Problem Set 1
Table of contents
Problem 1
Consider the following map: $ d : \mathbb{R}^2 \times \mathbb{R}^2 \to [0, \infty) $ where
\(d(x, y) = \begin{cases} \Vert x - y\Vert _{\mathbb{R}^2} & x, y, 0 \text{ collinear} \\ \Vert x\Vert _{\mathbb{R}^2} + \Vert y\Vert _{\mathbb{R}^2} & \text{otherwise.} \end{cases}\)
Here, I use $\Vert \cdot\Vert _{\mathbb{R}^2}$ to denote the Euclidean norm/magnitude of a vector in $\mathbb{R}^2$. Show that this map is a metric on $\mathbb{R}^2$. This is called the British Railway metric. (Try to figure out why!)
Hint: Try drawing a picture, and use the fact that $\Vert \cdot\Vert _{\mathbb{R}^2}$ is a metric.
(i) Positive definite:
\[\Vert x - y\Vert _{\mathbb{R}^2} \ge 0, \quad \Vert x\Vert _{\mathbb{R}^2} + \Vert y\Vert _{\mathbb{R}^2} \ge 0 \implies d(x, y) \ge 0\]If $d(x, y) = 0$, then if collinear, $\Vert x - y\Vert _{\mathbb{R}^2} = 0 \implies x = y.$ Otherwise:
\[\Vert x\Vert _{\mathbb{R}^2} + \Vert y\Vert _{\mathbb{R}^2} = 0 \implies \Vert x\Vert _{\mathbb{R}^2} = \Vert y\Vert _{\mathbb{R}^2} = 0 \implies x = y\]If $x = y$, then $d(x, y) = 0.$
(ii) Symmetry: If collinear, $\Vert x - y\Vert _{\mathbb{R}^2} = \Vert y - x\Vert _{\mathbb{R}^2}.$ Otherwise, $\Vert x\Vert _{\mathbb{R}^2} + \Vert y\Vert _{\mathbb{R}^2} = \Vert y\Vert _{\mathbb{R}^2} + \Vert x\Vert _{\mathbb{R}^2}.$
(iii) Triangle inequality:
Let $x, y, z \in \mathbb{R}^2.$ If $x, z$ are not collinear with $0$, and $x, y, 0$ are collinear, then $y, z, 0$ cannot be collinear. We conclude two cases.
(Case 1) Only one pair collinear, say $x, y, 0$:
\[d(x, y) = \Vert x - y\Vert \le \Vert x\Vert + \Vert y\Vert \le (\Vert x\Vert + \Vert z\Vert ) + (\Vert y\Vert + \Vert z\Vert ) = d(x, z) + d(y, z) \tag{*1}\] \[d(x, z) = \Vert x\Vert + \Vert z\Vert \le \Vert x - y\Vert + \Vert y\Vert + \Vert z\Vert = d(x, y) + d(y, z) \tag{*2}\](Case 2) No pair collinear. By $(*1)$, triangle inequality holds.
Problem 2
Is $ d : C^{1}([0,1]) \times C^{1}([0,1]) \to [0,\infty) $ defined by
\(d(f,g) = \sup_{x \in [0,1]} \vert f'(x) - g'(x)\vert\) a metric on $ C^{1}([0,1]) $? If so, prove it. If not, show what properties of a metric $d$ satisfies, and explain which properties of a metric $d$ fails.
(i) Positive definite: $d(f,g) \ge 0$. If $f = g$, then $f’(x) = g’(x), \, x \in [0,1]$, hence $d(f,g) = 0.$
If $d(f,g) = 0$, then $\forall x \in [0,1]$, $\vert f’(x) - g’(x)\vert = 0 \implies f’(x) = g’(x).$ But this does not lead to $f(x) = g(x)$. A counterexample is $f(x) := g(x) + C,$ where $C$ is a constant.
(ii, iii) Symmetry and Triangle inequality both hold.
Note: To make it a metric,
\[\tilde{d}(f,g)=\sup_{x \in [0,1]} \vert f(x) - g(x)\vert +\sup_{x \in [0,1]} \vert f'(x) - g'(x)\vert\]See Example 14 in Lecture 1.
Problem 3
Show that $ d : \mathbb{R} \times \mathbb{R} \to [0, \infty) $ where
\(d(x, y) = \frac{\vert x - y\vert }{1 + \vert x - y\vert }\) is a metric on $\mathbb{R}$.
Positive definite and symmetry are trivial to prove.
To prove triangle inequality, rewrite
\[d(x, y) = 1 - \frac{1}{1 + \vert x - y\vert }.\]Then
\[\begin{aligned} d(x, z) = 1 - \frac{1}{1 + \vert x - z\vert } &< 1 - \frac{1}{1 + \vert x - y\vert + \vert y - z\vert } \\ &= \frac{\vert x - y\vert }{1 + \vert x - y\vert + \vert y - z\vert } + \frac{\vert y - z\vert }{1 + \vert x - y\vert + \vert y - z\vert } \\ &\le \frac{\vert x - y\vert }{1 + \vert x - y\vert } + \frac{\vert y - z\vert }{1 + \vert y - z\vert } \\ &= d(x, y) + d(y, z). \end{aligned}\]
Problem 4
Define a semi-metric on $X$ as a metric that satisfies symmetry, the triangle inequality, and $d(x, y) \ge 0$ for all $x, y \in X$, but doesn’t necessarily satisfy $d(x, y) = 0 \implies x = y$. Specifically, $x = y \implies d(x, y) = 0$ but the opposite implication need not be true. Show that the sum of a metric and a semi-metric on $X$ is a metric on $X$. In other words, if $d$ is a metric on $X$, and $d’$ is a semi-metric on $X$, then $d + d’$ is a metric on $X$.
$d + d’$ satisfies symmetry, the triangle inequality, and $(d + d’)(x, y) \ge 0.$
Now show it also satisfies $(d + d’)(x, y) = 0 \implies x = y.$
If $(d + d’)(x, y) = 0$, suppose $x \ne y.$ Then $d(x, y) \ne 0$ since it is a metric. To satisfy $(d + d’)(x, y) = 0$, $d’(x, y) = -d(x, y) \ne 0.$ Thus $x \ne y \implies d’(x, y) \ne 0$ is equivalent to $d’(x, y) = 0 \implies x = y$, then $d’$ is a metric instead of a semi-metric. This contradicts the assumption.
Therefore, when $(d + d’)(x, y) = 0$, $x = y.$
Problem 5
Show that $ I_t : C^{0}([a,b]) \to C^{1}([a,b]) $ is a continuous map where
\(I_t(f) = \int_{a}^{t} f(x) \, dx\) for some $ t \in [a,b] $.
For $ f, g \in C^{0}([a,b]) $, define
\(d_{C^{0}}(f,g) = \sup_{x \in [a,b]} \vert f(x) - g(x)\vert\)For $ F, G \in C^{1}([a,b]) $, define
\(d_{C^{1}}(F,G) = \sup_{x \in [a,b]} \vert F(x) - G(x)\vert + \sup_{x \in [a,b]} \vert F'(x) - G'(x)\vert\)Let $\varepsilon > 0$. We want to find $\delta > 0$ such that if $ d_{C^{0}}(f,g) < \delta $, then $d_{C^{1}}(I_t(f), I_t(g)) < \varepsilon$.
Denote $ F = I_t(f) $, $ G = I_t(g) $. Then
\(F(x) - G(x) = \int_{a}^{x} (f(s) - g(s)) \, ds\)Thus,
\(\vert F(x) - G(x)\vert \le \int_{a}^{x} \vert f(s) - g(s)\vert \, ds \le (b - a) \sup_{x \in [a,b]} \vert f(x) - g(x)\vert = (b - a) d_{C^{0}}(f,g).\)Also, $\vert F’(x) - G’(x)\vert = \vert f(x) - g(x)\vert \le d_{C^{0}}(f,g).$ Then,
\(\begin{aligned} d_{C^{1}}(I_t(f), I_t(g)) &= \sup_{x \in [a,b]} \vert F(x) - G(x)\vert + \sup_{x \in [a,b]} \vert F'(x) - G'(x)\vert \\ &\le (b - a) d_{C^{0}}(f,g) + d_{C^{0}}(f,g)= (b - a + 1) d_{C^{0}}(f,g) \end{aligned}\)We find $\delta = \frac{\varepsilon}{b - a + 1}$, then if $ d_{C^{0}}(f,g) < \delta $,
\(d_{C^{1}}(I_t(f), I_t(g)) < (b - a + 1) \cdot \frac{\varepsilon}{b - a + 1} = \varepsilon.\)