In November of last year, I published a guide, if you will, on how to take an arbitrary metric space and obtain a complete metric space which contains a dense isometric “copy” of . This process is called completing a metric space, and the space is called the completion of .
The construction of that I gave involved defining an equivalence relation between Cauchy sequences in by
We then defined to be the set of equivalence classes endowed with the metric
where and are representative elements of the equivalence classes and , respectively. has a natural embedding into by sending each to the equivalence class containing the sequence which is identically .
Today, I want to give an entirely different construction using Lipschitz functions. This construction came to my attention by way of Halsey Royden’s textbook Real Analysis. The construction is actually part of Exercise 17 in Section 13.2.
Let be a nonempty metric space containing a point . Denote the space of Lipschitz functions which vanish at the point by . I claim that is a real normed space, with norm defined by
Indeed, fix real scalars and functions . It is clear that the function vanishes at the point . Since and are Lipschitz,there exist positive constants and such that
Set . Then
To see that is a norm, observe
Since we require that , we conclude that is the zero function. Positive homogeneity is obvious, and the triangle inequality is a consequence of the triangle inequality of absolute value.
Moreover, is a Banach space. Indeed, let be a Cauchy sequence in . I claim that, for any point , the real sequence is Cauchy. Indeed, let and be sufficiently large so that, for given,
where we use the hypothesis that . Hence,
So, the sequence has a pointwise limit . It remains for us to show that . Since for all , it is clear that . To see that is Lipschitz on the space , observe that
Above, we used the result that converges in , since the reverse triangle inequality implies it is a Cauchy sequence. We will later see that . Lastly, let be a natural number such that implies that . For any with ,
Since this bound is independent of and , we can take the supremum of the LHS above over all to obtain .
I claim that there is an isometric embedding of into the space of bounded linear functionals on (i.e. the dual space of ), which we denote by . Indeed, for each , let denote the linear functional which evaluates the function at the point . I claim that
Fix . Then by definition of the norm,
For the reverse inequality, observe that the map
is a Lipschitz function which vanishes at . Since if and only if , we see that the mapping is injective. Since any closed subset of a complete metric space is also complete, we conclude that the closure of is a completion of .
We actually didn’t need to show that is a Banach space in order to prove that has a completion in , since the space of bounded (real or complex) linear functionals is Banach space, even if is not complete.
Halsey Royden and Patrick Fitzpatrick, Real Analysis (Fourth Edition), Prentice Hall, New York, 2010.