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.
