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I'm trying to prove the following: Every metric space can be isometrically embedded in a Banach space, so that it's a linearly independent set.

I came up with the following idea:

Let $ (X,d) $ be a metric space. We can take the vector space of all functions from $ X $ to $ \mathbb{R} $. Now we can send an element $ x \in X $ to the function which takes the value $ 1 $ on $ x $ and $ 0 $ everywhere else. This would embed $ X $ in $ \mathbb{R}^X $, so that it would be a linearly independent set and actually a basis.

The problem seems to be finding a suitable norm, to make the embedding also an isometry. With that, we could just use the fact that every normed space can be isometrically embedded in a Banach space to get the desired result.

Any hints on constructing the metric? Or have I chosen a bad space?

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  • $\begingroup$ The set you constructed is NOT a basis for $\Bbb R^X$. $\endgroup$ Oct 29, 2014 at 18:52
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    $\begingroup$ Your embedding does not use the metric $d$. Try to embed $X$ into a space of functions by associating to every $x$ a function built from $x$ and $d$. $\endgroup$
    – polmath
    Oct 29, 2014 at 18:54
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    $\begingroup$ Probably useful: en.wikipedia.org/wiki/Kuratowski_embedding $\endgroup$ Oct 29, 2014 at 18:57
  • $\begingroup$ I've been reading about Kuratowsi embeddings, but can they/ something similar be used to densely embed any seperable metric space into a Banach space? $\endgroup$
    – ABIM
    Dec 17, 2018 at 22:27

2 Answers 2

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The metric space X embeds into a Banach space isometrically such that X is linearly independent:

Consider the vector space of all the Lipschitz functions from $X$ to $\mathbb R$, denoted by $L(X)$. Recall that a function $f:X\to\mathbb R$ is called Lipschitz if there exists constant $C$ such that $$ |f(x)-f(y)| \le C d(x,y) \text{ for all } x,y \in X $$ The infimum of $C$ is denoted by $L(f)$, which is not a norm on $L(X)$. (e.g. $L(1) = 0$).

However if we fix an element $z$ of $X$, then $L(X)$ has a norm given by $$ |f| = |f(z)| + L(f). $$ (To see this, both $L(f)$ and $|\text{point evaluation at} \ z|$ are seminorms. Thus the sum is a seminorm, so one needs to show that if $|f| = 0$, then $f=0$, which is easy.)

Now we wish to embed $X$ into $L(X)^*$, the continuous dual of $L(X)$. (Remark: continuous dual of a normed space is always a Banach space!).

The norm on $L(X)^*$ is the standard one: $$ |F|= \sup\bigg\{|F(f)|: f\ \text{is in}\ L(X)\ \text{with}\ |f(z)|+L(f) \le 1 \bigg\}. $$

We first observe that point evaluations are continuous:

if $a\in X$, then $\delta_{a}$ given by $\delta_{a} (f):=f(a)$ has norm $\le d(z,a)+1$. In fact

\begin{align} |\delta_{a}|&= \sup\{|f(a)|: |f(z)|+L(f) \le 1\},\\ &\le \sup\{|f(a)-f(z)|+|f(z)|:\ |f(z)|+L(f) \le 1\},\\ &\le \sup\{L(f)\cdot d(a,z) + |f(z)|:\ |f(z)|+L(f) \le 1\},\\ &\le d(a,z)+1. \end{align}

Therefore we have a well-defined map $\Phi$ from $X$ to $L(X)^*$ given by $\Phi(a) = \delta_a$.

Now we will show that $\Phi$ is an isometry:

Fix $a, b$. First note that

\begin{align} |\delta_a-\delta_b|&=\sup\{|f(a)-f(b)|: |f(z)|+L(f)\le1 \},\\ &\le \sup\{|f(a)-f(b)|: L(f)\le1\},\\ &\le d(a,b). \end{align}

Now, consider the function $f_a(x) = d(x,a)-d(z,a)$.

Then $f_a(z)=0$ and $L(f_a)\le1$. Thus $|f_a|\le1$. Moreover $|f_a(a)-f_a(b)| = d(a,b)$. Hence $|\delta_a-\delta_b| = d(a,b)$.

Finally given distinct $a_1,...,a_n$ in X one can show that $\delta_{a_1}, \dots ,\delta_{a_n}$ are linearly independent. If

$$c_1 \delta_{a_1} + \dots + c_n \delta_{a_n} = 0$$

then consider the function $g(x) = d(x,\{a_2,...,a_n\})$. Function $g$ is a Lipschitz function. Note that

$$(c_1 \delta_{a_1} + \dots + c_n \delta_{a_n}) (g) = 0,$$ $$c_1 g(a_1) + c_2 g(a_2) + \dots + c_n g(a_n) = 0,$$

$c_1 g(a_1) = 0$ which implies that $c_1=0$. In a similar fashion one can show that $c_2, \dots ,c_n$ must be all 0.

This finishes the proof.

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  • $\begingroup$ Welcome to MSE! Please see this tutorial on how to typeset your posts so they are easier to read. $\endgroup$
    – mrp
    Oct 4, 2016 at 8:56
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    $\begingroup$ Also standard Kuratowski's embedding theorem guarantees that the metric space X embeds into a Banach space, namely CB(X), continuous bounded functions from X to R with sup-norm. We first fix z. Given y, define phi_y from X to R by phi_y(x) = d(x,y) - d(z,x). One can easily verify that phi is an isometry. Also phi_z = 0. But the image of X-{z} in CB(X) is linearly independent. $\endgroup$
    – Ali Kavruk
    Oct 4, 2016 at 10:40
  • $\begingroup$ Compared to the Kuratowski embedding ($K:X\to \ell^\infty(X)$, $x\mapsto d(x,\cdot)- d(z,\cdot)$) the construction in Ali Kavruk's answer has the virtue of givng a functor from the category of pointed metric spaces (i.e., a metric space with a base point $z$ used to define the norm of L(X)) and base point preserving Lipschitz maps to the category of Banach spaces and continuous linear operators by assigning to a basepoint preserving Lipschitz function $h:X\to Y$ the transposed of the composition opeator $C_h:L(Y)\to L(X)$, $g\mapsto g\circ h$. $\endgroup$
    – Jochen
    Nov 7, 2023 at 16:55
  • $\begingroup$ I doubt that also the Kuratowski embedding can be made a functor. $\endgroup$
    – Jochen
    Nov 7, 2023 at 16:57
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Hint: Perhaps try the continuous real-valued functions on $X$ with a suitable norm.

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