Any metric spaces can be viewed as a subset of normed spaces

• Dec 15th 2011, 02:34 PM
DragonCryings
Any metric spaces can be viewed as a subset of normed spaces
Consider a metric space (X, d). Let (B(X), ||.||) be the vector space of bounded real valued functions on X with the sup norm. Denote fx(y) = d(x, y). Fix point a in X and let gx=fx – fa. Show that x-> gx is an isometric embedding of X into B(X). Note that d(x, y) = ||fx – fy||

I uploaded a picture of the problem since I was unable to introduce the subscripts properly (sorry still new in the forum).

If someone could give me an idea on how to solve it, I would be really greatful.

• Dec 15th 2011, 03:04 PM
emakarov
Re: Any metric spaces can be viewed as a subset of normed spaces
$$\|f_x-f_y\|$$ gives you $\|f_x-f_y\|$.

It seems that the claim is obvious once you know that $d(x, y) = \|f_x – f_y\|$. Do you understand why this fact holds?
• Dec 15th 2011, 03:40 PM
DragonCryings
Re: Any metric spaces can be viewed as a subset of normed spaces
Thank you for the tip :)

I do not understand why that becomes obvious. What does that property denote, or imply that can be related to the rest of the problem?
• Dec 15th 2011, 03:46 PM
emakarov
Re: Any metric spaces can be viewed as a subset of normed spaces
You need to show that for all $x,y\in X$, $d(x,y)=\|g_x-g_y\|$. But by the definition of $g_x$ and $g_y$, $\|g_x-g_y\|=\|(f_x-f_a)-(f_y-f_a)\|=\|f_x-f_y\|$.
• Dec 15th 2011, 03:54 PM
DragonCryings
Re: Any metric spaces can be viewed as a subset of normed spaces
I think I understood what you mean. Thank you very much for your help, time and patience ^^
• Dec 15th 2011, 04:00 PM
emakarov
Re: Any metric spaces can be viewed as a subset of normed spaces
It is just slightly more complicated to show $d(x, y) = \|f_x – f_y\|$.