I'm self-learning Real Analysis from Stephen Abbott's book, Understanding Analysis. Exercise 4.3.10 asks to prove certain facts about the supremum of a sequence of continuous functions. I'd like someone to verify my solution to part (a) of the problem, and supply any hints (spoilers), but not the complete proof to part (b) of the problem.
Exercise 4.3.10 Observe that if $a$ and $b$ are real numbers, then
$$\max \{a,b\}=\frac{1}{2}[(a+b)+\vert a – b \vert$$
(a) Show that if $f_1,f_2,\ldots,f_n$ are continuous functions, then
$$g(x)=\max \{f_1(x),f_2(x),\ldots,f_n(x)\}$$
is a continuous function.
(b) Let's explore whether the result in (a) extends to the infinite case. For each $n \in \mathbf{N}$, define $f_n$ on $\mathbf{R}$ by
$$f_n(x)=\begin{cases}1 & \text{ if } \vert x \vert \geq 1/n \\
n \vert x \vert & \text{ if } \vert x \vert < 1/n
\end{cases}$$Now explicitly compute $h(x)=\sup\{f_1(x),f_2(x),f_3(x),\ldots\}$
Proof.
(a) Assume that $\displaystyle g_{k}( x) =\max\{f_{1}( x) ,f_{2}( x) ,\dotsc ,f_{k}( x)\}$ is continuous. We are interested to prove that $\displaystyle g_{k+1}( x) =\max\{g_{k}( x) ,f_{k+1}( x)\}$ is also continuous.
We have:
\begin{equation*}
g_{k+1}( x) =\frac{1}{2}[( g_{k}( x) +f_{k+1}( x)) +| g_{k}( x) -f_{k+1}( x)|
\end{equation*}
We would like to prove that $\displaystyle g_{k+1}( x)$ is continuous.
By the Algebraic continuity theorem, since $\displaystyle g_{k}( x)$ and $\displaystyle f_{k+1}( x)$ are continuous functions, the sum and difference $\displaystyle g_{k}( x) +f_{k+1}( x)$ and $\displaystyle g_{k}( x) -f_{k+1}( x)$ are continuous functions.
Also, let $\displaystyle f( x)$ be an arbitrary continuous function. We are interested to prove that $\displaystyle | f( x)| $ is also continuous.
We are interested to make the distance $\displaystyle | | f( x)| -| f( c)| | $ as small as we please. Let $\displaystyle \epsilon >0$ be arbitrary. Let us explore the condition $\displaystyle | | f( x)| -| f( c)| | < \epsilon $. Since $\displaystyle | | f( x)| -| f( c)| | \leq | f( x) -f( c)| $, replacing $\displaystyle | | f( x)| -| f( c)| | $ by $\displaystyle | f( x) -f( c)| $ strengthens the condition we are interested to prove.
We want to show that $\displaystyle | f( x) -f( c)| < \epsilon $. But, by the definition of functional limits, for all $\displaystyle \epsilon >0$, there exists $\displaystyle \delta >0$, such that for all $\displaystyle | x-c| < \delta $, we have $\displaystyle | f( x) -f( c)| < \epsilon $.
Consequently, there exists $\displaystyle \delta >0$, such that for all $\displaystyle | x-c| < \delta $, the condition $\displaystyle | | f( x)| -| f( c)| | < \epsilon $ is satisfied. So, $\displaystyle \lim _{x\rightarrow c}| f( x)| =| f( c)| $. $\displaystyle | f( x)| $ is a continuous function.
We infer that, both $\displaystyle g_{k}( x) +f_{k+1}( x)$ and $\displaystyle | g_{k}( x) -f_{k+1}( x)| $ are continuous. Again by algebraic continuity theorem, $\displaystyle g_{k+1}( x) =\frac{1}{2}[( g_{k}( x) +f_{k+1}( x)) +| g_{k}( x) -f_{k+1}( x)| ]$ is continuous.
(b) Intuitively, the sequence of functions $\{f_1(x),f_(x),\ldots\}$ can be plotted as follows.
I am yet to study, what a sequence of functions mean (in chapter 6), but I think, if I take a point close enough to $0$, it's images under $f_1,f_2,f_3,\ldots$ form an increasing sequence. But, I am not exactly sure, what analysis-style arguments should I make here.
Best Answer
(a) What you did looks fine to me.
(b) For each $x\in\mathbf R$,$$\sup_{n\in\Bbb N}f_n(x)=\begin{cases}0&\text{ if }x=0\\1&\text{ otherwise.}\end{cases}$$Therefore, $\sup_{n\in\Bbb N}f_n$ is discontinuous.