How to think of homogeneous Sobolev spaces

Question

The homogeneous Sobolev space $\dot H^s(\mathbb{R}^d)$ can be defined as the completion of $\mathscr{S}(\mathbb{R}^d)$ (the space of Schwartz functions) under the norm $$ \|f\|_{\dot H^s(\mathbb{R}^d)} = \||\xi|^s\hat{f}\|_{\mathbb{R}^d}. $$ The corresponding homogeneous space can be defined as the closure of Schwartz functions under the norm $$ \|f\|_{H^s(\mathbb{R}^d)} = \|\langle\xi\rangle^s\hat{f}\|_{\mathbb{R}^d}. $$ The inhomogeneous space corresponds with its classical counterpart in the case $s = k \in \mathbb{Z}$, so that roughly it is the space of functions $f$ so that $f$ and its derivatives up to and including order $k$ are in $L^2$.

My question: Intuitively, is the inhomogeneous space detecting only top-order blowup/bad behavior/lack of sufficient decay? If so, how bad is lower-order bad behavior allowed to be? For instance, why is the constant 1 function not in the homogeneous space with $s = 1$, since its derivative is (very) integrable? I've seen it stated that membership in $\dot H^s$ is equivalent to having $f \in L^2$ and $\partial^\alpha f \in L^2$ for all $|\alpha| = s$ when $s$ is an integer. Is this true? (This would imply $\dot H^1 = H^1$?) And if so, is there a good reference for things like this?

Any general intuition for what the homogeneous spaces are capturing is much appreciated, as well. Thanks in advance.

Some more (not required) background/motivation for my question: My confusion stems from the homogeneous Sobolev embedding: for $1 < p < q < \infty$, if $s > 0$ and $q^{-1} = p^{-1} - s/d$, then there exists a constant $C$ such that $$ \|f\|_{L^q(\mathbb{R}^d)} \leq C\|f\|_{\dot W^{s, p}(\mathbb{R}^d)} $$ for any $f \in \dot W^{s, p}(\mathbb{R}^d)$. However, if (say) the constant 1 function were in $\dot H^1$, then we would get for some $q < \infty$ the statement $\|1\|_{L^q(\mathbb{R}^d)} < \infty$, which is not true. Furthermore, if the $\dot H^1$ norm is measuring the size of the gradient, then the right hand side should actually be zero; is this the case?

In Evans, for instance, whenever a Sobolev-type inequality is put forth bounding an $L^q$ norm of $f$ in terms of an $L^p$ norm of its derivative, he either always takes the domain to be bounded, or states that the inequality holds only for $f$ with compact support (both of these conditions rule out the constant 1 function being a counterexample). The fact that this homogeneous Sobolev embedding exists on all of $\mathbb{R}^d$ suggests to me that maybe constant functions are not in the homogeneous spaces, or that I'm missing some hypothesis. Either way, any clarification would be much appreciated.

Answer 1

The definitions are subtle here. You have defined $\dot H^1(\Bbb R^n)$ as the completion of $\cal S(\Bbb R^n)$ under the associated norm. Now we also know that for all $f \in \cal S(\Bbb R^n)$ we do have the Sobolev embedding theorem $$\lVert f \rVert_{L^q(\Bbb R^n)} \leq C \lVert f \rVert_{\dot H^1(\Bbb R^n)}, \tag{G-N} $$ where $q = \frac{2n}{n-2}$ and we assume $n \geq 3.$ Hence as $\cal S(\Bbb R^n)$ is dense in $\dot H^1(\Bbb R^n)$ by definition, the above inequality extends to all $f \in \dot H^1(\Bbb R^n).$ In particular, we see that $1 \not\in \dot H^1(\Bbb R^n).$

However, one can alternatively define the homogeneous Sobolev space $\dot V^1(\Bbb R^n)$ (non-standard notation!) as the space of tempered distributions $f \in \cal S'(\Bbb R^n)$ such that $\hat f$ is represented by a locally integrable function such that $$ \int_{\Bbb R^n} |\xi|^2 |\hat f(\xi)|^2 \,\mathrm{d} \xi < \infty.$$ Under the definition we see that $1 \in \dot V^1(\Bbb R^n),$ and the embedding result $(\text{G-N})$ fails in general for $f \in \dot V^1(\Bbb R^n).$

In general I've seen both of the above definitions used for homogeneous Sobolev spaces, as one may be more useful in a given context over the other. As such it's always important to refer back to the definition, and be wary of how things may differ.

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As it happens, in this special case the constant functions are the only obstruction. More precisely we have $$\dot H^1(\Bbb R^n) \cong \dot V^1(\Bbb R^n) / \{\text{ constant functions} \},$$ where we identify two functions which differ by a constant multiple of $1.$ An elementary proof of this result is presented in Theorem 2.1 of this paper by Ortner & Süli.

Also for the case of general domains, there is a mountain of literature on this subject. For instance you can refer to the first chapter in Sobolev Spaces by Maz'ya for a detailed discussion of related matters in the context of bounded domains.

Answer 2

The answer to your first question is yes. Nevertheless, I think you are getting confused with the definitions. Notice that constant functions do belong to $\dot{H}^1$, actually, they belong to any $\dot{H}^s$ for any integer $s\geq 1$. On the other hand, for functions $f:\mathbb{R}^n\to\mathbb{R}$ in order to belong to $\dot{H}^s$ it is not mandatory to satisfy $f\in L^2$, they only need to satisfy $\partial^\alpha f\in L^2$ for $\vert \alpha\vert=s$. Actually, by standard Fourier analysis, you can prove that if a function satisfies $f\in L^2$ and $\partial^\alpha f\in L^2$ for every $\vert \alpha\vert=s$, then $f\in H^s$ (use your norm definition in terms of $\xi$). Therefore, there is an inclusion, but they certainly are not equivalent.

Notice also that polynomials of order $s-1$ belong to $\dot{H}^s$, so yes, $\dot{H}^s$ only see regularity at the top-order level.