Show that C[0,1] and C[a,b] are isometric.
Solution:
A mapping T of X into X' is said to be isometric if for all x, y in X
d'(Tx,Ty)=d(x,y)
C is continous [0,1] and [a,b] closed interval..
The distance on Function space C[a,b] is d(x,y)=max|x(t)-y(t)|
I couldnt procced from here....i need some help....
thanks
This is what I did:
You are looking for a function, say F from C[a,b] into C[0,1], which is injective.
Choose F such that F(u)=u(b*t-(1-t)*a). Here you take a function u which is an element of C[a,b] and, using F this u becomes an element of C[0,1].
This function clearly preserves the distance propierty (isometric) which you are requiered. Indeed:
Lets take and allows to take , now
.
Finally is bijective from the form it was selected.
You were told to find a mapping from [a, b] to [0, 1]. There are many such mappings, some are bijective and some are not. Which did you use?
To prove that a mapping, f, is one-to-one, you need to show that "if f(x)= f(y) then x= y". HOW you do that depends upon exactly what f is. To prove that a mapping, from [a, b] to [0, 1] is onto, you need to show that "if y is in [0, 1] then there exist an x in [a, b] such that f(x)= y.