Martin could you advise me as an expert, would it make sense to start a bounty for this question? I want to do this, but the problem seems to be unsolvable. — Valery Saharov2 hours ago
There are some doubts to which extent bounties are effective, see How effective are bounties? on meta. I would not go as far as calling the problem unsolvable. (Well, if we calculate the value of $L$ for the above function, then we solved it at least for some configuration of points.) It is possible that similar problems were studied somewhere and somebody working in a related area would be able to give a good reference. — Martin Sleziak37 mins ago
Ok, I see. Should I try on MO maybe? This question, if I don't solve it, will cost me a good research work. Btw, I see the major problem in $X$ rather than in the dimension of the Euclidean space for some reason. And the main problem, why I doubt the solution exists, is because even for two points something like the above mentioned inequality arises. Moreover, even $X$ were itself a Euclidean space, there needs to be a whole lot of conditions satisfied to build up a piecewise linear function. We would need to separate simplices or something like that ... — Valery Saharov9 mins ago
I am hardly the correct person to say whether the question is appropriate for MO. You can try to simply post it there and you will see. (But it is common practice to indicate clearly both here and at MO, that the question was posted at the other site, too.) Or you can try to ask in their chatroom. (However, it seems to be almost always empty.) Or you can try to ask in our chatroom (which is rather active). — Martin Sleziak2 mins ago
@ValerySaharov We started to add here many comments which seem to be unrelated to the (mathematical content) of the question. If you wish to continue the discussion, let us do so in chat. — Martin Sleziak14 secs ago
Now I checked MO and I see you have already posted it there:
Suppose there is a compact metric space $(X,\rho)$ and a Euclidean space $\mathbb{R}^n$.
There is a sequence of unequal points $\{x_1,...x_N\}$ in $X$ such that all metrics $\rho(x_i,x_j)$ are known and $f(x_i)=a_i$ for some $a_i$ in $\mathbb{R}^n$ whereas:
$$ \forall x_i,x_j \in \{x_1,...x_N\...
Two points, neither of which address your question directly, but need to be thought about
Moderator supported is not synonymous with official. In fact, a better (albeit still not perfect) descriptor of "official" on the Stack network is "current community consensus".
Irony aside, this very que...
I might be missing something, but $L=1$ seems a bit two optimistic. What about $X=\mathbb R$ and $x_0=a_0=0$, $x_1=1$, $a_1=2$. I.e., we have prescribed values $f(0)=0$ and $f(1)=2$. Then the Lipschitz constant has to be at least $2$.
Without loss of generality you can assume that $L=1$.
By Kirszbraun theorem you can make $f$ to be 1-Lipschitz.
If $X\subset \mathbb{R}^n$ you can make it piecewise distance preserving (in particular piecewise linear).
See Brehm, U., Extensions of distance reducing mappings to piecewise
congrue...
Do I understand at least correctly that this is special case of your general problem: a) Find a Lipschitz function $f\colon[0,1]\to\mathbb R$ such that $f(0)=0$, $f(1)=2$. b) For what $L$ is it possible to get such function?
Of course, even if it turns out that my objection is valid, it does not change the fact that Anton Petrunin's answer could be useful for you.
He told you about Kirszbraun theorem. It seems that this result might be useful in your setting.
If I understood the OPs question correctly, one special case is $X=[0,1]$ with prescribed values $f(0)=0$ and $f(1)=2$. In such case we cannot hope to get $L=1$. I guess that choice of $x_i$'s and $a_i$'s will influence possibilities for $L$ somehow. (Of course, I might have misunderstood something there.) — Martin Sleziak3 mins ago
Anton Petrunin's answer is saying basically this: Define function $f$ just on the subset $\{x_1,...,x_n\}$. This function is Lipschitz with constant L.
Kirszbraun theorem then says that it can be extended to a function on $X$.
Without loss of generality you can assume that $L=1$.
By Kirszbraun theorem you can make $f$ to be 1-Lipschitz.
If $X\subset \mathbb{R}^n$ you can make it piecewise distance preserving (in particular piecewise linear).
See Brehm, U., Extensions of distance reducing mappings to piecewise
congrue...
