quad n \to \infty. \notag\]
Hence, it follows that \(\{f_n\}_{n\in \mathbb{N}} \) converges uniformly to \(f \) on \(D_1 \).
2. If \(t\in D_2 \), then
\[|f_n(t)-f(t)|=\frac{t^2-1}{n+t^2}. \notag\]
Assume that \(\{f_n\}_{n\in \mathbb{N}} \) converges uniformly to \(f \) on \(D_2 \). Then, for \(0<\varepsilon<\frac{1}{2} \), there existss \(N=N(\varepsilon) \) so that \(n>N \) implies
\[\frac{t^2-1}{n+t^2}<\varepsilon \quad\mbox{for any}\quad t\in D_2. \notag\]
We take \(t=n+2\in D_2 \). Then
\[\frac{(n+2)^2-1}{n+(n+2)^2}<\frac{1}{2}, \notag\]
so
\[\frac{n^2+4n+3}{2n^2+4n+4}<\frac{1}{2}, \notag\]
so
\[n^2+4n+3<n^2+2n+2, \notag\]
so
\[2n+1<0, \notag\]
which is a contradiction. Therefore, \(\{f_n\}_{n\in \mathbb{N}} \) does not converge uniformly to \(f \) on \(D_2 \).
Let
\[ \mathbb{T}=\{0\}\cup\left\{\frac{1}{n}:\; n\in \mathbb{N}\right\}\cup 2^{ \mathbb{N}_0}. \notag\]
Investigate the sequence
\[\left\{f_n(t)=\frac{n+t}{nt+1}\right\}_{n\in \mathbb{N}} \notag\]
for uniform convergence on \(D_1=[0,1] \) and \(D_2=2^{ \mathbb{N}_0} \).
Answer
The sequence is not uniformly convergent on \(D_1 \) and it is uniformly convergent on \(D_2 \).
If the function sequence \(\{f_n\}_{n\in \mathbb{N}} \) is pointwise convergent to \(f \) on \(D\subset \mathbb{T} \), then \(\{f_n\}_{n\in \mathbb{N}} \) is uniformly convergent to \(f \) on \(D \) if and only if for an arbitrary sequence \(\{t_n\}_{n\in \mathbb{N}} \), \(t_n\in D \), we have
\begin{equation}\label{192}
\lim_{n \to\infty}\left(f_n(t_n)-f(t_n)\right)=0.
\end{equation}
Proof
1. Suppose that \(\{f_n\}_{n\in \mathbb{N}} \) is uniformly convergent on \(D \). Then, we have
\[\sup_{t\in D}|f_n(t)-f(t)| \to 0 \quad\mbox{as}\quad n \to\infty. \notag\]
Hence, for any sequence \(\{t_n\}_{n\in \mathbb{N}} \), \(t_n\in D \), we have
\[|f_n(t_n)-f(t_n)|\leq\sup_{t\in D}|f_n(t)-f(t)| \to 0 \quad\mbox{as}\quad n \to\infty, \notag\]
i.e., \eqref{192} holds.
2. Assume that for any sequence \(\{t_n\}_{n\in \mathbb{N}} \), \(t_n\in D \), \eqref{192} holds and the sequence \(\{f_n\}_{n\in \mathbb{N}} \) does not converge uniformly to \(f \) on \(D \). Hence, there exists \(\varepsilon_0>0 \) such that for any \(N>0 \), there exist \(n>N \) and \(t\in D \) so that
\[|f_n(t)-f(t)|\geq\varepsilon_0. \notag\]
For \(N_1=1 \), there exist \(n_1>1 \) and \(t_{n_1}\in D \) such that
\[|f_{n_1}(t_{n_1})-f(t_{n_1})|\geq\varepsilon_0. \notag\]
For \(N_2=n_1 \), there exist \(n_2>n_1 \) and \(t_{n_2}\in D \) such that
\[|f_{n_2}(t_{n_2})-f(t_{n_2})|\geq\varepsilon_0, \notag\]
and so on. Thus, we get a sequence \(\{t_{n_k}\}_{k\in \mathbb{N}} \), \(t_{n_k}\in D \), such that
\[|f_{n_k}(t_{n_k})-f(t_{n_k})|\geq\varepsilon_0, \notag\]
which leads to a contradiction due to \eqref{192}.
The proof is complete.
Consider \(f_n(t)=nt(1-t)^n \) on \(D=\left\{\frac{1}{n}:n\in \mathbb{N}\right\}\cup\{0\} \). We have that \(\{f_n\}_{n\in \mathbb{N}} \) is pointwise convergent to \(0 \) on \(D \). Suppose that \(\{f_n\}_{n\in \mathbb{N}} \) is uniformly convergent to \(0 \) on \(D \). Then, applying Theorem \mathbb{R}ef{theorem190} for \(t_n=\frac{1}{n} \), we have
\[f_n\left(\frac{1}{n}\right)=\left(1-\frac{1}{n}\right)^n \mathbb{N}ot \to 0, \notag\]
which is a contradiction. Therefore, \(\{f_n\}_{n\in \mathbb{N}} \) is not uniformly convergent to \(0 \) on \(D \).
Consider \(f_n(t)=\frac{1}{1+nt} \) on \(D=\left\{\frac{1}{n}:n\in \mathbb{N}\right\}\cup\{0\} \). We have that \(\{f_n\}_{n\in \mathbb{N}} \) is pointwise convergent to \(0 \) on \(D \). Assume that \(\{f_n\}_{n\in \mathbb{N}} \) is uniformly convergent to \(0 \) on \(D \). Then, using Theorem \mathbb{R}ef{theorem190} for \(t_n=\frac{1}{n} \), we have
\[f_n\left(\frac{1}{n}\right)=\frac{1}{2} \mathbb{N}ot \to 0, \notag\]
which is a contradiction. Therefore, \(\{f_n\}_{n\in \mathbb{N}} \) is not uniformly convergent to \(0 \) on \(D \).
Consider \(f_n(t)=1-(1-t^2)^n \) on \(D=\left\{\frac{1}{n}:n\in \mathbb{N}\right\}\cup\{0\} \). We have that \(\{f_n\}_{n\in \mathbb{N}} \) is pointwise convergent to \(1 \) on \(D \). Assume that \(\{f_n\}_{n\in \mathbb{N}} \) is uniformly convergent to \(1 \) on \(D \). Then, for \(t_n=\frac{1}{n} \), we have
\[f_n\left(\frac{1}{n}\right)-1=-\left(1-\frac{1}{n^2}\right)^n \mathbb{N}ot \to 0, \notag\]
which is a contradiction. Therefore, \(\{f_n\}_{n\in \mathbb{N}} \) is not uniformly convergent to \(1 \) on \(D \).
Consider \(f_n(t)=\frac{nt+2}{3+4n^2t^2} \) on \(D=\left\{\frac{1}{n}:n\in \mathbb{N}\right\}\cup\{0\} \). Prove that \(\{f_n\}_{n\in \mathbb{N}} \) is not uniformly convergent to \(0 \) on \(D \).
Let \(D\subset \mathbb{T} \). If \(f_n:D \to\mathbb{R} \), \(n\in \mathbb{N} \), are rd-continuous and \(\{f_n\}_{n\in \mathbb{N}} \) is uniformly convergent to \(f:D \to\mathbb{R} \) on \(D \), then \(f \) is rd-continuous on \(D \) and
\[\int_a^bf(t) ^ \Delta t=\lim_{n \to\infty}\int_a^bf_n(t) ^ \Delta t \notag\]
for every \([a,b]\subset D \).
Proof
Let \(t_0\in D \) be left-dense and right-scattered. Since \(\{f_n(t)\}_{n\in \mathbb{N}} \) is uniformly convergent to \(f \) on \(D \), for any given \(\varepsilon>0 \), there exists \(M\in \mathbb{N} \) such that
\[|f_M(t)-f(t)|<\frac{\varepsilon}{3} \quad\mbox{for all}\quad t\in D. \notag\]
Because \(f_M \) is rd-continuous on \(D \), there exists \( ^ \Delta>0 \) such that
\[|f_M(t')-f_M(t'')|<\frac{\varepsilon}{3} \quad\mbox{for any}\quad t',t''\in(t_0- ^ \Delta,t_0). \notag\]
If \(t_n \to t_0^- \), \(n \to\infty \), \(n\in \mathbb{N} \), then there exists \(N\in \mathbb{N} \) such that \(m,n>N \) imply \(t_m,t_n\in(t_0- ^ \Delta,t_0) \) and
\[|f_M(t_n)-f_M(t_m)|<\frac{\varepsilon}{3}. \notag\]
Hence, for \(m,n>N \), we have
\[ \begin{aligned}
|f(t_n)-f(t_m)|
&=& |f_M(t_n)-f(t_n)-f_M(t_n)-f_M(t_m)+f_M(t_m)-f(t_m)| \\ \\
&\leq& |f_M(t_n)-f(t_n)|+|f_M(t_n)-f_M(t_m)| \\ \\
&& +|f_M(t_m)-f(t_m)| \\ \\
&<& \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3} \\ \\
&=& \varepsilon.
\end{aligned} \notag\]
Therefore, \(f \) is rd-continuous on \(D \). Hence, \(f \) is integrable on every \([a,b]\subset D \). For every \(n>M \), we have
\[ \begin{aligned}
\left|\int_a^bf_n(t) ^ \Delta t-\int_a^bf(t) ^ \Delta t\right|
&=& \left|\int_a^b(f_n(t)-f(t)) ^ \Delta t\right| \\ \\
&\leq& \int_a^b|f_n(t)-f(t)| ^ \Delta t \\ \\
&<& \frac{\varepsilon}{3}(b-a),
\end{aligned} \notag\]
which completes the proof.
Suppose that the function sequence
\[\{f_n(t)\}_{n\in \mathbb{N}},\quad f_n(t):[a,b] \to\mathbb{R},\quad n\in \mathbb{N}, \notag\]
satisfies the following conditions.
1. \(f_n \), \(n\in \mathbb{N} \), is differentiable on \([a,b] \), and its derivative \(f_n ^\Delta \) is rd-continuous on \([a,b] \),
2. \(f_n \) converges pointwise to \(f \) on \([a,b] \),
3. \(\left\{f_n ^\Delta\right\}_{n\in \mathbb{N}} \) is uniformly convergent to \(g \) on \([a,b] \).
Then \(f \) is differentiable on \([a,b] \) and
\[f ^\Delta(t)=g(t) \quad\mbox{for any}\quad t\in[a,b]. \notag\]
Proof
By Theorem \mathbb{R}ef{theorem196}, we have that \(g \) is rd-continuous on \([a,b] \). Therefore, \(g \) is integrable on \([a,b] \). Hence, we get
\[ \begin{aligned}
\int_a^tg(s) ^ \Delta s
&=& \lim_{n \to\infty}\int_a^tf_n ^\Delta(s) ^ \Delta s \\ \\
&=& \lim_{n \to\infty}\left(f_n(t)-f_n(a)\right) \\ \\
&=& f(t)-f(a) \quad\mbox{for any}\quad t\in[a,b].
\end{aligned} \notag\]
The left-hand side of the above formula is differentiable, so the right-hand side is also differentiable, and this leads to
\[f ^\Delta(t)=g(t) \quad\mbox{for all}\quad t\in[a,b], \notag\]
completing the proof.
Assume that the function sequence \(\{f_n\}_{n\in \mathbb{N}} \), \(f_n:[a,b] \to\mathbb{R} \), converges pointwise to the function \(f \) on \([a,b] \). If the conditions
1. \(f_n \), \(n\in \mathbb{N} \), are rd-continuous on \([a,b] \),
2. \(f \) is rd-continuous on \([a,b] \),
3. for any given \(t\in[a,b] \), \(\{f_n(t)\}_{n\in \mathbb{N}} \) is monotone with respect to \(n\in \mathbb{N} \)
hold, then \(\{f_n\}_{n\in \mathbb{N}} \) is uniformly convergent to \(f \) on \([a,b] \).
Proof
Suppose that the sequence \(\{f_n\}_{n\in \mathbb{N}} \) is not uniformly convergent to \(f \) on \([a,b] \). Then there exists \(\varepsilon_0>0 \) such that for any given \(N\in \mathbb{N} \), there exist \(n>N \), \(t\in[a,b] \), implying
\begin{equation}\label{196}
|f_n(t)-f(t)|\geq\varepsilon_0.
\end{equation}
For \(N=1 \), there exist \(n_1>1 \), \(t_1\in[a,b] \), such that
\[|f_{n_1}(t_1)-f(t_1)|\geq\varepsilon_0. \notag\]
For \(N=n_1 \), there exist \(n_2>n_1 \), \(t_2\in[a,b] \), such that
\[|f_{n_2}(t_2)-f(t_2)|\geq\varepsilon_0, \notag\]
and so on. For \(N=n_k \) there exist \(n_{k+1}>n_k \), \(t_{k+1}\in[a,b] \), such that
\[|f_{n_{k+1}}(t_{k+1})-f(t_{k+1})|\geq\varepsilon_0. \notag\]
Hence, we obtain a point sequence \(\{t_k\}_{k\in \mathbb{N}} \), \(t_k\in[a,b] \). This sequence has a convergent subsequence. Let \(\{t_{k_l}\}_{l\in \mathbb{N}} \) be a convergent subsequence of the sequence \(\{t_k\}_{k\in \mathbb{N}} \) and \(t_{k_l} \to\xi \) as \(l \to\infty \). We have that \(\xi\in[a,b] \). Because \(f_n(\xi) \to f(\xi) \) as \(n \to\infty \), for the above \(\varepsilon_0 \), there exists \(N\in \mathbb{N} \) such that
\[|f_N(\xi)-f(\xi)|<\frac{\varepsilon_0}{2}. \notag\]
Suppose that \(\xi \) is left-dense and right-scattered. Then, for the above sequence \(\{t_{k_l}\}_{l\in \mathbb{N}} \), \(t_{k_l}\leq\xi \) and \(t_{k_l} \to\xi \) as \(l \to\infty \). Thus, for \(\varepsilon_0>0 \), as above, there exists \(L\in \mathbb{N} \) such that \(l>L \) implies
\[|\xi-t_{k_l}|<\varepsilon_0. \notag\]
Since \(f_N \) and \(f \) are rd-continuous on \([a,b] \), we have that
\[\left|\left(f_N(t_{k_l})-f(t_{k_l})\right)
-\left(f_N(\xi)-f(\xi)\right)\right|
<\frac{\varepsilon_0}{2}. \notag\]
Hence,
\[ \begin{aligned}
\left|f_N(t_{k_l})-f(t_{k_l})\right|
&<& \frac{\varepsilon_0}{2}+|f_N(\xi)-f(\xi)| \\ \\
&<& \frac{\varepsilon_0}{2}+\frac{\varepsilon_0}{2} \\ \\
&=& \varepsilon_0.
\end{aligned} \notag\]
Suppose that \(\xi \) is not left-dense and right-scattered. Then \(\xi \) is a point of continuity of \(f_N \) and \(f \). Because \(t_{k_l} \to\xi \), \(l \to\infty \), there exists \(L_1\in \mathbb{N} \) such that \(l>L_1 \) implies
\[|f_N(t_{k_l})-f(t_{k_l})|<\varepsilon_0. \notag\]
By using the monotonicity condition, we get
\[|f_n(t_{k_l})-f(t_{k_l})|\leq|f_N(t_{k_l})-f(t_{k_l})|<\varepsilon_0 \notag\]
with \(n>N \), \(l>\max\{L,L_1\} \). So when \(n \) is sufficiently large, \(n_l>N \) and \(l>\max\{L,L_1\} \) are satisfied. Thus,
\[|f_{n_{k_l}(t_{k_l})}-f(t_{k_l})|<\varepsilon_0, \notag\]
which contradicts to \eqref{196}. This completes the proof.