Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 1 / 3E (Analysis I)

A continuous function at $x_0$ satisfies the following quantified condition:The choice of $\delta$ may depend on $x_0$ and $\epsilon$.

For the bounded example, takeThis is a continuous function on its domain and $|f(x)|\le 1-x<1$. Along $x_n=(\pi /2+2\pi n{)}^{-1}$ its values tend to $1$, while along $y_n=(3\pi /2+2\pi n{)}^{-1}$ its values tend to $-1$. Hence the supremum is $1$ and the infimum is $-1$, and neither is attained. “Upper and lower bound” here means the least upper and greatest lower bounds.

For the nonnegative function on the whole real line, if $f\equiv 0$ the conclusion is immediate. Otherwise choose $x_{\ast }$ with $m=f(x_{\ast })>0$. The limits at both infinities give $R>|x_{\ast }|$ such that $f(x)<m/2$ whenever $|x|>R$. By the extreme value theorem, a continuous function on the compact interval $[-R,R]$ has a maximum $M=f(\alpha )$ there. Since $M\ge m$, it also exceeds every value outside this interval. Thus $f$ is bounded above and attains its global maximum. Nonnegativity ensures that either the function is zero or such a positive comparison value exists; without it, a function such as $-e^{-x^2}$ would have an unattained supremum of zero.