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:

Originally Posted by kinkong
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 $[a,b]\to[0,1]$ in an increasing way, call this $g$, and you considered $T:C[a,b]\to C[0,1]$ defined by $Tf(x)=f(g(x))$?
• 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:

Originally Posted by 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...

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...