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Nov 28, 2019 at 13:01 comment added ABIM Fair enough, then I will restrict the supremum to a compact set. This takes care of the problem.
Nov 28, 2019 at 13:01 history edited ABIM CC BY-SA 4.0
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Nov 27, 2019 at 12:55 review Close votes
Dec 2, 2019 at 3:05
Nov 27, 2019 at 12:46 comment added Mateusz Kwaśnicki @N00ber: Your edit changes nothing: every "unusual" solution will grow even faster. See this article by Tychonoff, and this answer by George Lowther.
Nov 27, 2019 at 10:54 comment added ABIM We can assume that $p$ has finite sup and that it is strictly non-negatively valued.
Nov 27, 2019 at 10:53 history edited ABIM CC BY-SA 4.0
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Nov 26, 2019 at 17:59 comment added Christian Remling The fact that $p$ is completely arbitrary as a function of $t$ seems a strange feature of the set-up. For example, for many $p$'s the supremum will just be infinite for any $u$.
Nov 26, 2019 at 15:59 comment added Dirk If you want to project, you should minimize instead of maximize, right? To clarify: $p(t,x)$ is given and you want to find $u\in X$ which minimizes the distance $\|u-p\|$? (Which norm, by the way…)
Nov 26, 2019 at 12:38 comment added Mateusz Kwaśnicki I mean: if $u_1$ and $u_2$ are solutions and both $u_1-p$ and $u_2-p$ are bounded, then $u_1-u_2$ is a bounded solution with zero initial data, and hence $u_1=u_2$.
Nov 26, 2019 at 12:36 comment added ABIM But p does not solve the heat equation so the difference $u-p$ is not a solution to the heat equatoin, in general.
Nov 26, 2019 at 10:56 comment added Mateusz Kwaśnicki If there is any such solution, it is automatically a minimizer. Indeed: the difference of any two solutions is a solution of the heat equation with zero initial data, and hence it is either unbounded or identically zero.
Nov 26, 2019 at 10:42 history asked ABIM CC BY-SA 4.0