Thread: Sup norm is metric proof.

Consider the set of all continuous functions on [0,1], . Define .

Prove that

Proof so far.

Let and

As well as

Then , there exists such that:





So now, I have ... I'm stuck here...

How should I go on from here? Thanks!

Originally Posted by tttcomrader

Consider the set of all continuous functions on [0,1], . Define .

Prove that

Proof so far.

Let and

As well as

Then , there exists such that:









So now, I have ... 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 :

sup{|f(x)-g(x)|:xε[0,1]} and

sup{|g(x)-h(x)|: xε[0,1]}

BUT:

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]} p(f,g) + p(g,h) ,
thus p(f,h) p(f,g) + p(g,h)

But does necessarily takes on one of the value of for some ? If the sup is outside of this set, then the inequality doesn't hold.

Thanks!