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Aug 11, 2021 at 22:07 history edited YCor CC BY-SA 4.0
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Aug 11, 2021 at 17:50 vote accept Peter Wacken
Aug 11, 2021 at 16:32 answer added Christian Remling timeline score: 2
Aug 11, 2021 at 16:07 comment added Peter Wacken @ChristianRemling I don't see how to derive a contradiction, because $g$ depends on $n$ here. Why can't $g$ be close to $\sin 2^n x$ while at the same time have a moderately growing $H^1_0$-norm?
Aug 11, 2021 at 14:54 comment added Christian Remling I think for a counterexample you could consider something like $P(\sin nx)= \sin 2^n x$. Now an approximation of the desired type would have to map $\sin nx$ to functions $g$ with $\|g-\sin 2^n x\|_2<\epsilon$, $\|g'\|_2\le Cn$, which seems impossible (fix $\epsilon$, send $n\to\infty$).
Aug 11, 2021 at 14:24 history asked Peter Wacken CC BY-SA 4.0