Why Embeddings Matter
Smoothness classes describe related ways of controlling how a function changes, but their definitions are not interchangeable. An embedding says that every function in one class belongs to another, with a norm bound. It does not usually say that the two classes contain exactly the same functions or that their norms are equivalent. This distinction matters when a statistical result is proved for one class but the assumed regression function is described using another.
We work on a bounded Lipschitz domain \(\Omega\subset\mathbb{R}^d\). The domain assumption is important: it supports bounded extension operators for the classes below and makes finite-measure comparisons between norms possible. Constants in embeddings may depend on \(\Omega\), the dimension, and the indices, but not on the function.
Classes and the Meaning of an Embedding
For a nonnegative integer \(m\), \(W^{m,p}(\Omega)\), \(1\le p\le\infty\), is the Sobolev space of functions whose weak derivatives through order \(m\) belong to \(L^p(\Omega)\). Its norm combines the \(L^p\) norms of those derivatives. For \(0<s<1\) and \(1\le p<\infty\), the fractional Sobolev space \(W^{s,p}(\Omega)\) has norm given by the \(L^p\) norm plus the Gagliardo seminorm
For \(k\) a nonnegative integer and \(0<\alpha<1\), the Hölder space \(C^{k,\alpha}(\Omega)\) consists of functions with continuous derivatives through order \(k\), whose derivatives of order \(k\) satisfy a uniform \(\alpha\)-Hölder condition. In particular, for \(k=0\), this condition is \(|f(x)-f(y)|\le C|x-y|^\alpha\).
To specify Besov spaces, for \(h\in\mathbb{R}^d\) define the \(r\)-th difference by \(\Delta_h^r f(x)=\sum_{j=0}^r(-1)^{r-j}\binom{r}{j}f(x+jh)\). On \(\mathbb{R}^d\), write \(\omega_r(f,t)_p=\sup_{|h|\le t}\|\Delta_h^r f\|_p\). Choose an integer \(r>s>0\). The Besov norm is equivalent to
with the integral term replaced by \(\sup_{0<t\le1}t^{-s}\omega_r(f,t)_p\) when \(q=\infty\). On \(\Omega\), define the class by restriction from \(\mathbb{R}^d\), with norm the infimum of the extension norms. Different admissible choices of \(r>s\) give equivalent norms. A continuous embedding \(A\hookrightarrow B\) means \(A\subseteq B\) and \(\|f\|_B\le C\|f\|_A\) for a fixed constant \(C\).
Hölder and Besov: The Noninteger Match
For \(k\ge0\) and \(0<\alpha<1\), the classical Hölder–Besov identification on a suitable extension domain is
The difference definition explains the direction from Hölder to Besov: the Hölder bound controls differences at scale \(t\) by a constant times \(t^{k+\alpha}\), so the Besov supremum term is bounded. The converse identification recovers the corresponding Hölder regularity from difference bounds. At integer smoothness, the endpoint \(B_{\infty,\infty}^{m}\) is generally a Zygmund class, not simply \(C^m\); replacing the noninteger statement by an integer-order equivalence is therefore unsafe.
On a bounded domain, finite measure also gives \(L^\infty(\Omega)\hookrightarrow L^p(\Omega)\) for finite \(p\). In particular, the noninteger Hölder class embeds continuously into \(B_{p,\infty}^{k+\alpha}(\Omega)\). This is a one-way inclusion: finite-\(p\) Besov control does not in general give a uniform Hölder bound.
Worked Example: A Hölder Function in a Finite-\(p\) Besov Class
Let \(\Omega=[0,1]\), \(f(x)=x^{3/2}\), and consider \(C^{1,1/2}([0,1])\). We verify the relevant regularity. The derivative is \(f'(x)=\tfrac32\sqrt{x}\), and
The inequality follows by squaring \(|\sqrt{x}-\sqrt{y}|\le\sqrt{|x-y|}\). Also, \(\|f\|_\infty=1\) and \(\|f'\|_\infty=3/2\), so the \(C^{1,1/2}\) norm is finite. The Hölder–Besov identification gives \(f\in B_{\infty,\infty}^{3/2}\). Since \([0,1]\) has measure one, every \(g\) on this interval satisfies \(\|g\|_2\le\|g\|_\infty\); applying this bound to the differences yields \(f\in B_{2,\infty}^{3/2}\). The argument establishes containment, not equivalence of the \(B_{\infty,\infty}^{3/2}\) and \(B_{2,\infty}^{3/2}\) norms.
Besov Index Inclusions
Two useful inclusions follow from the Besov norm. First, at fixed \(s\) and \(p\), a smaller summability index gives a smaller class:
To see why, the Besov norm has an equivalent dyadic form consisting of the \(L^p\) norm and the \(\ell^q\) norm of the scale sequence \(2^{js}\omega_r(f,2^{-j})_p\), \(j=0,1,\ldots\). For sequences, \(\|a\|_{\ell^{q_2}}\le\|a\|_{\ell^{q_1}}\) when \(q_1\le q_2\): if \(q_1<\infty\), each \(|a_j|\le\|a\|_{\ell^{q_1}}\), and summing \(|a_j|^{q_2}\le\|a\|_{\ell^{q_1}}^{q_2-q_1}|a_j|^{q_1}\) gives the claim (with the equal-index case immediate). The case \(q_2=\infty\) follows directly from \(\sup_j|a_j|\le\|a\|_{\ell^{q_1}}\). Applying this inequality to the scale sequence proves the embedding.
Second, on a bounded Lipschitz domain, a Besov class embeds into a class of strictly lower smoothness:
For the scale sequence, the \(s_2\)-weighted term is \(2^{-j(s_1-s_2)}\) times the \(s_1\)-weighted term. The geometric factor is summable to every positive finite power and bounded, so the sequence belongs to every \(\ell^{q_2}\) when the \(s_1\)-weighted sequence is bounded, as it is for every \(q_1\). The \(L^p\) term is unchanged. This proves the stated inclusion. It does not imply an embedding in the reverse direction.
Worked Example: Changing the Besov Summability Index
Suppose \(f\in B_{2,1}^{1/2}(\Omega)\) on a bounded Lipschitz domain. Since \(1\le2\), the summability embedding gives \(f\in B_{2,2}^{1/2}(\Omega)\) and
This conclusion keeps both \(p=2\) and smoothness \(s=1/2\) fixed; only the scale-summability index changes. If instead we know \(f\in B_{2,1}^{3/4}(\Omega)\), the loss-of-smoothness embedding gives \(f\in B_{2,5}^{1/2}(\Omega)\), because \(3/4>1/2\). The strict inequality in smoothness is what allows the target summability index \(5\) here.
Sobolev Classes and Besov Classes
For integer \(m\ge1\), an elementary difference estimate gives a useful inclusion without asserting an unjustified equivalence of norms:
Here is the estimate. Take an extension \(F\in W^{m,p}(\mathbb{R}^d)\) of \(f\), and choose an integer \(r>m\). For smooth \(F\), repeated use of the fundamental theorem of calculus gives
where \(D_h\) is differentiation in the direction \(h/|h|\) and the constant includes the directional derivative bound. Since \(\Delta_h^r=\Delta_h^{r-m}\Delta_h^m\), and each difference operator has \(L^p\) operator norm at most \(2\), we obtain
The same inequality holds for Sobolev functions by approximation in \(W^{m,p}\) (or the corresponding translation identities). Taking the supremum over \(|h|\le t\) bounds \(t^{-m}\omega_r(F,t)_p\) by a constant times \(\|F\|_{W^{m,p}}\). The \(L^p\) term is also controlled, so the Besov norm is bounded by the Sobolev norm. Taking the infimum over extensions proves the domain inclusion. \(\square\)
For fractional orders \(0<s<1\) and \(1\le p<\infty\), the Gagliardo seminorm is the \(p\)-integral of first differences against the scale weight. On extension domains this gives the standard norm equivalence \(W^{s,p}(\Omega)=B_{p,p}^s(\Omega)\), with equivalent norms. This is stronger than a mere embedding, but it concerns this fractional Sobolev convention and this range of indices; it should not be transferred automatically to integer orders or other definitions of Sobolev spaces.
Worked Example: An Integer Sobolev Function
Take \(f(x)=x^2\) on \([0,1]\). Its weak derivatives are its ordinary derivatives: \(f'(x)=2x\) and \(f''(x)=2\). Thus
The integer Sobolev embedding therefore places \(f\) in \(B_{2,\infty}^{1}([0,1])\), with its Besov norm bounded by a constant times \(\sqrt{23/15}\). This conclusion follows for the whole space \(W^{1,2}\), not just this polynomial; the calculation verifies that the example satisfies the theorem's hypothesis.
Worked Example: Checking a Fractional Sobolev Seminorm
Let \(f(x)=x\) on \([0,1]\). For \(s=1/2\) and \(p=2\), the Gagliardo seminorm is directly calculable:
The \(L^2\) norm squared is \(\int_0^1x^2\,dx=1/3\), so the full norm squared is \(4/3\). The fractional Sobolev–Besov equivalence gives \(f\in B_{2,2}^{1/2}([0,1])\) with equivalent norm. It also gives \(B_{2,2}^{1/2}\hookrightarrow B_{2,\infty}^{1/2}\) by the summability embedding, but does not claim equivalence with that latter Besov norm.
Common Mistakes and Why the Distinctions Matter
- Reversing an embedding. From \(A\hookrightarrow B\), one may conclude that every member of \(A\) is in \(B\), not that every member of \(B\) is in \(A\).
- Calling a continuous inclusion an equivalence. Equivalence requires the same set and two-sided norm bounds. The integer-order result above supplies only \(W^{m,p}\hookrightarrow B_{p,\infty}^m\).
- Ignoring endpoints. The Hölder–Besov identification stated here uses noninteger smoothness \(k+\alpha\), \(0<\alpha<1\). At integer smoothness the \(B_{\infty,\infty}^m\) endpoint has a different characterization.
- Changing indices without checking them. At equal smoothness and \(p\), the direct summability inclusion requires \(q_1\le q_2\). A strict smoothness loss permits more flexibility, but does not justify an arbitrary same-smoothness change.
- Dropping the domain condition. Finite-measure norm comparisons and extension arguments used here rely on the stated bounded-domain setting.
These embeddings let an analyst translate regularity assumptions while preserving the direction of every implication. In nonparametric regression, that can identify which established approximation or risk bound applies to a function class, but the inclusion alone does not say that the resulting bound is sharp for both classes.
Check Your Understanding
For each question, state the relevant indices and the direction of any claimed inclusion.
- Why does \(B_{p,1}^s(\Omega)\hookrightarrow B_{p,2}^s(\Omega)\) hold, and what index is unchanged?
- What restriction on smoothness is needed to infer \(B_{p,q_1}^{s_1}(\Omega)\hookrightarrow B_{p,q_2}^{s_2}(\Omega)\) for arbitrary \(q_1,q_2\)?
- What Besov class is guaranteed by the integer-order inclusion for \(W^{m,p}(\Omega)\)? Does that result assert norm equivalence?
- For which fractional Sobolev range stated here is \(W^{s,p}(\Omega)\) norm-equivalent to \(B_{p,p}^s(\Omega)\)?
- Why should the identification between \(C^{k,\alpha}\) and \(B_{\infty,\infty}^{k+\alpha}\) not be stated unchanged at integer smoothness?