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The answer is negative anyway. Take $\mathbb R=(-\infty,a)\cup(-a,+\infty)$ with small $a>0$. Take $f=e^{-|x|}$. Then $\widehat f(y)\approx \frac 1{1+y^2}$. Now take $s=3-\delta$. $\|f\|_{H^s(\mathbb …

Maybe I'm missing something but it looks to me that is false in general. Take $f(x,y)=\max[(1+y)(1-Ax)_+,A^{-1}(1-x)]$ with big $A>0$. Then the LHS is about $A^{-1}$ while the RHS is about $A^{-2}$.

I'll try to keep your notation except for three things: I prefer to have all parameters non-negative not to get confused myself, I'd rather have $z=(x,y)\in\mathbb R^2$ coordinates on the plane to avo …

It looks like you are totally new to such estimates, so let me show you the old $\varepsilon$ vs $C_\varepsilon$ trick. If you need to estimate some product $xy$, you can write $|xy|\le |x|^2+|y|^2$, …

Just use the Wirtinger's inequality that says that if $u=0$ at the center of an interval $I\subset \mathbb R$, then $\int_I|u|^2\le (|I|/\pi)^2\int_I|u'|^2$. Thus, if the dimension is $>n$, we can cre …

You can do it the way described in the previous answer, no question, but technically, once you knew the result about $D_1D_2$ and $D_1^2,D_2^2$ and also knew that the support does not matter as long a …