Exercise 4.3.10

Question

Observe that if $a$ and $b$ are real numbers, then
  $$
  \max \{a, b\}=\frac{1}{2}[(a+b)+|a-b|]
  $$
  \begin{enumerate}[label=(\alph*)] 
  \item Show that if $f_{1}, f_{2}, \ldots, f_{n}$ are continuous functions, then
    $$
    g(x)=\max \left\{f_{1}(x), f_{2}(x), \ldots, f_{n}(x)ight\}
    $$
    is a continuous function.
  \item 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
    $$
    f_{n}(x)= \begin{cases}1 & 	ext { if }|x| \geq 1 / n \ n|x| & 	ext { if }|x|<1 / n\end{cases}
    $$
    Now explicitly compute $h(x)=\sup \left\{f_{1}(x), f_{2}(x), f_{3}(x), \ldotsight\}$
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item We will prove this by induction. The base case is $$\max\{f_1(x), f_2(x)\} = \frac 12\left[(a+b) + |a-b| ight]$$ Which is obviously continuous. Now assume $\max\{f_1, \dots, f_{n-1}\}$ is continuous, letting $m(x) = \max\{f_1, \dots, f_{n-1}\}$ we can write $$ g(x) = \max\{f_1, \dots, f_{n-1}, f_n\} = \max\{f_n, m(x)\} $$ Now since $f_n(x)$ and $m(x)$ are continuous functions $\max\{f_n,m\}$ is continuous by the base case! \item We can reason by cases. if $x = 0$ then $f_n(0) = 0$ for all $n$ so $h(0) = 0$. If $x e 0$ then $|x| > 1/n$ for all $n > N$ meaning we have $h(x) = \max\{f_1(x), \dots, f_N(x), 1\}$. Since $n|x| < 1$ for all $|x| < 1/n$ we have $h(x) = 1$ and so $$ h(x) = \begin{cases} 0 & ext{ if } x = 0 \ 1 & ext{ if } x e 0 \end{cases} $$ Which is not continuous at $x = 0$, therefore (a) does not hold in the infinite case. \end{enumerate}

Exercise 4.3.10

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Exercise 4.3.10

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