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Sep 19, 2020 at 14:28 comment added an_ordinary_mathematician So that, combined with the observation about the Herglotz transform above is a quite roundabout proof that the Hilbert transform is unbounded on $L^1(\mathbb{T})$ !
Sep 19, 2020 at 14:25 history bounty ended an_ordinary_mathematician
Sep 19, 2020 at 14:16 comment added fedja @an_ordinary_mathematician The square root of a function with non-negative real part has non-negative real part, doesn't it? $|\Im g|\le|\Re g|$ is insufficient because it isn't preserved under weak limits. $\sqrt 2$ is there, of course, but it is just absorbed into $C$.
Sep 19, 2020 at 9:12 comment added an_ordinary_mathematician Thanks a lot, this is very nice ! Doing the calculations I don't see whe $\Re g$ is positive, I get instead $| \Im g| \leq |\Re g | $ and $\sqrt{2} |\Re g| \geq f$. Probably I am missing something.
Sep 19, 2020 at 8:54 vote accept an_ordinary_mathematician
Sep 19, 2020 at 0:15 history edited fedja CC BY-SA 4.0
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Sep 18, 2020 at 18:54 history answered fedja CC BY-SA 4.0