Editing Uniform continuity (section)

== Relations with the extension problem ==

Let <math>X</math> be a metric space, <math>S</math> a subset of <math>X</math>'','' <math>R</math> a complete metric space, and <math>f: S \rightarrow R</math> a continuous function. A question to answer: ''When can <math>f</math> be extended to a continuous function on all of <math>X</math>?''

If ''<math>S</math>'' is closed in <math>X</math>, the answer is given by the [[Tietze extension theorem]]. So it is necessary and sufficient to extend ''<math>f</math>'' to the closure of <math>S</math> in <math>X</math>: that is, we may assume without loss of generality that ''<math>S</math>'' is dense in <math>X</math>, and this has the further pleasant consequence that if the extension exists, it is unique. A sufficient condition for <math>f</math> to extend to a continuous function <math>f: X \rightarrow R</math> is that it is [[Cauchy-continuous function|Cauchy-continuous]], i.e., the image under ''<math>f</math>'' of a Cauchy sequence remains Cauchy. If <math>X</math> is complete (and thus the completion of ''<math>S</math>''), then every continuous function from ''<math>X</math>'' to a metric space ''<math>Y</math>'' is Cauchy-continuous. Therefore when ''<math>X</math>'' is complete, ''<math>f</math>'' extends to a continuous function <math>f: X \rightarrow R</math> if and only if ''<math>f</math>'' is Cauchy-continuous.

It is easy to see that every uniformly continuous function is Cauchy-continuous and thus extends to ''<math>X</math>''. The converse does not hold, since the function <math>f: R \rightarrow R, x \mapsto x^2</math> is, as seen above, not uniformly continuous, but it is continuous and thus Cauchy continuous. In general, for functions defined on unbounded spaces like ''<math>R</math>'', uniform continuity is a rather strong condition. It is desirable to have a weaker condition from which to deduce extendability.

For example, suppose <math>a > 1</math> is a real number. At the precalculus level, the function <math>f: x \mapsto a^x</math> can be given a precise definition only for rational values of ''<math>x</math>'' (assuming the existence of qth roots of positive real numbers, an application of the [[Intermediate value theorem|Intermediate Value Theorem]]). One would like to extend <math>f</math> to a function defined on all of <math>R</math>. The identity

: <math>f(x+\delta)-f(x) = a^x\left(a^{\delta} - 1\right)</math>

shows that ''<math>f</math>'' is not uniformly continuous on the set ''<math>Q</math>'' of all rational numbers; however for any bounded interval ''<math>I</math>'' the restriction of ''<math>f</math>'' to <math>Q \cap I</math> is uniformly continuous, hence Cauchy-continuous, hence <math>f</math> extends to a continuous function on ''<math>I</math>''. But since this holds for every ''<math>I</math>'', there is then a unique extension of ''<math>f</math>'' to a continuous function on all of ''<math>R</math>''.

More generally, a continuous function <math>f: S \rightarrow R</math> whose restriction to every bounded subset of ''<math>S</math>'' is uniformly continuous is extendable to ''<math>X</math>'', and the converse holds if ''<math>X</math>'' is [[locally compact]].

A typical application of the extendability of a uniformly continuous function is the proof of the inverse [[Fourier transformation]] formula. We first prove that the formula is true for test functions, there are densely many of them. We then extend the inverse map to the whole space using the fact that linear map is continuous; thus, uniformly continuous.