# Ngô Quốc Anh

## April 14, 2019

### Extending functions between metric spaces: Continuity, uniform continuity, and uniform equicontinuity

Filed under: Giải Tích 3, Giải tích 8 (MA5206) — Tags: — Ngô Quốc Anh @ 15:02

This topic concerns a very classical question: extend of a function $f : X \to Y$ between two metric spaces to obtain a new function $\widetilde f : \overline X \to Y$ enjoying certain properties. I am interested in the following three properties:

• Continuity,
• Uniformly continuity,
• Pointwise equi-continuity, and
• Uniformly equi-continuity.

Throughout this topic, by $X$ and $Y$ we mean metric spaces with metrics $d_X$ and $d_Y$ respectively.

CONTINUITY IS NOT ENOUGH. Let us consider the first situation where the given function $f : X \to Y$ is only assumed to be continuous. In this scenario, there is no hope that we can extend such a continuous function $f$ to obtain a new continuous function $\widetilde f : \overline X \to Y$. The following counter-example demonstrates this:

Let $X = [0,\frac 12 ) \cup (\frac 12, 1]$ and let $f$ be any continuous function on $X$ such that there is a positive gap between $f(\frac 12+)$ and $f(\frac12-)$. For example, we can choose

$\displaystyle f(x)=\begin{cases}x^2&\text{ if } x<\frac 12,\\x^3 & \text{ if } x>\frac 12.\end{cases}$

Since $f$ is monotone increasing, we clearly have

$\displaystyle f(\frac12-)-f(\frac 12+)=\frac18.$

Hence any extension $\widetilde f$ of $f$ cannot be continuous because $\widetilde f$ will be discontinuous at $x =\frac 12$. Thus, we have just shown that continuity is not enough. For this reason, we require $f$ to be uniformly continuous.