Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $S$ be the set of continuous function $f$ defined on some segment $[0,a_f], a_f \ge 0$, and such that $f(0)=0$. For $f$ and $g$ in $S$, let $$ c_{fg}=\max\{z \in [0,a_f \land a_g] : f(x)=g(x) \text{ for all } x \in [0,z] \}, $$ then, define the metric, \begin{align} d(f,g) &= a_f - c_{fg} + a_g - c_{fg} \\ &= a_f + a_g -2c_{fg}, \end{align} Is $S$ complete?

I don't really know.
On the yes side, I tried this. That is, if $\left\{ f_n \right\}$ is Cauchy, then $$ \lim_{n,m \to \infty} a_n - c_{n,m} + a_m - c_{n,m} = 0, $$ where $f_n$ is defined on $[0,a_n]$ and $f_m$ is defined on $[0,a_m]$.
I don't see why the $c_{n,m}$ should go to a limit.

On the no side, I tried this. That is the function sequence $$ f_n(0)=0,\\ f_n\left(1/n\right)=n, $$ which converges to the map $\{0\} \mapsto \{0\}$ in $S$.

share|improve this question
1  
So the domain depends on the function; is that right? Which is to say I can even have the function $g : \{0\} \rightarrow \{0\}$ for which $g(0) = 0$? If so, you should clean up the definition of $c_{fg}$ to reflect which interval the maximum is taken over (i think $[0,a_f \wedge a_g]$, right?). –  A Blumenthal Feb 5 '13 at 18:24
1  
Also, in the definition of your metric $d(\cdot, \cdot)$, do you mean $a_f + a_g - 2 c_{fg}$? –  A Blumenthal Feb 5 '13 at 18:26
    
@ABlumenthal I specified over which interval the sup is taken for $c$. Yes, I mean $a_f+a_g-2c_{fg}$. –  Nicolas Essis-Breton Feb 5 '13 at 20:21
1  
Thanks for the revisions, that's a lot clearer. I kinda get the feeling that the space is complete, for the following reason. The problem feels like it comes from $a_{f_n} \rightarrow \infty$, but in order for the sequence to be Cauchy, you'd be forced to have $c_{f_n, f_m}$ growing just as much as $a_{f_n}$. I'm having trouble, though, with the intermediate case that $a_{f_n}$ oscillates wildly. –  A Blumenthal Feb 5 '13 at 23:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.