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S Dec 7, 2023 at 1:08 history bounty ended CommunityBot
S Dec 7, 2023 at 1:08 history notice removed CommunityBot
S Nov 28, 2023 at 23:59 history bounty started user479223
S Nov 28, 2023 at 23:59 history notice added user479223 Draw attention
Nov 24, 2023 at 12:52 comment added Nate River You probably also want a uniform in $t$ Lipschitz condition in $x$ for both $b_1$ and $b_2$, with respect to the above pointwise distance.
Nov 24, 2023 at 12:48 comment added Nate River If we set up $b_1$ and $b_2$ so that the greedy choice is also globally optimal, we see that the bound given by the above is sharp. (2/2)
Nov 24, 2023 at 12:47 comment added Nate River I feel like the appropriate distance between $b_1 (x, t)$ and $b_2 (x, t)$ should just be the simple $\sup_{a_1 \in b_1 (x, t), a_2 \in b_2(x, t)} |a_1 - a_2|$. And then you probably want $\text{dist}(b_1, b_2)$ to be the sup over $x, t$ of the above distance. The reason being that someone who wanted to ruin our day could just choose for $dY_1/dt$ and $dY_2/dt$ the worst possible choice (i.e. the one that maximises the difference). This is the locally optimal way to maximise the distance. (1/2)
Nov 23, 2023 at 0:35 history edited user479223 CC BY-SA 4.0
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Nov 22, 2023 at 23:59 history asked user479223 CC BY-SA 4.0