Let $a$ and $b$ be points in $ \mathbb R^2$ and let $X$ be the set of all smooth parametrized curves joining $a$ to $b$ in $\mathbb R^2$, with parameter interval $[0,1]$. That is, $X$ is the set of all continuously differentiable functions $\gamma: [0,1] \rightarrow \mathbb R^2$, with $\gamma(0)=a $ and $\gamma(1) = b$. Show that if $\delta(\gamma_1, \gamma_2)$ = $sup\{||\gamma_1(t)-\gamma_2(t)||: t \in [0,1]\},$ then $\delta$ is a metric on $X$.
I'm am so lost as to where to start with this problem. My first thought was to manipulate $||\gamma_1(t)-\gamma_2(t)||$ and possibly start with $||a - b||$ then use the triangle inequality some how and showing that this satisfies the properties of a metric.
For the triangle inequality , take $\gamma_{3} \in X$. Fix $t \in [0,1]$, then use the triange inequality for
$\gamma_1 (t), \gamma_2 (t), \gamma_3 (t)$ to get $||\gamma_1(t) - \gamma_2(t)|| \leq ||\gamma_1 (t) -\gamma_3(t)|| +||\gamma_3(t)-\gamma_2(t)||(*)$.
Now $||\gamma_1(t)-\gamma_3(t)||\leq \delta(\gamma_1,\gamma_3)$ and $||\gamma_3(t)-\gamma_2(t)||\leq \delta (\gamma_3,\gamma_2)$. Plug in (*) to get
$\forall t \in [0,1]$ , $||\gamma_1(t)-\gamma_3(t)||\leq \delta(\gamma_1,\gamma_3)+\delta(\gamma_3,\gamma_2)$. So by taking sup over all $t$ you get
the triangle inequality for $\delta$.