Can we say $\| f\|_{\dot H^s(\mathbb R^n)} \le \| f \|_{\dot H^q(\mathbb R^n)}$ if $s\le q$?

Question

If $s\le q$, then can we say that $$ \| f\|_{\dot H^s(\mathbb R^n)} \le \| f \|_{\dot H^q(\mathbb R^n)} $$ holds? Here the homogeneous Sobolev seminorm $\|f\|_{\dot H^s(\mathbb R^n)} = \|(-\Delta)^{s/2}f\|_{L^2(\mathbb R^n)}$, where $\Delta$ is the Laplacian in $\mathbb R^n$.

Answer 1

No, in general this is false. Consider as an elementary example $s=0$, $q=2$ (so that we don't have to deal with fractional derivatives or anything like this). We are inspired by the following observation: if $f(x)=C$ is a nonzero constant function, then $\|f\|_{\dot{H}^2} = \|-\Delta f\|_{L^2} = 0$, while naively "$\|f\|_{L^2} = +\infty$". There are scare quotes here for a reason; typically (as @user190080 mentioned) we define $\dot{H}^s$ as the closure in the $\dot{H}^s$ norm of smooth test functions (which have compact support).

So instead we argue by scaling. Let $\phi$ be any smooth test function on $\mathbb{R}^n$, and define $\phi_r(x) = \phi(rx)$. Then $$ \|\phi_r\|_{L^2(\mathbb{R}^n)} = r^{-n/2} \|\phi\|_{L^2(\mathbb{R}^n)}, $$ while $$ \|\phi_r\|_{\dot{H}^2(\mathbb{R}^n)} = r^{2-n/2}\|\phi\|_{\dot{H}^2(\mathbb{R}^n)}. $$ Sending $r\to 0$, we see that this contradicts the inequality you want.

Two more comments. An intuitive way to test out these relationships is to use localized wavepackets as examples. See Terry Tao's answer at https://mathoverflow.net/questions/17736/way-to-memorize-relations-between-the-sobolev-spaces for an explanation of how to do this.

Second, there are spaces where the inequality you want is true. In particular this works on hyperbolic space, or any space with infinite volume and a spectral gap. The reason I mention this is to point out that the scaling argument really is needed to make the example rigorous, and the simplistic example of a constant function can lead you astray.