Math Help - Sup norm is metric proof.

1. Sup norm is metric proof.

Consider the set of all continuous functions on [0,1], $C[0,1]$. Define $p(f,g) = \sup _{x \in [0,1] } \mid f(x)-g(x) \mid$.

Prove that $p(f,g)+p(g,h) \geq p(f,h)$

Proof so far.

Let $a= \sup _{x \in [0,1] } \mid f(x)-g(x) \mid$ and $b= \sup _{x \in [0,1] } \mid g(x)-h(x) \mid$

As well as $c= \sup _{x \in [0,1] } \mid f(x)-h(x) \mid$

Then $\forall \epsilon > 0$, there exists $x-1 , x_2 , x_3 \in [0,1]$ such that:

$a- \epsilon < \mid f(x_1)-g(x_1) \mid$
$b- \epsilon < \mid g(x_2)-h(x_2) \mid$
$c- \epsilon < \mid f(x_3)-h(x_3) \mid$

So now, I have $a+b = a - \epsilon + b - \epsilon + 2 \epsilon$... I'm stuck here...

How should I go on from here? Thanks!

Consider the set of all continuous functions on [0,1], $C[0,1]$. Define $p(f,g) = \sup _{x \in [0,1] } \mid f(x)-g(x) \mid$.

Prove that $p(f,g)+p(g,h) \geq p(f,h)$

Proof so far.

Let $a= \sup _{x \in [0,1] } \mid f(x)-g(x) \mid$ and $b= \sup _{x \in [0,1] } \mid g(x)-h(x) \mid$

As well as $c= \sup _{x \in [0,1] } \mid f(x)-h(x) \mid$

Then $\forall \epsilon > 0$, there exists $x-1 , x_2 , x_3 \in [0,1]$ such that:

$a- \epsilon < \mid f(x_1)-g(x_1) \mid$
$b- \epsilon < \mid g(x_2)-h(x_2) \mid$
$c- \epsilon < \mid f(x_3)-h(x_3) \mid$

So now, I have $a+b = a - \epsilon + b - \epsilon + 2 \epsilon$... I'm stuck here...

How should I go on from here? Thanks!
if a+b = a -ε +b -ε +2ε then 0=0

For all xε[0,1] we have :

$|f(x)-g(x)|\leq$ sup{|f(x)-g(x)|:xε[0,1]} and

$|g(x)-h(x)|\leq$sup{|g(x)-h(x)|: xε[0,1]}

BUT:

$|f(x)-h(x)|\leq |f(x)-g(x)| + |g(x)-h(x)|\leq$ sup{|f(x)-g(x)|:xε[0,1]} + sup{|g(x)-h(x)|:xε[0,1]} = p(f,g) + p(g,h).

Hence sup{|f(x)-h(x)|:xε[0,1]} $\leq$ p(f,g) + p(g,h) ,
thus p(f,h) $\leq$ p(f,g) + p(g,h)

3. But does $sup \mid f(x)-h(x) \mid$ necessarily takes on one of the value of $\mid f(x_0) - h(x_0) \mid$ for some $x_0 \in [0,1]$? If the sup is outside of this set, then the inequality doesn't hold.

Thanks!