Let $S$ be a set, $B(S) = \{ f: S \rightarrow \mathbb{R} | f $ limited$ \}$. $B(S)$ is a normed vector space with the norm $||f|| = \sup_{x \in S} |f(x)|$. If $(S,d)$ is a limited metric space, for each $ x \in S$ we define $f_x \in B(S)$ by $f_x(y) = d(x,y)$.

Prove the function $ x \in S \mapsto f_x \in B(s)$ is an isometric immersion

(An application $f:M \rightarrow N$ between two metric spaces is an isometric immersion if $d(f(x),f(y)) = d(x,y) \forall x,y \in M$)

My attempt: I'll call the function $g$.

$d(g(x),g(y)) = d(f_x , f_y)$, by definition.

Since it is a normed space,

$d(f_x, f_y) = ||f_x - f_y|| $

Also, by definition:

$||f_x - f_y|| = \sup_{x \in S}|f_x(x) - f_y(x)| = \sup_{x \in S}|d(x,x) - d(y,x)| = d(x,y)$.

Can you please verify this? Thanks.

  • $\begingroup$ You have to use a different variable to take the $\sup$. It can't be $x$ again. $\endgroup$ – Aweygan Aug 21 '16 at 19:42
  • $\begingroup$ Also, can you say what you mean by "$f$ limited" and a limited metric space? $\endgroup$ – Aweygan Aug 21 '16 at 19:44

It is better not to use the same symbols for different meanings. Like the $f:M\rightarrow N$. Better call it $g$ right away. Also $d$ denotes a metric with two different meanings. The main estimate should be done in more details

$$\| f_x-f_y\|=\sup_{u\in S} |f_x(u)-f_y(u)|=\sup_{u\in S} |d(x,u)-d(y,u)|$$ Note that the triangle inequality implies that $|d(x,u)-d(y,u)|\leq d(x,y)$ (first prove this) and that equality is attained when setting $u=x$ (or $y$).

  • $\begingroup$ $|d(x,u) - d(y,u)| \leq d(x,y) \iff d(x,u)-d(y,u) \leq d(x,y) $ and $ d(x,u) - d(y,u) \geq -d(x,y) $ $;;;$ $d(x,u) \leq d(u,y) + d(y,x)$ (by t.i) $d(x,u) - d(y,u) \leq d(x,y)$ the first one is proved. suppose by contradiction: $d(x,u) - d(y,u) < -d(x,y)$ then $d(x,u) + d(x,y) < d(y,u) (1)$ but t.i. implies that: $d(y,u) \leq d(x,y) + d(x,u)$ then $(1)$ can't happen. $\endgroup$ – user286485 Aug 22 '16 at 19:26
  • $\begingroup$ That looks fine. Perhaps slightly more direct: $d(x,u) \leq d(x,y)+d(y,u)$ shows that $d(x,u)-d(y,u)\leq d(x,y)$. Interchanging $x$ and $y$ we get: $d(y,u)-d(x,u)\leq d(y,x)$ and the two together shows $-d(x,y)\leq d(x,u)-d(y,u)\leq d(x,y)$ and then the claim. Anyway your argument is fine. $\endgroup$ – H. H. Rugh Aug 22 '16 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy