Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 3 / 12G (Analysis II)

For a metric space $(X,d)$, uniform continuity meansThe choice of $\delta$ is independent of the points. A Lipschitz continuous function satisfies $|f(x)-f(y)|\le Ld(x,y)$ for one finite constant $L\ge 0$. Choosing $\delta =\epsilon /(L+1)$ proves Lipschitz continuity implies uniform continuity.

Let $(x_n)$ be a Cauchy sequence. For a given $\epsilon >0$, uniform continuity supplies $\delta >0$, and the Cauchy property supplies $N$ with $d(x_m,x_n)<\delta$ for $m,n\ge N$. Then $|f(x_m)-f(x_n)|<\epsilon$, so $(f(x_n))$ is Cauchy in $\mathbb{R}$. By completeness of the real numbers, $(f(x_n))$ converges, even if the original sequence has no limit in $X$.

Suppose the continuous $f$ were not uniformly continuous. There would be an ${\epsilon }_0>0$ and pairs $x_n,y_n\in X$ with $d(x_n,y_n)<1/n$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0$. Sequential compactness gives a subsequence $x_{n_j}\to x\in X$. The triangle inequality gives $y_{n_j}\to x$ as well. Continuity at $x$ then gives $f(x_{n_j})-f(y_{n_j})\to 0$, a contradiction. Thus $f$ is uniformly continuous.

It need not be Lipschitz continuous. On the sequentially compact interval $[0,1]$, the continuous function $f(x)=\sqrt{x}$ hasNo finite Lipschitz constant exists.

For each $x\in X$, density of $Y$ allows a sequence $y_n\in Y$ with $d(y_n,x)<1/n$. If $f_1,f_2$ are two continuous extensions, continuity givesThus a continuous extension is unique if it exists.

Now assume $g$ is uniformly continuous. Each chosen $y_n\to x$ is a Cauchy sequence, so part (a), applied to $Y$, shows that $g(y_n)$ converges in $\mathbb{R}$. DefineThis is independent of the choice: for another $z_n\to x$ in $Y$, $d(y_n,z_n)\to 0$, and uniform continuity gives $g(y_n)-g(z_n)\to 0$. At $x\in Y$, comparison with the constant sequence $x$ proves $f(x)=g(x)$.

To prove uniform continuity of the extension, choose $\delta >0$ so that $|g(y)-g(z)|<\epsilon /2$ whenever $y,z\in Y$ and $d(y,z)<\delta$. If $d(x,x^{\prime })<\delta /3$, choose sequences $y_n\to x$ and $z_n\to x^{\prime }$ in $Y$. For large $n$, both approximation distances are less than $\delta /3$, so $d(y_n,z_n)<\delta$. Passing to the limit gives $|f(x)-f(x^{\prime })|\le \epsilon /2<\epsilon$. The same $\delta /3$ works for all $x,x^{\prime }$. Therefore the extension exists and is uniformly continuous. No completeness of $X$ is needed; completeness of the target $\mathbb{R}$ is the relevant requirement.