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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.

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  • $\begingroup$ I follow this because sooner or later I’d have asked this, too $\endgroup$
    – tommy1996q
    Commented Dec 21, 2019 at 0:06
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    $\begingroup$ Related, hope this helps. $\endgroup$ Commented Dec 22, 2019 at 16:30
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    $\begingroup$ You might find interesting the description of these spaces of the book by Lieb and Loss. They speak of "D^1" and "D^{1/2}" spaces. They are defined as those spaces of functions that vanish at infinity (in an appropriate sense) and that satisfy $\|\nabla f\|_{L^2}<\infty$ and, respectively, $\|D^{1/2} f\|_{L^2}<\infty$. $\endgroup$ Commented Dec 22, 2019 at 18:06
  • $\begingroup$ Yes, related to this, math.stackexchange.com/questions/4584970/… $\endgroup$
    – LL 3.14
    Commented Nov 6, 2023 at 16:21

2 Answers 2

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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.


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.

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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.

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  • $\begingroup$ Thanks for the answer! My confusion that constant functions are not in $\dot H^s$ comes from the homogeneous Sobolev embedding, which states that $\|f\|_{L^p} \leq C\|f\|_{\dot H^s}$ (for certain $s,p$). But if $f$ is the constant one function, then this cannot be true, since the left hand side is infinite while the right hand side would be finite? $\endgroup$
    – Chris
    Commented Dec 22, 2019 at 16:49
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    $\begingroup$ Exactly, that's why that inequality does not hold in general. You have to assume, for instance, that $f$ is compactly supported. Notice that all of this is assuming your domain is the whole space $\mathbb{R}^n$. If you consider a bounded open set $\Omega$ you can obtain different hypotheses on $f$ in order to satisfy your inequality. I strongly recommend you to check chapter 5 of Evans' book $\endgroup$
    – Sharik
    Commented Dec 22, 2019 at 16:57
  • $\begingroup$ Sorry to add this later, but I've added an edit to my question - so this homogeneous-type Sobolev embedding requires additional assumptions on the functions (like you mentioned in this last comment)? $\endgroup$
    – Chris
    Commented Dec 22, 2019 at 16:59
  • $\begingroup$ The answer to your second question is exactly the same as the first one. Look at the proof of those theorems, it is crucial to assume that $f$ has a compact support. They do not hold in general. You are not missing anything, you are just trying to apply the theorem without the hypothesis. $\endgroup$
    – Sharik
    Commented Dec 22, 2019 at 18:19

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