Timeline for Sobolev embedding for a specific family of weighted Sobolev spaces

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Oct 5, 2017 at 20:28	comment	added	Christian Remling		These $f$ are locally absolutely continuous, so $f(x)=f(1)-\int_x^1 f'(t)\, dt$, and if now $f'\in L^1(0,1)$ (as it will be when $\alpha<1/2$, by CS), then it follows that $f$ is bounded (in fact, continuous on $[0,\infty)$).
Oct 5, 2017 at 19:18	comment	added	Delio Mugnolo		@ChristianRemling This is true: one certainly has $f\in W^{1,1}_{loc}$ and, as you observe, even a bit more for certain ranges of $\alpha$. But how does this help to say anything, near 0, about $f$ (rather than $f'$)? Can you please elaborate on your "of course"? Are you thinking of some interpolation inequality between $L^2(0,\infty)$ and $\{f\in L^2(0,\infty):f'\in L^1(0,\infty)$?
Oct 4, 2017 at 18:44	comment	added	Christian Remling		If $\alpha<1/2$, then $f'\in L^1(0,1)$, so there are no problems near $x=0$ (and weaker assumptions would suffice of course to get $f\in L^p$), and for large $x$ we never have any problems since $f,f'\in L^2$ already imply that $f\in L^p$ for all $p\ge 2$.
Oct 4, 2017 at 9:08	history	asked	Delio Mugnolo	CC BY-SA 3.0