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%\header{MATH 11}{Exam 1}{Fall 2016}
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\begin{center}\huge MATH 4590 - EXAM 3 - SPRING 2018 \\
SOLUTION
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\begin{flushleft}\vspace{0.5in}
20 April 2018\\
Instructor: Tom Cuchta
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\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
\vspace*{10pt}
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\begin{questions}
\question[15] Write the definition.
\begin{parts}
\part[3] Define what it means for $P,T$ to be a partition pair of the interval $[a,b]$. \\
\textit{Solution}: $P=\{x_0,x_1,\ldots,x_n\} \subset [a,b]$ and $T=\{t_1,\ldots,t_n\} \subset [a,b]$ with $x_0 < t_1 < x_1 < t_2 < x_3 < \ldots < t_n < x_n$

\part[3] Define what it means for a function $f$ to be Riemann integrable on $[a,b]$ with integral $I$.  \\
\textit{Solution}: $\forall \epsilon > 0 \exists \delta > 0$ such that if $P,T$ is a partition pair of $[a,b]$ with $\mathrm{mesh}(P)<\delta$, then $|R(f,P,T) - I|<\epsilon$

\part[3] Let $f_n \colon [a,b] \rightarrow \mathbb{R}$ be a sequence of functions. Define what it means for $f_n \rightarrow f$ (i.e. pointwise convergence).\\
\textit{Solution}: $\forall x \in [a,b] \forall\epsilon >0 \exists N \in \mathbb{N}$ so that if $n \geq N$, then $|f_n(x)-f(x)|<\epsilon$ 

\part[3] Let $f_n \colon [a,b] \rightarrow \mathbb{R}$ be a sequence of functions. Define what it means for $f_n \rightrightarrows f$ (i.e. uniform convergence).\\
\textit{Solution}: $\forall \epsilon>0 \exists N \in \mathbb{N}$ so that if $n \geq N$ and $x \in [a,b]$, then $|f_n(x)-f(x)|<\epsilon$

\part[3] Define what a zero set is.\\
\textit{Solution}: $\forall \epsilon > 0$ there is a sequence of open sets $(a_n,b_n)$ so that $Z \subset \displaystyle\bigcup_{n=1}^{\infty} (a_n,b_n)$ and $\displaystyle\sum_{n=1}^{\infty} b_n-a_n < \epsilon$
\end{parts}

\question[15] Let $f_n \colon [a,b] \rightarrow \mathbb{R}$ be a sequence of functions defined for $n=1,2,3,\ldots$, suppose that $f_n \rightrightarrows f$, and let $\alpha \in \mathbb{R} \setminus \{0\}$. Prove that $\alpha f_n \rightrightarrows \alpha f$. \\
\textit{Solution}: Let $\epsilon>0$. Since $f_n \rightrightarrows f$, we know there is $N \in \mathbb{N}$ so that if $n \geq N$ and $x \in [a,b]$, $|f_n(x)-f(x)|<\dfrac{\epsilon}{\alpha}$. So, calculate
$$|\alpha f_n(x) - \alpha f(x)| = |\alpha| |f_n(x) - f(x)| < |\alpha| \dfrac{\epsilon}{|\alpha|} = \epsilon,$$
completing the proof. $\blacksquare$
\question[14] Let $f_n \colon [0,1] \rightarrow \mathbb{R}$ be defined for $n=1,2,3,\ldots$ by $f_n(x)=\dfrac{x^n}{n}$. 
\begin{parts}
\part[3] Find the pointwise limit of $f_n$.  \\
\textit{Solution}: $f_n(x) \rightarrow 0$ for all $x \in [0,1]$
\part[4] Is the convergence in (b) uniform? \\
\textit{Solution}: Yes. 
\part[3] Compute $f_n'$ and find its pointwise limit.  \\
\textit{Solution}: $f_n'(x) = x^{n-1}$ and its pointwise limit is 
$$x^{n-1} \rightarrow \left\{ \begin{array}{ll}
1, & \quad x=1 \\
0, & \quad x=0.
\end{array} \right.$$
\part[4] Is the convergence in (c) uniform? \\
\textit{Solution}: No -- the behavior at $x=1$ prevents this (draw an $\epsilon$-tube to observe).
\end{parts}

\newpage 

\question[15] Consider the sequence $f_n \colon [0,\infty) \rightarrow \mathbb{R}$ defined for $n=1,2,3,\ldots$ by $f_n(x)=e^{-nx}$. 
\begin{parts}
\part[6] What does $f_n$ converge pointwise to?\\
\textit{Solution}: $f_n(x) \rightarrow \left\{ \begin{array}{ll}
1, & \quad x=0 \\
0, & \quad x \in (0,\infty)
\end{array} \right.$
\part[6] Does $f_n$ converge uniformly?  \\
\textit{Solution}: No -- the behavior at $x=0$ prevents it.
\part[3] If the domain is modified to be instead $[1,\infty)$, does this change your conclusion in part (b)? \\
\textit{Solution}: Yes.
\end{parts}

\question[15] Consider the sequence $f_n \colon [0,\infty) \rightarrow \mathbb{R}$ defined for $n=1,2,3,\ldots$ by $f_n(x)=xe^{-nx}$. 
\begin{parts}
\part[6] What does $f_n$ converge pointwise to? \\
\textit{Solution}: $f_n(x) \rightarrow 0$ for all $x\ in [0,\infty)$
\part[6] Does $f_n$ converge uniformly? \\
\textit{Solution}: Yes
\part[3] If the domain is modified to be instead $[1,\infty)$, does this change your conclusion in part (b)? \\
\textit{Solution}: No -- it will still converge uniformly.
\end{parts}

\question[13] Suppose that $f_n \colon [a,b] \rightarrow \mathbb{R}$ is a sequence of Riemann integrable functions and let $\alpha_1, \alpha_2, \ldots$ be a sequence of real numbers. Prove using induction that
$$\displaystyle\int_a^b \displaystyle\sum_{k=1}^n f_k(x) \mathrm{d}x = \displaystyle\sum_{k=1}^n \displaystyle\int_a^b f_k(x) \mathrm{d}x.$$ 
\textit{Proof}: The case $n=1$ is a basic property of integrals (integral of the sum of two functions is the sum of the integrals). Assume it holds for $n=N$. Now calculate
$$\begin{array}{ll}
\displaystyle\int_a^b \displaystyle\sum_{k=1}^{N+1} f_k(x) \mathrm{d}x &= \displaystyle\int_a^b \left( \displaystyle\sum_{k=1}^N f_k(x)\right) + f_{N+1}(x) \mathrm{d}x \\
&=\underbrace{\displaystyle\int_a^b \displaystyle\sum_{k=1}^N f_k(x) \mathrm{d}x}_{\text{use induction hyp.}} + \displaystyle\int_a^b f_{N+1}(x) \mathrm{d}x \\
&=\displaystyle\sum_{k=1}^N \left( \displaystyle\int_a^b f_k(x) \mathrm{d}x \right) + \displaystyle\int_a^b f_{N+1}(x) \mathrm{d}x \\
&=\displaystyle\sum_{k=1}^{N+1} \displaystyle\int_a^b f_k(x) \mathrm{d}x,
\end{array}$$
completing the proof. $\blacksquare$
\newpage
\question[13] Suppose that $f \colon [0,1] \rightarrow \mathbb{R}$ is continuous and the following formula holds for all $x \in [0,1]$:
$$\displaystyle\int_0^x f(t) \mathrm{d}t = \displaystyle\int_x^1 f(t) \mathrm{d}t.$$
Explain why $f(x)=0$ for all $x \in [0,1]$. (\textit{hint}: using the properties of integrals is essential here!) \\
\textit{Solution}:  Add $\displaystyle\int_x^1 f(t) \mathrm{d}t$ to both sides to get
$$\underbrace{\displaystyle\int_0^x f(t) \mathrm{d}t + \displaystyle\int_x^1 f(t) \mathrm{d}t}_{=\displaystyle\int_0^1 f(t) \mathrm{d}t} =2\displaystyle\int_x^1 f(t) \mathrm{d}t.$$
Thus we have 
$$\underbrace{\displaystyle\int_0^1 f(t)\mathrm{d}t}_{\text{a constant}} = -2 \underbrace{\displaystyle\int_1^x f(t) \mathrm{d}t}_{\text{function of }x}$$
Now differentiate both sides with respect to $x$ to get
$$0 = -2f(x),$$
and hence $f(x) = 0$, as was to be shown.
\question[3] (\textbf{Bonus}) What is the ``refinement principle" in Darboux integrability? Draw a picture to illustrate it. \\
\textit{Solution}: The lower sum gets bigger and the upper sum gets smaller.
\question[3] (\textbf{Bonus}) Draw a picture of a sequence of functions that is uniformly convergent on $[\epsilon,1-\epsilon]$ for any $0<\epsilon<\dfrac{1}{2}$, but not uniformly convergent on $[0,1]$. \\
\textit{Solution}: 
\question[3] (\textbf{Bonus}) Prove that if $Z_1, Z_2, Z_3, \ldots$ is a sequence of zero sets, then $Z=Z_1 \cup Z_2 \cup Z_3 \cup \ldots$ is also a zero set.  \\
\textit{Solution}: 
\end{questions}
\iffalse
\newpage
\noindent A sequence $f_n \colon [a,b] \rightarrow \mathbb{R}$ of functions is called pointwise convergent to the function $f \colon [a,b] \rightarrow \mathbb{R}$ provided that $\forall x \in [a,b] \forall \epsilon > 0 \exists N \in \mathbb{N}$ so that if $n \geq N$, then $|f_n(x)-f(x)| < \epsilon$. \\

\noindent A sequence $f_n \colon [a,b] \rightarrow \mathbb{R}$ of functions is called uniformly convergent to the function $f \colon [a,b] \rightarrow \mathbb{R}$ provided that $\forall \epsilon > 0 \exists N \in \mathbb{N}$ so that if $n \geq N$ and $x \in [a,b]$, then $|f_n(x)-f(x)| < \epsilon$. \\

\noindent Recall basic properties of integrals: if $f$ and $g$ are integrable on $[a,b]$ and $\alpha, \beta \in \mathbb{R}$ are constants, then \\
$$\displaystyle\int_a^b \alpha f(x) + \beta g(x) \mathrm{d}x = \alpha \displaystyle\int_a^b f(x) \mathrm{d}x + \beta \displaystyle\int_a^b g(x) \mathrm{d}x$$
$$\displaystyle\int_a^b f(x) \mathrm{d}x = \displaystyle\int_a^c f(x) \mathrm{d}x + \displaystyle\int_c^b f(x) \mathrm{d}x$$
$$\dfrac{\mathrm{d}}{\mathrm{d}x} \displaystyle\int_a^x f(t) \mathrm{d}t = f(x)$$
\fi
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