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Willie Wong
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The key computation is the commutator $$ \Lambda^s (xf) - x \Lambda^s f $$ You can check this "classically" in the case $s = 4$ to find $$ (1 - \Delta)^2 (xf) = x (1-\Delta)^2 f - 4 (1-\Delta) f'$$ which differs from was found in the paper (their computation dropped the final commutator term).

In fact, your computations imply the following result $$ \Lambda^s(xf) = x\Lambda^s f - s \Lambda^{s-2} f' $$ from which you'd get (returning to the author's original term) $$ \int \Lambda^s f \Lambda^s(xf') = \int \Lambda^s f\Lambda^s(xf)' - \Lambda^s f \Lambda^s f = - \int \Lambda^s f' \Lambda^s (xf) - \int (\Lambda^s f)^2 $$ $$ = - \frac12 \int x [(\Lambda^s f)^2]' + s \int (\Lambda^{s-1} f')^2 - \int (\Lambda^s f)^2 = s \int (\Lambda^{s-1} f')^2 - \frac12 \int (\Lambda^s f)^2 $$

(The fact that the extra term can be written as something with a sign means that this cannot just disappear due to some strange cancellations.)

This may have an impact on the main results of the paper; I haven't looked too carefully, but it appears that in deriving the Bernoulli type differential inequality after equation (3.21), the authors relied on this (and similar terms) to be signed to drop them from consideration. The corrected expression adds a term with the wrong sign. There may be a way to absorb this, but it is not immediately obvious.

Willie Wong
  • 39.1k
  • 4
  • 94
  • 176