Answer to the F puzzle
At firts I'm going to describe a continuous game where the maximum speed of fugitive has norm $1$, while the officer's sum of speed is not more then $1$. (c.f. the question above for the formal definitions) This will exempt us from considerate the "delta" in the firts place. We will just be concerned by escaping (gradually($**$)) the convex hull of officers, but we will be concerned as well by not losing to much movement during the run ($****$), so that we stay in a compact area and we will deduce a "delta" depending on the initial disposition, witch will be a first step, then we will finish the proof (we want the existence of a "general delta" only depending on $N$) exploiting the idea that if an officer is insanally far he won't have any relevant incidence on the "delta" that win the original dicrete game.
($**$) the fugitive will aim to increase the number of officer whom he is not in the convex hull
($****$) at most $N$ "straight runs" will be required
Let $conv(W)$ be the convex hull of any $W\subset \mathbb R^2$, and for all $a\in \mathbb R^2$ $d(W,a):=min_{w\in W}||w-a||$, also $d(W,V)=min_{w\in W,v\in V} ||w-v||$ I will sometime use the notation $(u_1,...,u_k)^*=:\left\{u_1,...,u_k\right\}$ but I might also identify them as soon as there is no possible confusion.
I) Escaping the convex hull continuoussly without being caught and with no consideration on the minimal distances yet
$\mathcal X(t)=\left\{X_1(t),...,X_N(t)\right\}$ is the set of officers and $F(t)$ is the fugitive at time $t$. Let's denote by $K(t)$ the maximal cardinality that can have some $\mathcal Y\subset \mathcal X$ such that $F(t)\notin conv(\mathcal Y)$. $F$ will escape the convex hull of officer and then win, iff $K$ is by some time, stationnary equal to $N$. So the aim of F will be to increse $K$. (see ($**$) a bit upper)
When $F\notin conv(\mathcal Y)$ and $Y$ is maximal (wrt inclusion) for this propoerty (this is indeed the case when it is maximal in cardinality and $|\mathcal Y|=K$) , $F\in \bigcap_{\mathcal Z\subset \mathcal X\setminus\mathcal Y} conv(\mathcal Y \cup \mathcal Z)$ , so that no officer from $\mathcal X\setminus \mathcal Y$ can be in the interior of this convex, that is equal (and this is the TRICK of the proof) to $conv(\mathcal Y\cup\left\{V\right\})=:[V]$ for some point $V\in \mathbb R^2$ that speed is less than $1$ (it is $1$ iff an officer is exactly at the place of $V$). The important point is At the very moment that $F$ get out of $[V]$ he will still be closer to $V$ than to any officer of$\mathcal X\setminus \mathcal Y$ .
Phase 1 : (straight) foward run
Let $(F_mF_M)=(FF_m)$ the line that is perpendicular to line $(FV)$ and intersectes the frontier of $[V]$ in $F_m$ and $F_M$, where $F_m$ is closer to $F$ than $F_M$. If $F$ is the middle, it does not mater as soon as we are going to consider $r_1\,:\, t=||F_m(t)-F(t)||$ that is continuous. Let $r_2(t)=d((F_mF),\mathcal Y)$ at time $t$. During the foward run $r_1/r_2$ can be as small as we want. So we will eventually get $r_1<r_2/2$ at some point, then we go to phase 2 at time $t_0$.
phase 2 : (straight) side run
$F$ will have a strait run in full motion on the semi line $[FF_m)$. I claim that he will never cross any officer, indeed it is a simple exercice that $||F_m-F||$ is decreasing, and if any officer in $\mathcal Y$ arrives to $F$ he will have run $r_2(t_0)>||F_m-F||$ and $F$ will be already out of $[V]$.
There will not be any member of $\mathcal X\setminus \mathcal Y$ closer than $F$ is to $V$ by definition of $V$.
When we cross $[V]$, $\mathcal Y$ is not anymore maximal, and we reset the porcess until we get tout of $conv(\mathcal X)$.
For each "maximal $\mathcal Y$", the total run of $F$ will not be more than twice the diameter of $\mathcal Y=max_{y,y'\in\mathcal Y\cup \left\{V\right\}}||y-y'||$ (to be very large and secure). So there exists a compact $C=C_{\mathcal X}$ such that when $F$ will be out of $C$ he will be out of $conv(\mathcal X)$ as well. Let's call this fact $(a)$
II) Compacity argument
Then we have for all $w=(F_0,\mathcal X_0)\in \mathbb R^2\times (\mathbb R^2)^N$ a continuous $\phi_w : (\mathbb R^2)^N\to \mathbb R^2$ such that $\phi_w(X_1,...,X_N)\notin \left\{X_1,...,X_N\right\}:=\mathcal X^*$, for any $\mathcal X=(X_1,...,X_N)\in (\mathbb R^2)^N$. Witch is equivalent to the fact that for any $\mathcal X\in (\mathbb R^2)^N$, $d(\phi_w(\mathcal X),\mathcal X^*)>0$. Then we can deduce by a), that there exists $\delta=\delta_{\phi_w}>0$, s.t. $d(\phi_w(\mathcal X),\mathcal X^*)\geq\delta$.
(Let's call any $\phi_w$ of this form a $(\delta,w)$-strategy, and $\delta(\phi_w)=\inf_{\mathcal X\in \mathbb R^{2N}} d(\phi_w(\mathcal X),\mathcal X^*)$
Then :
For any compact $C\subset \mathbb R^2\times (\mathbb R^2)^N$, there exists $\delta_C>0$, such that for all valid $w\in C$ (valid means that the fugitive is at at least distance $1$ from officers) ,there exists a $(\delta_C,w)$-strategy.
Indeed, if it was'nt the case, there would exist a sequance ${{w_k)}_{k\in \mathbb N}$ that tends to $w\in C$, such that for all $i\in \mathbb N$, and all strategies $\phi_{w_k}$, $\delta(\phi_{w_k})$ tends to $0$. But this is impossible because Any $(\delta,w)$-strategy induce the existence of a $\mathcal W=\mathcal W_{w}$ neigbourhood of $w$ s.t. for all $w'\in\mathcal W$, there exists a $(\delta(\phi_w)/4,w')$ strategy $\phi_{w'}$. (take everybody in a circle that radius is $\delta(\phi_w)/4)$)
III) Far officers are not a problem
So now we have to prove that $\delta_C$ is bounded when $C$ runs in compacts.
Let's call $D(A,r)$ (resp. $C(A,r)$) the disk (resp. circle)centered in $A\in \mathbb R^2$ that radius is $r\in \mathbb R$ If $\mathcal X^*=\left\{X_1,...,X_N\right\}\subset C(F,1)$, then by what we said before, there exists $\phi_0$ ,$\delta_0,(F,X_1,...,X_N))$ winning strategy and $R\in \mathbb R_+$ such that $\phi(\mathcal X) \notin D(F,R/2)\Rightarrow d(\phi(\mathcal X),\mathcal X^*)\geq\delta_0$.
I claim that we can assume that the fugitive and officers are in $\delta_{D_N(F,R^{N+1})}$ wlog. (If any $\delta_N$ is winning for F, then F will be able to win the game with this delta regardless the initial disposition)
(See the conclusion for the smallest compact that does the job for any other)
The proof is almost contain in the construction and is easy by by induction, say on $\mathcal X\setminus \mathcal Y$ where $conv(\mathcal Y)$ is a $\phi_0$-maximal convex defined in the I)
(We suppose it is true for all $2<K<N$, by the pigeon hole principle there exists $k\in \mathbb N$ such that $(\mathcal X\setminus \mathcal Y)^*\cap (D(F,R^{k+1}\setminus D(F,R^k))$ is empty. We then apply a first time the induction hypothesis to $0<M<N$ officers that are in $D(F,R^k)$, and when the fugitive is out of it, then another time up to a normalisation that is a $\delta_M$-homotethy.)
Conclusion and improvement:
We proved that there exists $\delta_N$ such that F can win with it regardless the initial disposition, and that it can be done in not more then $N-1$ straight moves (because fwe can chose $\mathcal Y$ s.t. $|\mathcal Y|\geq (N-1)/2$. The prove is also working for any $c$-game (defined in Pace Nielsen answer) that would work for the case $N=3$ with a "forbiden edge", it is always working if $c<2$ , indeed it is working as soon as the two officer of the "forbiden edge" are not able to maintain a constant distance to F, like it could be the case when $c\geq 2$)
Also there is a smart way to say that the game can be reduced to the case where officers are all at distance exactly $1$ from the fugitive.
Indeed, suppose that $\phi$ is working to escape such an initial configuration with $\delta_N$ as the minimum distance, and consider the stereographic projection of $\mathcal X=\left\{X_1,...X_N\right\}$ on $C(F,\delta_N)$ (intersection of the segments $[FX_i]$ with the circle) where $(F,\mathcal X)\in \mathbb R^2\times \mathbb R^{2N}$ is the initial configuration. Then there exists $\mathcal X'=\left\{{X'}_1,...,{X'}_N\right\}$ on the unit circle centered in $F$ such that $F$ $\delta_N$-wins against $\mathcal X'$ implies that $F$, $\delta_N$-wins against $\mathcal X$ (just take any $X'_i\in [FX_i]$ that officer want and that obei to the rules)
Maybe to finish with something a bit complete, it would be nice to state that the hardest initial case for fugitive is the initial regular polygon - I did not investigate futher, but it would suprise me that it is not the case. Of course the best achievement would be to give the maximum $\delta_N$, I would pronosticate that it should not be far from $2^{-N}$, but not sure yet...
By now, there is no "need" to read, the proof is finished (even since before the "conclusion"), and I'm going to get less formal, so don't worry if you don't understand where I'm going, you can still read in "diagonal", and pick-up one of the arguments that you would like to eventually investigate deeper...
There can also be improvement of the proof itself, that can be shorter, the "stereographic idea" might be one way, another can be to find a induction argument that capture the essence of something deep and decisive about the problem but does not demand construction details. I would call this a logical improvement. I've got some ideas, one of them use the fact that each $P\in \mathbb R^2$, can be associated to $I(P)$, an ideal (as a lattice that order is inclusion), $I(P)\subset \mathcal X^*$ such that $\Sigma_{W\in I(P)}||W-P||<||P-F||$. Say $|I(P)|$ is the weight of $P$. A notable fact is that if the fugitive is running in a straight motion according to a line $\lambda$, then for each $I(P)$ there is $[P_{min}P_{max}]\subset \lambda$ such that the fonction $I$ is constant on it. What is notable is that when $F$ is running on this line the $P_min$ that are in the front are always "running away" in the sens of the fugitive run, except in one only case, witch is that all the officers of $I(P_{min})$ are running in a straight full motion in the direction of $P_{min}$.Then an idea could be to import the idea "maximal convex" of the precedent proof, in a strongest but simpler action : we could say that a particular officer $X_0$ (we do the induction hypothesis that anyone can be taken for the job) is carying a line $\gamma(0)$ at time $0$, and that at any time $t$, where the officer is located in $X_0(t)$, theire is a line $\gamma(t)\ni X_0(t)$ that is parallele to $\gamma(0)$ such that if this line crosses the fugitive then he will lose (this is like if in the other proof we had took some officer "at infinity"). But instead of getting rid of the far officer like we did in III), we would conclude w.r.t. the fonction $I$. For example the fugitive canrun straight perpendiculary to $\gamma$, and then if all the $P_{min}$ are running away to, no problem, he will win. If not, he will just have to continue the run until he's much closer than some $I(P)$ than from any other officers. Then two facts are going to lead us to apply the induction hypothesis
- $I(P)\ne \mathcal X\setminus\left\{X_0\right\}$ (if it is the case, it is very easy to win
- $I(P)$ are kind of stucked all together in a small circle
Then all we have to do is put at least one officer behind the parallele to $\gamma$ that contain $F$, and apply the induction hypothesis. The only obtruction to this would come iff an officer $X_1$ does not let $F$ go futher, so that he cannot pass throught the parallele to $\gamma$ that contain $X_1$, but then $X_1$ would have to run at full motion and the other officer would not move. $F$ would then do the same thing but with $X_1$ at the place of $X_0$, and according to a different $\delta$ (that I thing can be taken as $2^{|I(P_min)|}$ (where $I(P_{min})\ni X_1$ at the time $F$ was running the first strait run). Then officer will have to chose either to keep the $X_1$ running either to mobilise another ideal.
But this is also a bit too constructive to enter in the kind of proof that I would qualify of essentially logical. I would rather see some induction hypothesis that only capture some wise relative magnitudes... I'm sure that something short can be found...
In the kind of opposite, there could be some physical proofs. For exmpale the weights that I've defined as $|I(P)|$ could be related to some distribution, and/or some "altitude", like a third coordinate, and $F$ would try to avoid, when it is possible the upper areas, it is quite intuitive that the less effort run (with the idea that climbing is more effort then going down).
Another physical ispired idea can be to consider officer as positive charges, and $F$ also positive, and $F$ would pick some field line that goes out, then what is going behind (downstreem)has no importance, and if some officer intent to cross the field line upstream than $F$ would have more than the time to change the field line. It seems very intuitive, but maybe a bit anoying to establish properly.
I also had a nice idea about considering all the segment that link officer, we have then a tilling of convex polygones, that stay continuossly convex (so it is easy for the fugitive to escape each one of them), some "real officer" might eventually be vertices of such polygon, but something nice is happening :
if in a polygon there is one vertex that is not an officer, then there is at most two officers on the other verteces
if each vertex of a polygone is an officer, then it is a triangle.
if there is only triangles, or if all the neighbors of a triangle is a triangle, then it is a very easy winning configuration.
So I didn't investigate more, but, this sounds kind of pretty to me... even if it is neither shorter to establish properly, either giving better boundings.
