# Sup norm is metric proof.

• Sep 10th 2009, 04:03 PM
Sup norm is metric proof.
Consider the set of all continuous functions on [0,1], $\displaystyle C[0,1]$. Define $\displaystyle p(f,g) = \sup _{x \in [0,1] } \mid f(x)-g(x) \mid$.

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

Proof so far.

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

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

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

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

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

How should I go on from here? Thanks!
• Sep 10th 2009, 06:33 PM
xalk
Quote:

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

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

Proof so far.

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

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

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

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

So now, I have $\displaystyle a+b = a - \epsilon + b - \epsilon + 2 \epsilon$... 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 :

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

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

BUT:

$\displaystyle |f(x)-h(x)|\leq |f(x)-g(x)| + |g(x)-h(x)|\leq$ 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]}$\displaystyle \leq$ p(f,g) + p(g,h) ,
thus p(f,h)$\displaystyle \leq$ p(f,g) + p(g,h)
• Sep 11th 2009, 05:46 PM
But does $\displaystyle sup \mid f(x)-h(x) \mid$ necessarily takes on one of the value of $\displaystyle \mid f(x_0) - h(x_0) \mid$ for some $\displaystyle x_0 \in [0,1]$? If the sup is outside of this set, then the inequality doesn't hold.