Show that $\max_x |f(x) - g(x)|$ is a metric on $C[a,b]$ taking Analysis for the first time and I'm not that great at doing stuff like this.
The question is to show that: $d(f,g) = max_{a \leq t \leq b}\mid$$f(t) - g(t) \mid$ defines a metric on $C[a,b]$ on the closed interval $[a,b]$.
I have an idea how to show that this is a metric, but I don't fully understand the distance function. Is it just the maximum distance between the two functions $f(t)$ and $g(t)$ based on the interval: $[a, b]$?
Here's my take on this question:
$(i)$ $0 \leq d(f,g) < \infty$
This one seems sort of obvious, this distance function will never be negative and will always give return a number... I just don't really don't know how to show it... (Help?)
$(ii)$ $d(f, g) = 0$ if and only if $f(t) = g(t)$ 
This is pretty direct. If $d(f, g) = 0$ that just implies that $f(t) = g(t)$ and the reverse is direct by just plugging in $f(t) = g(t)$
$(iii)$ $d(f, g) = d(g, f)$
This is pretty direct as well.
$(iv)$ $d(f,h) \leq d(f,g) + d(g,h)$
This one doesn't really seem really obvious to me. Any hints for this?
 A: First note that the metric is well defined because if $f$ and $g$ are continuous, then so is $|f-g|$, so the latter does have a finite maximum value on $[a,b]$. This, along with the fact that $|f-g|$ is nonnegative, proves (i).
For (ii), it's clear that if $f = g$ then $d(f,g) = 0$. For the converse, note that if $d(f,g) = 0$, then $\max_x |f(x) - g(x)| = 0$, so $|f(x) - g(x)| = 0$ for all $x$, hence $f(x) = g(x)$ for all $x$.
Part (iii) is clear since $|f(x) - g(x)| = |g(x) - f(x)|$.
For the triangle inequality (iv), note that for all $x\in [a,b]$,
$$\begin{align}
|f(x) - h(x)|  &= |f(x) - g(x) + g(x) - h(x)| \\
&\leq |f(x) - g(x)| + |g(x) - h(x)| \\
&\leq \max_x |f(x) - g(x)| + \max_x |g(x) - h(x)| \\
&= d(f,g) + d(g,h)\end{align}$$
Therefore, for all $x \in [a,b]$ we have
$$|f(x) - h(x)| \leq d(f,g) + d(g,h)$$
and so if we take the max of the left hand side over $[a,b]$, it also must be bounded above by $d(f,g) + d(g,h)$. We conclude that
$$d(f,h) = \max_x |f(x) - h(x)| \leq d(f,g) + d(g,h)$$
as desired.
