Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics II / 4

A function $f:E\to \mathbb{R}$ is continuous on $E$ if it is continuous at every $p\in E$, meaningThe function has uniform continuity on $E$ whenFor continuity, $\delta$ may depend on both $p$ and $\epsilon$. For uniform continuity, one $\delta$ must work for every pair of points in $E$.

Suppose the continuous function $f$ did not have uniform continuity on $[a,b]$. Negating its definition gives ${\epsilon }_0>0$ such that for every $n\ge 1$ there are $x_n,y_n\in [a,b]$ withThe Bolzano-Weierstrass theorem, which states that every bounded sequence of real numbers has a convergent subsequence, gives $x_{n_k}\to p$. The inequalities $a\le x_{n_k}\le b$ imply $p\in [a,b]$. By the triangle inequality,Since continuous functions preserve convergent sequences, both $f(x_{n_k})$ and $f(y_{n_k})$ tend to $f(p)$. Their difference therefore tends to zero, contradicting its absolute value being at least ${\epsilon }_0$. This proves the Heine-Cantor theorem in the required setting: $f$ has uniform continuity on $[a,b]$.

For any $x,y\in \mathbb{R}$, rationalizing the difference of square roots givessince $f(x),f(y)\ge 1$. Given $\epsilon >0$, use uniform continuity of $f$ with tolerance $2\epsilon$ to obtain $\delta >0$. Whenever $|x-y|<\delta$, the displayed estimate yields $|g(x)-g(y)|<\epsilon$. Thus $g$ has uniform continuity on $\mathbb{R}$. The square root function is in fact Lipschitz continuous with constant $1/2$ on $[1,\infty )$.

TakeBoth functions have values in $[1,\infty )$. The reverse triangle inequality gives $|g(x)-g(y)|\le |x-y|$, so $g$ is Lipschitz continuous and hence has uniform continuity.

For $x_n=n$ and $y_n=n+1/n$, the distances satisfy $|x_n-y_n|=1/n\to 0$, whereasIn particular tolerance $\epsilon =1$ defeats every proposed $\delta$: choose $n$ with $1/n<\delta$. Thus $f$ does not have uniform continuity on $\mathbb{R}$.