Isometric

• May 23rd 2011, 11:01 PM
kinkong
Isometric
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
• May 23rd 2011, 11:14 PM
Drexel28
Quote:

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

What if you did a mapping that bijectively took $\displaystyle [a,b]\to[0,1]$ in an increasing way, call this $\displaystyle g$, and you considered $\displaystyle T:C[a,b]\to C[0,1]$ defined by $\displaystyle Tf(x)=f(g(x))$?
• May 23rd 2011, 11:54 PM
kinkong
Thank u very much...
But how can we prove that the map is bijective...i mean how can we see that it is one-to-one and onto...
• May 24th 2011, 12:05 AM
Drexel28
Quote:

Thank u very much...
But how can we prove that the map is bijective...i mean how can we see that it is one-to-one and onto...

Have you tried to prove it?
• May 24th 2011, 12:23 AM
kinkong
yes i thought about it..but i couldnt find a way...please can u help me...
• Aug 26th 2014, 12:35 PM
fabri
Re: Isometric
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 $\displaystyle u,w \in C[a,b]$ and $\displaystyle F$ allows to take $\displaystyle F(u),F(w) \in C[0,1]$, now
$\displaystyle d_{C[a,b]}(u,w)=\max_{t\in[a,b]} |u(t)-w(t)|=\max_{t\in[0,1]}|u(bt-(1-t)a)-u(bt-(1-t)a)|=d_{C[0,1]}(F(u),F(w))$.

Finally $\displaystyle F$ is bijective from the form it was selected.
• Aug 26th 2014, 03:33 PM
HallsofIvy
Re: Isometric
Quote:

Thank u very much...
But how can we prove that the map is bijective...i mean how can we see that it is one-to-one and onto...

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.
• May 12th 2017, 02:25 PM
Hernan
Re: Isometric
Excuseme, in this moment i trying resolve the same problem, but i dont understand why the norm are the same?. is it neccesary to show it by continuity of functions?. are there anything that i dont understand completely?. is it just a simple geometric notion, or is it something that i must prove?.

Thanks very much.