Let $1/2 < s<1$. I got a question about the fractional Sobolev space $H^s(\mathbb R^n)$.
It is well known that, if $2s>n$, then $H^s(\mathbb R^n)$ is continuously embedded in $L^\infty(\mathbb R^n)$, namely $$\|u\|_{L^\infty(\mathbb R^n)}\le c \|u\|_{H^s(\mathbb R^n)},$$ for $c$ depending only on $n, s$. This result comes from Theorem 8.2 in the Hitchhiker’s guide to the fractional Sobolev spaces.
Also, again if $s>2n$ one has that $H^s(\mathbb R^n)$ id continuously embedded in $L^2(\mathbb R^n)$, namely $$\|u\|_{L^2(\mathbb R^n)}\le c_2 \|u\|_{H^s(\mathbb R^n)},$$ for $c_2$ depending only on $n, s$. Is that result true? How one can deduce that? The Hitchhiker’s guide to the fractional Sobolev spaces provides this result only for $2s<n$, but many people use it also when $2s>n$ (see e.g. the first answer to If $u \in H^s(\mathbb{R}^n)$ for $s > n/2$, then $u \in L^\infty(\mathbb{R}^n)$?).
Anyone could please help with this thing?
For all $s\geq 0$, we have $H^s(\Bbb{R}^n)$ continuously embedded into $L^2(\Bbb{R}^n)$, with constant $c_2=1$. This is pretty obvious from the Fourier-transform definition of everything, coming down to the trivial algebraic inequality that $1\leq (1+|\xi|^2)^{s/2}$ for all $s\geq 0$ and $\xi\in\Bbb{R}^n$.
If you restrict yourself to integer $s$, this is obvious even from the physical space definition since $H^s$ is the space of functions such that the function and all (weak) partials up to order $s$ belong to $L^2$, so almost by definition, $H^s\subset L^2$.
Sobolev’s inequality however is a non-trivial result which tells us that if you’re in a sufficiently high-order $L^2$-based Sobolev space $H^s$, then you can ‘trade in’ some of that regularity to give you extra ‘integrability’, i.e if you sacrifice some of the $s$, then you can increase the $L^p$ space to which you belong.