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May 30, 2018 at 20:57 comment added Jochen Glueck Yes, restricting the support of the Fourier transform of $u$ to a bounded set is one possibility to obtain a non-trivial example of closed $X$. (It is actually the same example as mentioned by the OP who suggests to apply the spectral measure of the Laplacian to a bounded set). I agree with your assessment that there might not be many more examples of closed spaces $X$ which fulfil the uniform continuity condition. Still, I have no idea how to prove this.
May 30, 2018 at 20:02 comment added Bazin @Jochen Glueck Yes, this is correct. However, I have strong doubts on the existence of a non-trivial $X$, so I mention what I qualified as a crude answer. Note that you can be logarithmically close to $L^2$, playing a bit with the second integral in the series of inequalities. Also you may bluntly assume some condition of support for the Fourier transform of $u$, providing a closed space of very regular functions (in that case, you are done with the first integral, with a proper choice of $\lambda$).
May 30, 2018 at 14:55 comment added Jochen Glueck But actually the space $H^s$ is not a closed subspace of $L^2$; it is another Banach space which is continuously embedded in $L^2$, so this is a somewhat different setting, isn't it?
May 30, 2018 at 13:16 history edited Bazin CC BY-SA 4.0
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May 30, 2018 at 11:46 history answered Bazin CC BY-SA 4.0