Exercise 2.11

Question

Exercise 11:
    For $x\in\mathbf R$ and $y\in\mathbf R$, define
    \begin{align*}
        d_1(x,y) &= (x-y)^2, \
        d_2(x,y) &= \sqrt{|x-y|}, \
        d_3(x,y) &= |x^2-y^2|, \
        d_4(x,y) &= |x-2y|, \
        d_5(x,y) &= \frac{|x-y|}{1+|x-y|}. 
    \end{align*}
    Determine, for each of these, whether it is a metric or not.

ghostofgarborg · Accepted Answer

$d_2$ and $d_5$ are metrics, the others are not. 
We note  that $d_3(1,-1) = 0$, and therefore violates condition (a). $d_4(0,1) \neq d_4(1,0)$, and violates
condition (b). We further note that 
$$ d_1(1, -1) = 4 > 2 = d_1(1,0) + d_1(0,-1) $$
which violates (c).

We now note the following:

Let $f : \mathbb R^{\geq 0} \to \mathbb R^{\geq 0}$ be a strictly increasing function such that $f(0) = 0$, which is subadditive, i.e.:
$$ f(a + b) \leq f(a) + f(b) $$
and let $d$ be a metric.
Then $f \circ d$ is a metric. That $f \circ d$ satisfies condition (a) follows from the injectivity of $f$,
and from the fact that $f(0) = 0$. That it satisfies $(b)$ follows because $d$ does. Lastly, if
$$ d(a,c) \leq d(a,b) + d(b,c) $$
then by virtue of being increasing and subadditive: 
$$ f \circ d(a,c) \leq f(d(a,b) + d(b,c)) \leq f \circ d(a,b) + f \circ d(b,c) $$
which establishes (c).

To show that $d_2$ and $d_5$ are metrics, it therefore suffices to show that $ \sqrt{x} $ and $ \frac{x}{1+x} $
satisfy the criteria on $f$ above.

We know from chapter 1 that $\sqrt{x}$ is strictly increasing, and that $\sqrt{0} = 0$. Subadditivity follows
by noting that
$$ a+b \leq a + b + 2\sqrt{ab} = (\sqrt a + \sqrt b)^2 $$ 
Therefore, $d_2$ is a metric.

The function
$$ \frac{x}{1+x} = 1 - \frac{1}{1+x} $$
is clearly strictly increasing, and $0/(1+0) = 0$.
Moreover, 
$$ \frac{a+b}{1+a+b} = \frac{a}{1+a+b} + \frac{b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b} $$ 
Consequently, $d_5$ is a metric.

Exercise 2.11

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Exercise 2.11

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