Section 1.7 Isometries
As in other mathematical theories, we are interested in maps that preserve the given structure. In the case of metric spaces, this means the following.
Definition 1.62.
Let \((X,d_X)\) and \((Y,d_Y)\) be two metric spaces. If \(f \maps X
\to Y\) is a map such that
\begin{equation*}
d_Y(f(x_1),f(x_2)) = d_X(x_1,x_2)
\end{equation*}
for all \(x_1,x_2 \in X\text{,}\) then \(f\) is called an isometry. The two metric spaces are called isometric if there exists a bijective isometry between them.
Example 1.63.
Consider the set \(\{1, 2, 3\}\) with the discrete metric. Define \(f
\maps \{1, 2, 3\} \to \R^2\) by
\begin{equation*}
f(1) = (-1/2, 0),
\quad
f(2) = (1/2, 0),
\quad
f(3) = (0, \sqrt 3/2)\text{.}
\end{equation*}
(These are the corners of an equilateral triangle with side length \(1\text{.}\)) If \(\R^2\) is equipped with the Euclidean metric, then \(f\) is an isometry. But the two spaces are not isometric. Indeed, the first is finite while the second is infinite, and so there cannot exist a bijection between them.
When two metric spaces are isometric, we think of them as essentially the same.
Exercises Exercises
1. An isometry between \(\R^2\) with two different norms.
Consider the normed spaces \((\R^2,\n\blank_1)\) and \((\R^2,\n\blank_\infty)\) from Exercise 1.1.5. Let \(I\) be the \(2 \times 2\) identity matrix, and let
\begin{equation*}
A =
\begin{pmatrix}
1 \amp -1 \\
1 \amp 1
\end{pmatrix},
\end{equation*}
thought of as mappings \(\R^2 \to \R^2\text{.}\)
(a)
Show that \(I\) is not an isometry from \((\R^2,\n\blank_1)\) to \((\R^2,\n\blank_\infty)\text{.}\)
Hint.
It may be helpful to look at your pictures from Exercise 1.1.5. Or else you can just start picking points \(x,y \in \R^2\) and checking the definition.
(b)
Show that \(\n{Ax}_\infty = \n x_1\) for any \(x \in \R^2\text{.}\)
(c)
Use the previous part and the linearity of \(A\) to conclude that \(A\) is an isometry from \((\R^2,\n\blank_1)\) to \((\R^2,\n\blank_\infty)\text{.}\) Conclude that these two normed spaces are isometric.
2. Evaluation map on \(B(S)\).
Let \(S\) be a set and \(s_0 \in S\text{.}\) Consider the function \(\Phi \maps B(S) \to \R\) defined by \(\Phi(f) = f(s_0)\) for \(f \in B(S)\text{.}\)
(a)
Show that \(\Phi\) is Lipschitz continuous.
Solution.
For \(f, g \in B(S)\text{,}\) we have the estimate
\begin{equation*}
|\Phi(f) - \Phi(g)| = |f(s_0) - g(s_0)| \le \|f - g\|_{\sup}\text{,}
\end{equation*}
which shows that \(\Phi\) is Lipschitz continuous (with Lipschitz constant \(1\)).
(b)
Is \(\Phi\) an isometry?
Solution.
In general, \(\Phi\) is not an isometry. Suppose that there exists another point \(s_1 \in S\) with \(s_1 \ne s_0\) and consider a function \(f \in B(S)\) with \(f(s_0) = 0\) and \(f(s_1) \ne 0\text{.}\) Then \(|\Phi(f)| = 0\text{,}\) while \(\n f_{\sup} \ne 0\text{.}\) So \(\Phi\) does not preserve the distance between \(f\) and \(0\text{.}\)
But if \(S = \{s_0\}\) is a single point, then \(\Phi\) is an isometry. Indeed, any \(f,g \in B(S)\) will satisfy
\begin{equation*}
\n{f-g}_{\sup} = \sup_{s \in S} |f(s)-g(s)| = |f(s_0)-g(s_0)| = |\Phi(f)-\Phi(g)|\text{.}
\end{equation*}
Comment.
While the counterexample in the official solution is perhaps the simplest, there is nothing wrong with using a more complicated one. In previous years, for instance, several students working together took \(S=\R\) and \(s_0=0\) and considered a pair of trigonometric functions, say \(f=\cos\) and \(g = \sin\text{.}\) They then calculated
\begin{equation*}
\abs{\Phi(f)-\Phi(g)} = \abs{1-0} = 1
\end{equation*}
and
\begin{equation*}
\n{f-g}_{\sup} = \cdots = \sqrt 2 \text{.}
\end{equation*}
One minor comment I will make on this is that, rather than explicitly calculating \(\n{f-g}_{\sup}\text{,}\) it is enough to find a point \(s \in
\R\) where \(\abs{f(s)-g(s)} \gt 1\text{,}\) since this will force \(\n{f-g}_{\sup} \gt 1\text{.}\)
