How to Complete Your Metric Space II

In November of last year, I published a guide, if you will, on how to take an arbitrary metric space (X,d) and obtain a complete metric space (Y,\rho) which contains a dense isometric “copy” of (X,d). This process is called completing a metric space, and the space (Y,\rho) is called the completion of (X,d).

The construction of (Y,\rho) that I gave involved defining an equivalence relation between Cauchy sequences in X by

\displaystyle(x_{n})_{n=1}^{\infty}\sim(y_{n})_{n=1}^{\infty}\Longleftrightarrow\lim_{n\rightarrow\infty}d(x_{n},y_{n})=0

We then defined Y to be the set of equivalence classes \tilde{x} endowed with the metric

\displaystyle\rho(\tilde{x},\tilde{y})=\lim_{n\rightarrow\infty}d(x_{n},y_{n}),

where (x_{n})_{n=1}^{\infty} and (y_{n})_{n=1}^{\infty} are representative elements of the equivalence classes \tilde{x} and \tilde{y}, respectively. X has a natural embedding into Y by sending each x to the equivalence class containing the sequence which is identically x.

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 AnalysisThe construction is actually part of Exercise 17 in Section 13.2.

Let (X,d) be a nonempty metric space containing a point x_{0}. Denote the space of Lipschitz functions f which vanish at the point x_{0} by \text{Lip}_{0}(X). I claim that \text{Lip}_{0}(X) is a real normed space, with norm defined by

\displaystyle\left\|f\right\|=\sup_{x\neq y}\dfrac{\left|f(x)-f(y)\right|}{d(x,y)}

Indeed, fix real scalars \alpha,\beta and functions f,g\in\text{Lip}_{0}(X). It is clear that the function \alpha f+\beta g vanishes at the point x_{0}. Since f and g are Lipschitz,there exist positive constants C_{f} and C_{g} such that

\displaystyle\left|f(x)-f(y)\right|\leq C_{f}d(x,y)\text{ and }\left|g(x)-g(y)\right|\leq C_{g}d(x,y),\indent\forall x,y\in X

Set C:=\max\left\{C_{f},C_{g}\right\}. Then

\begin{array}{lcl}\displaystyle\left|(\alpha f+\beta g)(x)-(\alpha f+\beta g)(y)\right|&=&\displaystyle\left|\alpha\right|\left|f(x)-f(y)\right|+\left|\beta\right|\left|g(x)-g(y)\right|\\[1 em]&\leq&\displaystyle\left|\alpha\right|C_{f}d(x,y)+\left|\beta\right|C_{g}d(x,y)\\[1 em]&\leq&\displaystyle C(\left|\alpha\right|+\left|\beta\right|)\cdot d(x,y)\end{array}

To see that \left\|\cdot\right\| is a norm, observe

\displaystyle0=\left\|f\right\|\Longleftrightarrow \dfrac{\left|f(x)-f(y)\right|}{d(x,y)}=0\text{ }\forall x\neq y\in X\Longleftrightarrow f \equiv f(x_{0})

Since we require that f(x_{0})=0, we conclude that f is the zero function. Positive homogeneity is obvious, and the triangle inequality is a consequence of the triangle inequality of absolute value.

Moreover, \text{Lip}_{0}(X) is a Banach space. Indeed, let (f_{n})_{n=1}^{\infty} be a Cauchy sequence in \text{Lip}_{0}(X). I claim that, for any point x\in X, the real sequence (f_{n}(x))_{n=1}^{\infty} is Cauchy. Indeed, let n and m be sufficiently large so that, for \varepsilon>0 given,

\displaystyle\varepsilon>\left\|f_{n}-f_{m}\right\|=\sup_{x\neq y}\dfrac{\left|(f_{n}-f_{m})(x)-(f_{n}-f_{m})(y)\right|}{d(x,y)}\geq\sup_{x}\dfrac{\left|f_{n}(x)-f_{m}(x)\right|}{d(x,x_{0})},

where we use the hypothesis that f_{n}(x_{0})=f_{m}(x_{0})=0. Hence,

\displaystyle\left|f_{n}(x)-f_{m}(x)\right|<\varepsilon\cdot d(x,x_{0}),\indent\forall x\in X

So, the sequence f_{n} has a pointwise limit f. It remains for us to show that f\in\text{Lip}_{0}(X). Since f_{n}(x_{0})=0 for all n, it is clear that f(x_{0})=0. To see that f is Lipschitz on the space X, observe that

\begin{array}{lcl}\displaystyle\left|f(x)-f(y)\right|&=&\displaystyle\lim_{n,m\rightarrow\infty}\left|f_{n}(x)-f_{m}(y)\right|\\[1 em]&\leq&\displaystyle\lim_{n,m\rightarrow\infty}\left|f_{n}(x)-f_{n}(y)\right|+\left|f_{n}(y)-f_{m}(y)\right|\\[1 em]&\leq&\displaystyle\left(\lim_{n\rightarrow\infty}\left\|f_{n}\right\|\right)\cdot d(x,y)+\lim_{m\rightarrow\infty}\left|f_{n}(y)-f_{m}(y)\right|\\[1 em]&=&\displaystyle\left(\lim_{n\rightarrow\infty}\left\|f_{n}\right\|\right)\cdot d(x,y)\end{array}

Above, we used the result that \left\|f_{n}\right\| converges in \mathbb{R}, since the reverse triangle inequality implies it is a Cauchy sequence. We will later see that \left\|f_{n}\right\|\rightarrow\left\|f\right\|. Lastly, let N be a natural number such that n,m\geq N implies that \left\|f_{n}-f_{m}\right\|<\varepsilon. For any x,y\in X with x\neq y,

\begin{array}{lcl}\displaystyle\dfrac{\left|(f_{n}-f)(x)-(f_{n}-f)(y)\right|}{d(x,y)}&=&\lim_{m\rightarrow\infty}\dfrac{\left|(f_{n}-f_{m})(x)-(f_{n}-f_{m})(y)\right|}{d(x,y)}\\[1 em]&\leq&\displaystyle\limsup_{m\rightarrow\infty}\left\|f_{n}-f_{m}\right\|\\[1 em]&<&\displaystyle\varepsilon\end{array}

Since this bound is independent of x and y, we can take the supremum of the LHS above over all x\neq y to obtain \left\|f_{n}-f\right\|\leq\varepsilon.

I claim that there is an isometric embedding of X into the space of bounded linear functionals on \text{Lip}_{0}(X) (i.e. the dual space of \text{Lip}_{0}(X)), which we denote by \mathcal{L}(\text{Lip}_{0}(X),\mathbb{R}). Indeed, for each x\in X, let F_{x} denote the linear functional which evaluates the function f at the point x. I claim that

\displaystyle\left\|F_{x}-F_{y}\right\|=d(x,y),\indent\forall x,y\in X

Fix x,y\in X. Then by definition of the norm,

\begin{array}{lcl}\displaystyle\left\|F_{x}-F_{y}\right\|=\sup_{\left\|f\right\|=1}\left|f(x)-f(y)\right|&=& d(x,y)\cdot\sup_{\left\|f\right\|=1}\dfrac{\left|f(x)-f(y)\right|}{d(x,y)}\\[1 em]&\leq&\displaystyle d(x,y)\cdot\sup_{\left\|f\right\|=1}\left\|f\right\|\\[1 em]&=&\displaystyle d(x,y)\end{array}

For the reverse inequality, observe that the map

\displaystyle d(\cdot,x_{0}): X\rightarrow\mathbb{R},\text{ }x\mapsto d(x,x_{0})

is a Lipschitz function which vanishes at x_{0}. Since d(x,y)=0 if and only if x=y, we see that the mapping x\mapsto F_{x} is injective. Since any closed subset of a complete metric space is also complete, we conclude that the closure of X\hookrightarrow\mathcal{L}(\text{Lip}_{0}(X),\mathbb{R}) is a completion of X

We actually didn’t need to show that \text{Lip}_{0}(X) is a Banach space in order to prove that X has a completion in \mathcal{L}(\text{Lip}_{0}(X),\mathbb{R}), since the space of bounded (real or complex) linear functionals is Banach space, even if X is not complete.

References

Halsey Royden and Patrick Fitzpatrick, Real Analysis (Fourth Edition),  Prentice Hall, New York, 2010.

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3 Responses to How to Complete Your Metric Space II

  1. Reblogged this on Being simple and commented:
    very interesting post about completation of a metric space.

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