Prove that Supremum Metric is less than

Question

Let $C'([0,1])$ be the set of differentiable functions $f:[a,b]\to \Bbb R$, having continuous derivatives

The supremum metric is $$d(f,g)= \sup_{t\in [0,1]}|f(t)-g(t)|$$

and we are given

$$p(f,g)=|f(0)-g(0)|+ \sup_{t\in [0,1]}|f'(t)-g'(t)|$$

Show that

$$d(f,g) \leq p(f,g)$$

My Attempt:

I have proven that $p(f,g)$ is a metric on $C'([0,1])$, then we can define a new function $h(t)=f(t)-g(t)$ and prove that

$$\sup_{t\in [0,1]}|h(t)| \leq |h(0)|+ \sup_{t\in [0,1]}|h'(t)|$$

But I'm stuck here

Answer 1

We have (Mean value theorem)

$|h(t)|-|h(0)| \le |h(t)-h(0)|=t|h'(s)| \le |h'(s)| \le \sup_{t \in [0,1]} |h'(t)|$

for some $s \in [0,t]$.

Hence

$|h(t)| \le |h(0)|+\sup_{t \in [0,1]} |h'(t)|$ .

This gives

$\sup_{t\in [0,1]}|h(t)| \leq |h(0)|+ \sup_{t\in [0,1]}|h'(t)|$

Answer 2

Use the fundamental theorem of calculus: $$h(t) = h(0) + \int_0^t h'(s) \, ds$$ This leads to $$|h(t)| \le |h(0)| + \int_0^1 |h'(s)| \, ds \le |h(0)| + \sup_{t \in [0,1]} |h'(t)|.$$ Now take the supremum on the left.