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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 $V$), so we are reduced to study the almost obvious case where $K=N-1$ i.e. free up to one more officer (that will be necessarilly $V$) because at the very moment that $F$ get out of $[V]$ he will still be closer to $V$ than to any officer of$\mathcal X\setminus \mathcal Y$ . Let's first see intuitivally : it is not more difficult to escape $[V]$ than to escape $\mathcal X$ in the $K=N-1$ case where $[V]$ is the whole convex hull , if we are not in this easy case, the aim of the fugitive should be to get away from $\mathcal Y$ and when he get close to one of the $2$ edges that are adjacent vertices $[V]$ (and out of reach of $\mathcal Y$) he can pass thought this edge and increase $K$ of one unit, such that $\mathcal Y$ is far compare to $V$, where all other officer are locked in the "opposite angle" of the angle that define the vertex $V$ of $[V]$, that motion in continuous lipschitzian, just like a real officer. I said the $[V]=conv(\mathcal X) $ case in not more difficult because officers in $\mathcal X\setminus \mathcal Y$ dose not influence in any way the escape of $[V]$, that only dépend of This is for the intuition, now let's be precize, de are going to see that the case K=N-1 itselfs is not more difficult than the case of a triangle with a forbiden edge to cross for the fugitive. (Thus this rules the case $c<2$ of Pace's Nielen answer!)

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 close as we want. So we will eventually get $r_1<r_2/2$ at some point, then we go to phase 2 at time $t_0$.

Then I claim that :

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 $\phi_{w_k}$, $\delta(\phi_{w_k})$ 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 $\delta(\phi_w)/4)$)

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)

The proof is obvious by induction 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.

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 $(F_mF_M)$ is not able to maintain a constant distance to F, like it could be the case when $c\geq 2$)

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$\mathcal X\setminus \mathcal Y$ .

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 $t_0$.

Then :

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 $\delta(\phi_w)/4)$)

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)

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.)

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 $c\geq 2$)

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 $V$), so we are reduced to study the almost obvious case where $K=N-1$ i.e. free up to one more officer (that will be necessarilly $V$) because at the very moment that $F$ get out of $[V]$ he will still be closer to $V$ than to any officer of$\mathcal X\setminus \mathcal Y$ . Let's first see intuitivally : it is not more difficult to escape $[V]$ than to escape $\mathcal X$ in the $K=N-1$ case where $[V]$ is the whole convex hull , if we are not in this easy case, the aim of the fugitive should be to get away from $\mathcal Y$ and when he get close to one of the $2$ edges that are adjacent vertices $[V]$ (and out of reach of $\mathcal Y$) he can pass thought this edge and increase $K$ of one unit, such that $\mathcal Y$ is far compare to $V$, where all other officer are locked in the "opposite angle" of the angle that define the vertex $V$ of $[V]$, that motion in continuous lipschitzian, just like a real officer. I said the $[V]=conv(\mathcal X) $ case in not more difficult because officers in $\mathcal X\setminus \mathcal Y$ dose not influence in any way the escape of $[V]$, that only dépend of This is for the intuition, now let's be precize, de are going to see that the case K=N-1 itselfs is not more difficult than the case of a triangle with a forbiden edge to cross for the fugitive. (Thus this rules the case $c<2$ of Pace's Nielen answer!)

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", I think it is always working if $c<2$ but I did not think so far)

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 $V$), so we are reduced to study the almost obvious case where $K=N-1$ i.e. free up to one more officer (that will be necessarilly $V$) because at the very moment that $F$ get out of $[V]$ he will still be closer to $V$ than to any officer of$\mathcal X\setminus \mathcal Y$ . Let's first see intuitivally : it is not more difficult to escape $[V]$ than to escape $\mathcal X$ in the $K=N-1$ case where $[V]$ is the whole convex hull , if we are not in this easy case, the aim of the fugitive should be to get away from $\mathcal Y$ and when he get close to one of the $2$ edges that are adjacent vertices $[V]$ (and out of reach of $\mathcal Y$) he can pass thought this edge and increase $K$ of one unit, such that $\mathcal Y$ is far compare to $V$, where all other officer are locked in the "opposite angle" of the angle that define the vertex $V$ of $[V]$, that motion in continuous lipschitzian, just like a real officer. I said the $[V]=conv(\mathcal X) $ case in not more difficult because officers in $\mathcal X\setminus \mathcal Y$ dose not influence in any way the escape of $[V]$, that only dépend of This is for the intuition, now let's be precize, de are going to see that the case K=N-1 itselfs is not more difficult than the case of a triangle with a forbiden edge to cross for the fugitive. (Thus this rules the case $c<2$ of Pace's Nielen answer!)

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 $(F_mF_M)$ is not able to maintain a constant distance to F, like it could be the case when $c\geq 2$)

When $|\mathcal Y|=K$, $F\in \bigcap_{\mathcal Z\subset \mathcal X\setminus\mathcal Y} conv(\mathcal Y \cup \mathcal Z)$ ($*$), so that no officer from $\mathcal X\setminus \mathcal Y$ can be in the interior of this convex, that is equal 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 $V$), so we are reduced to study the almost obvious case where $K=N-1$ i.e. free up to one more officer (that will be necessarilly $V$) because at the very moment that $F$ get out of $[V]$ he will still be closer to $V$ than to any officer of$\mathcal X\setminus \mathcal Y$ so, like the obvious way to deal with in the $K=N-1$ case suggests it, the aim of the fugitive is to get away from $\mathcal Y$ and when he get close to one of the $2$ edges that are adjacent vertices $[V]$ (and out of reach of $\mathcal Y$) he can pass thought this edge and increase $K$ of one unit, such that $\mathcal Y$ is far compare to $V$, where all other officer are locked in the "opposite angle" of the angle that define the vertex $V$ of $[V]$, that motion in continuous lipschitzian, just like a real officer. This is for the intuition, now let's be precize :

($*$) note that this is the case iff $\mathcal Y$ is "maximal" for the inclusion (for the property that its convex hull does not contain $F$). So all this argument is valid even if $\mathcal Y$ is not maximal in cardinality (i.e. $|\mathcal Y|=K$) but still maximal for inclusion.

Let $(F_mF_M)$ 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 close as we want. So we will eventually get $r_1<r_2/2$ at some point, then we go to phase 2 at time $t_0$.

For any compact $C\subset \mathbb R^2\times (\mathbb R^2)^N$, that interior is not empty, there exists $\delta_C>0$, such that for all $w\in C$,there exists a $(\delta_C,w)$-strategy.

So now we have to prove that $\delta_C$ is bounded when $C$ runs in compacts that contain a valid initial disposition i.e. $C$ such that ($(F,\mathcal X)\in C$ iff $d(F,\mathcal X)\geq 1$)

I claim that $\delta_N:=\delta_{D_N(F,R^{N+1})}>\delta_C$ for all compact $C$ that contain $D_N(F,R^{N+1}):=\left\{(F,\mathcal X), \mathcal X^*\subset D(F,R^N)\right\}$

The proof is obvious by induction on $\mathcal X\setminus \mathcal Y$ where $conv(\mathcal Y)$ is a $\phi_0$-maximal convex defined in the I) ("maximal" wrt inclusion see in particular the note $(*)$)

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 $V$), so we are reduced to study the almost obvious case where $K=N-1$ i.e. free up to one more officer (that will be necessarilly $V$) because at the very moment that $F$ get out of $[V]$ he will still be closer to $V$ than to any officer of$\mathcal X\setminus \mathcal Y$ . Let's first see intuitivally : it is not more difficult to escape $[V]$ than to escape $\mathcal X$ in the $K=N-1$ case where $[V]$ is the whole convex hull , if we are not in this easy case, the aim of the fugitive should be to get away from $\mathcal Y$ and when he get close to one of the $2$ edges that are adjacent vertices $[V]$ (and out of reach of $\mathcal Y$) he can pass thought this edge and increase $K$ of one unit, such that $\mathcal Y$ is far compare to $V$, where all other officer are locked in the "opposite angle" of the angle that define the vertex $V$ of $[V]$, that motion in continuous lipschitzian, just like a real officer. I said the $[V]=conv(\mathcal X) $ case in not more difficult because officers in $\mathcal X\setminus \mathcal Y$ dose not influence in any way the escape of $[V]$, that only dépend of This is for the intuition, now let's be precize, de are going to see that the case K=N-1 itselfs is not more difficult than the case of a triangle with a forbiden edge to cross for the fugitive.   (Thus this rules the case $c<2$ of Pace's Nielen answer!)

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 close as we want. So we will eventually get $r_1<r_2/2$ at some point, then we go to phase 2 at time $t_0$.

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.

So now we have to prove that $\delta_C$ is bounded when $C$ runs in compacts.

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)

The proof is obvious by induction on $\mathcal X\setminus \mathcal Y$ where $conv(\mathcal Y)$ is a $\phi_0$-maximal convex defined in the I)