I am doing some proofs in real analysis with the sup-norm metric.
In my arguments, I have that $sup\: x\in[0,1] \big\{\int_{0}^{x}|f(t)-g(t)|dt\big\}\leq sup\: x\in[0,1] \big\{|f(t)-g(t)|\big\}$. My question; is this true for all continuous functions on $[0,1]$?
Yes it is true. Note you can just show it for $h(t)$ rather than $f(t)-g(t)$ (since the difference of continuous functions is a continuous function). I.e. if $h$ is a continuous function on $[0,1]$, then $$\sup\limits_{x\in[0,1]} \int_0^x |h(t)|\, dt \leq \sup\limits_{t\in [0,1]} |h(t)|.$$
You may want to try and prove this as an exercise. The main idea to use is that the integral of the function will be less than or equal to if we integrated its maximum value (a constant) over the interval.
Well, really as $\int_0^x |f(t)-g(t)|dt$ is the integral of a positive continuous function, then it is monotone so $sup_{x\in [0,1]} \int_0^x |f(t)-g(t)|dt = \int_0^1 |f(t)-g(t)|dt$. And as $|f(x)-g(x)|\leq sup_{x \in [0,1]} |f(x)-g(x)|$ then $\int_0^1 |f(t)-g(t)|dt \leq \int_0^1 sup_{x \in [0,1]} |f(t)-g(t)|dt= sup_{x \in [0,1]} |f(x)-g(x)|$. Of course you can repeat the same proof for any continuous function in $[0,1]$ so your thoughts were correct.
