# Metric Spaces: Basic Concepts

**Definition 1**
A metric space is a pair \((X, \rho)\) of a set \(X\) and a *distance* (a
single-valued nonnegative real function) \(\rho\) such that for all \(x, y \in X\):

\(\rho(x, y) = 0\) iff \(x = y\);

\(\rho(x, y) = \rho(y, x)\) [symmetry];

\(\rho(x, z) \leq \rho(x, y) + \rho(y, z)\) [triangle inequality].

Examples of different metric spaces are:

discrete space;

\(R^1\);

n-dimensional Euclidean space;

\(R_1^n\);

\(R_0^n\);

\(C_{[a, b]}\);

\(l_2\);

\(C_{[a, b]}^2\);

\(m\);

\(R_p^n\);

\(l_p\).

Consult the book for more detail.

*Homogeneous* if it holds for two points \((a_1, \ldots, a_n)\) and \((b_1,
\ldots, b_2)\), then it holds for \((\lambda a_1, \ldots, \lambda a_n)\) and
\((\mu b_1, \ldots, \mu b_n)\) where \(\lambda\) and \(\mu\) are arbitrary real
numbers.

When showing the triangle inequality, several inequalitites are used and named.

*Cauchy-Schwarz inequality*:
\[(\Sigma_{k=1}^n a_k b_k)^2 \leq \Sigma_{k=1}^n {a_k}^2 \Sigma_{k=1}^n
{b_k}^2\]

*Schwarz’s inequality*:
\[(\int_a^b x(t) y(t) dt)^2 \leq \int_a^b x^2(t) dt \int_a^b y^2(t) dt\]

*Minkowski’s inequality*:
\[(\Sigma_{k=1}^n |a_k + b_k|^p)^{\frac{1}{p}} \leq
(\Sigma_{k=1}^n |a_k|^p)^{\frac{1}{p}} + (\Sigma_{k=1}^n |b_k|^p)^{\frac{1}{p}}\]

*Hölder’s inequality*:
\[\Sigma_{k=1}^n |a_k b_k| \leq (\Sigma_{k=1}^n |a_k|^p)^{\frac{1}{p}}
(\Sigma_{k=1}^n |b_k|^q)^{\frac{1}{q}}\] where \(p > 1\), \(q > 1\) and
\(\frac{1}{q} + \frac{1}{q} = 1\)

Let \(f\) be a mapping of one metric space \(X\) into another metric space \(Y\). \(f\)
*continuous at the point* \(x_0 \in X\) if, given any \(\epsilon > 0\), there
exists a \(\delta > 0\) such that \(\rho\prime(f(x), f(x_0) < \epsilon\) wherever
\(\rho(x, x_0) < \delta\).

If \(f\) bijective and both \(f\) and \(f^{-1}\) are continuous, \(f\) is called a
*homeomorphism*. \(X\) and \(Y\) are homeomorphic if there exists a homeomorphism
between them.

**Definition 2**
A one-to-one mapping \(f\) of one metric space \(R=(X, \rho)\) onto another metric
space \(R\prime = (Y, \rho\prime)\) is said to be an *isometric mapping* (or *isometry*) if
\[\rho(x_1, x_2) = \rho\prime(f(x_1), f(x_2))\]
for all \(x_1\), \(x_2 in R\). Correspondingly, \(R\) and \(R\prime\) are said to be
*isometric*.

## Problems

Given a metric space \((X, \rho)\), prove that:

\(|\rho(x, z) - \rho(y, u)| \leq \rho(x, y) + \rho(z, u)\) where \(x, y, z, u \in X\)

\(|\rho(x, z) - \rho(y, z)| \leq \rho(x, y)\) where \(x, y, z \in X\)

Use the triangle inequality. Also take absolutes.

Verify that \[(\Sigma_{k=1}^n a_k b_k)^2 = \Sigma_{k-1}^n {a_k}^2 \Sigma_{k-1}^n {b_k}^2 - \frac{1}{2}\Sigma_{i=1}^n\Sigma_{j=1}^n (a_i b_j - b_i a_j)^2\] and deduce the Cauchy-Schwarz inequality from this identity.

Trivial to deduce Cauchy-Schwarz from this identity, so just need to verify. Expand and plod through.

Verify that \[(\int_a^b x(t) y(t) dt)^2 = \int_a^b x^2(t) dt \int_a^b y^2(t) dt - \frac{1}{2} \int_a^b \int_a^b [x(s)y(t) - y(s)x(t)]^2 ds dt\] and deduce Schwarz’s inequality from this identity.

Again, easy to deduce, just need to verify. Maybe use Lagrange identity and Fubini-Tonelli theorem?

What goes wrong in Example 10, page 41, if \(p < 1\)?

*Hint.*Show that Minkowski’s inequality fails for \(p < 1\).Find counterexample.

Prove that the metric \[\rho_0 (x, ) = \max_{1 \leq k \leq n} |x_k - y_k|\] is the limiting case of the metric \[\rho_p(x, y) = (\Sigma_{k=1}^n |x_k - y_k|^p)^\frac{1}{p}\] in the sense that \[\rho_0(x, y) = \max_{1 \leq k \leq n} |x_k - y_k| = \lim_{\rho \rightarrow \infty} (\Sigma_{k=1}^n |x_k - y_k|^p)^\frac{1}{p}\]

The proof by CoveredInChocolate is great. Write out the sum. Choose the max of the \(|x_k - y_k|\) and take it out as a factor. Then every other term tends to 0 as \(p\) approaches infinity.

Starting from Young’s inequality \[ab \leq \frac{a^p}{p} + \frac{b^q}{q}\] which applies for arbitrary positive \(a\) and \(b\) and whenever \(p > 1\) and \(q > 1\) and \[\frac{1}{p} + \frac{1}{q} = 1\] deduce Hölder’s integral inequality.

Define \[ a = \frac{|x(t)|}{(\int_a^b |x(t)|^p dt)^{1/p}}, \quad b = \frac{|y(t)|}{(\int_a^b |y(t)|^q dt)^{1/q}}\] then apply Young’s inequality, and take integrals, noting that the right side is equal to \[\frac{1}{p} + \frac{1}{q} = 1\] and then we’re almost there.

Use Hölder’s integral inequality to prove Minkowski’s integral inequality. \[(\int_a^b|x(t) + y(t)|^p dt)^{1/p} \leq (\int_a^b|x(t)|^p)^{1/p} + (\int_a^b|y(t)|^p)^{1/p},\quad (p \geq 1)\]

See CoveredInChocolate’s solution.

Exhibit an isometry between \(C_{[0,1]}\) and \(C_{[1,2]}\).

Too easy.