EDIT: I don't know!
The thief can evade the cop.
Let's call the thief's position as a function of time $\Theta(t)$, and the cop's position as a function of time $C(t)$. We know each function is continuous. We don't know if they're differentiable, but based on the maximum speed restriction we can say $|\Theta(b)-\Theta(a)| < b-a$ and $|C(b)-C(a)| < b-a$ for any two times $a<b$.
One important thing to note: The point-people are described as able to react instantly according to their "positions". But because of that same instant reaction, neither can assume any knowledge about how the other will move in the future.
The cop's strategy of "move at full speed until our positions coincide" ($C(t) = \max(\frac{2}{3}-t, \Theta(t))$) is not a valid strategy because it depends on the movement of the thief in the future. The cop can't move left at full speed until an unknown time any more than move left at full speed for a full second.
Also, a thief's strategy of "move at half the speed of the cop" ($\Theta'(t) = C'(t)$) is not valid, since speed is a derivative involving a prediction into the future and is not really based on the cop's current position.
But the thief does still have a valid strategy using only current positions as input: Label the "Zeno points" $Z_n = 2^{-n}: \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots.$ Whenever the thief is between Zeno points, the thief will move left at speed 1 until the next Zeno point. When a thief is at a Zeno point $Z_n$ and the cop is to the right of $Z_{n-1}$, the thief stands still. When the thief is at a Zeno point $Z_n$ and the cop is at or to the left of Zeno point $Z_{n-1}$, the thief moves left at speed 1. That is,
$$ \Theta'(t) = \begin{cases}
0 & \exists n: \Theta(t)=2^{-n} \mathrm{\ and\ } C(t)>2^{1-n} \\
-1 & \mathrm{otherwise}
\end{cases} $$
This is a valid strategy only if for every $T \geq 0$,
$$\Theta(0) + \int_0^T \Theta'(t)\, dt \in (0,1)$$
(including the requirement that $\Theta'$ is integrable in the first place.)
Given any continuous $C(t): [0, \infty) \to (0,1)$, we can show by induction that when the cop arrives at any Zeno point for the first time, the thief's strategy so far is valid, the thief is currently at the next Zeno point, and the thief has not yet been caught.
The initial thief movement, after integration, is $\Theta(t) = \frac{1}{3} - t$ for $0 \leq t \leq \frac{1}{12}$ to reach $\Theta(t) = Z_2 = \frac{1}{4}$ at time $\frac{1}{12}$. When $t < \frac{1}{12}$, $C(t) > C(0) - \frac{1}{12} = \frac{7}{12} > Z_1$. Since $\Theta(t) < \frac{1}{3} < Z_1$ during this entire time, the thief was not caught during the first $\frac{1}{12}$ second. So if the cop reaches $Z_1$ for the first time (if ever) at time $t_1$, $t_1 > \frac{1}{12}$, the thief is already at $\Theta(t_1) = Z_2$ and has still not been caught.
Now suppose the induction hypothesis that $C(t_n) = Z_n$, and for all $0 \leq t < t_n$ we have $C(t) > Z_n$, and the thief's strategy has produced valid movement up to $\Theta(t_n) = Z_{n+1}$. The thief will now move to the next Zeno point: if $t_n \leq t \leq t_n + 2^{-n-2}$ then $\Theta(t) = Z_{n+1} - (t-t_n)$, ending with $\Theta(t_n + 2^{-n-2}) = Z_{n+2}$. In this same time interval, $C(t) \geq Z_n - (t-t_n) \geq 2^{-n} - 2{-n-2} > 2^{-n-1} = Z_{n+1}$, so the thief is not caught during that time. Then if there is then a time $t_{n+1}$ such that $C(t_{n+1}) = Z_{n+1}$, $t_{n+1} > t_n$ and $|C(t_{n+1})-C(t_n)| \leq t_{n+1}-t_n$ implies $t_{n+1} \geq t_n + Z_n - Z_{n+1} = t_n + 2^{-n-1} > t_n + 2^{-n-2}$, so the thief is already at $\Theta(t_{n+1}) = Z_{n+2}$ and has not been caught before time $t_{n+1}$.
So if the cop visits only a finite number of Zeno points, the thief reaches the first unvisited Zeno point without being caught, moves to the second unvisited Zeno point without being caught, and safely stays there forever. If the cop visits all the Zeno points, the thief can stay "a step ahead" and avoid capture forever.
Supposing there could be a point $t_e$ where $C(t_e) = \Theta(t_e)$ leads to a contradiction: On the closed interval $t \in [0, t_e]$, a continuous function $C$ must attain its minimum: there is at least one time $t_m \leq t_e$ with $C(t) \geq C(t_m)$ whenever $0 \leq t \leq t_e$. Let $Z_n$ be the last Zeno point with $Z_n \geq C(t_m)$. ($n = \lceil - \log_2 C(t_m) \rceil$.) The thief has position $\Theta(t) \leq Z_{n+1}$ whenever $t \geq t_m$. But since $t_e \geq t_m$, this means $\Theta(t_e) \leq Z_{n+1} < C(t_m) \leq C(t_e)$; contradiction.
These Zeno points also hint at an understanding of the problem with the cop's strategy. Claiming "I will certainly catch the thief before time $\frac{2}{3}$" is somewhat like claiming "I can visit all the Zeno points before time $\frac{2}{3}$". But the cop needs to be somewhere at the actual time $t=\frac{2}{3}$, besides the times before that. And a continuous function from closed interval $[0,\frac{2}{3}]$ to a universe $(0,1)$ must attain a minimum within that universe, and so can only visit a finite number of Zeno points. The cop can visit an unlimited number of Zeno points in less than $\frac{2}{3}$ seconds, but can never visit all of them in any duration of time.
c(t)means the position of the cop at time t? – justhalf Nov 04 '21 at 05:44