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!

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- September 10th 2009, 05:03 PMtttcomraderSup 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! - September 10th 2009, 07:33 PMxalk
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) - September 11th 2009, 06:46 PMtttcomrader
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!