# showing $W^{s,p}(\mathbb{R}^n) \subseteq W^{r,p}(\mathbb{R}^n)$ for $r < s$

I'm aware of a way of doing this using pseudo-differential operator theory. One can easily reduce to showing that $W^{t,p}(\mathbb{R}^n) \subseteq L^p$ for $t > 0$. This in turn follows because the identity map is a symbol operator of order zero, and hence also of order $t$, and so will map $W^{t,p}(\mathbb{R}^n)$ to $L^p$ (and also isomorphically to itself), by a general boundedness result for symbol operators.

I was wondering if anyone knows a more direct proof which doesn't rely on the general (and nontrivial) boundedness result for symbol operators.

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Isn't this just Holder? – JSchlather Mar 19 '13 at 6:29
Where would you use Holder? I want to control $||\varphi||_{L^p}$ by $||\mathcal{F}^{-1}(1 + |\xi|^2)^{t/2}\mathcal{F}\varphi||_{L^p}$; I don't see an obvious application of Holder's inequality. – Jens Mar 19 '13 at 6:36

Thanks for the link, Tomas. I didn't specify this in the original post, but I was thinking of $W^{s,p}$ as being defined via the Fourier transform as $$W^{s,p}(\mathbb{R}^n) = \{\varphi \in \mathcal{S}^\prime : \mathcal{F}^{-1}(1 + |\xi|^2)^{s/2}\mathcal{F}\varphi \in L^p\}$$ with norm $||\varphi||_{W^{s,p}} = ||\mathcal{F}^{-1}(1 + |\xi|^2)^{s/2}\mathcal{F}\varphi||_{L^p}$. The definition in the link you posted is of course a bit different. (continued below) – Jens Mar 19 '13 at 22:48
I am not good with Fourier transform, but I remember that in case $p=2$, we have that $\|\mathcal{F}^{-1} u\|_2=\|u\|_2$, hence in the case $p=2$, we have that $(1+|\xi|^2)^{\frac{s}{2}}\mathcal{F}\phi\in L^2$. If $s>0$, then $(1+|\xi|^2)^{\frac{s}{2}}\geq 1$, therefore $\mathcal{F}\phi\in L^2$ which implies that $\phi\in L^2$. My guess is that if $u\in L^p$, then $\mathcal{F}u \in L^q$, where $\frac{1}{p}+\frac{1}{q}=1$. If this is true then you can apply the same argument to prove that $\phi\in L ^p$ if $s>0$. – Tomás Mar 20 '13 at 0:24
If $1 \leq p \leq 2$ and $\phi \in L^p$ then $\mathcal{F}\phi \in L^{p^\prime}$ (where $\frac{1}{p} + \frac{1}{p^\prime} = 1$) and $||\mathcal{F}\phi||_{p^\prime} \leq ||\phi||_{p}$ (known as the Hausdorff-Young inequality, and proven, for example, by interpolation between the well-known $L^\infty - L^1$ and $L^2 - L^2$ bounds). However we don't get the other direction, i.e. $||\mathcal{F}\phi||_{p^\prime}$ is not bounded below by (a constant times) $||\phi||_{p}$ unless $p = 2$. – Jens Mar 20 '13 at 1:51
To clarify, I should have said that the only value of $p$ $\textbf{between 1 and 2}$ for which $||\phi||_p \leq C||\mathcal{F}\phi||_{p^\prime}$ holds is $p = 2$ (in fact it holds for all $2 \leq p \leq \infty$, with $C = 1$, which follows from interchanging $p$ and $p^\prime$ in the Hausdorff-Young theorem, and replacing each occurrence of $\phi$ with $\mathcal{F}\phi$). – Jens Mar 20 '13 at 2:01