Every metric space can be isometrically embedded in a Banach space, so that it's a linearly independent set

Question

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?

Answer 1

Hint: Perhaps try the continuous real-valued functions on $X$ with a suitable norm.

Answer 2

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.