As a family, cusps demonstrate that a domain can be \(L^{s}\)-averaging for some *s* and not others. The cusps analyzed here were explored in [8]. We confirm those results using essential tubes and generalized Whitney subdivision.

### Theorem 5.1

*For* \(\alpha \geq 0\), *let* \(\Omega _{\alpha} \in \mathbb{R}^{n}\) *be the following domain*:

$$ \Omega _{\alpha} = \bigl\{ (x,y_{1}, y_{2}, \ldots , y_{n-1}): 0 < x < 1, \bigl(y_{1}^{2} + y_{2}^{2} + \cdots + y_{n-1}^{2} \bigr)^{\frac{1}{2}} < x^{\alpha}\bigr\} . $$

*Then*, \(\Omega _{\alpha}\) *is an* \(L^{s}\)-*averaging domain if and only if*

$$ (\alpha - 1) (s - n + 1) < n. $$

Note that if \(0 \leq \alpha \leq 1\) then \(\Omega _{\alpha}\) is a John domain and hence \(L^{s}\)-averaging for all \(s \geq 1\). At the same time, it is straightforward to check that the inequality is satisfied. With these observations in mind, we restrict to \(\alpha > 1\) for the remainder of this section.

The proof is broken into parts. In the first part, essential tubes are used to show that a cusp cannot be \(L^{s}\)-averaging if the given inequality is not satisfied. In the second part, a generalized Whitney subdivision is used to show the converse.

### Proof (part one)

Let \(0 < a < b < 1\) and consider the tube centered along the *x*-axis with ends at *a* and *b*, and its radius equal to \(b^{\alpha}\). This is an essential tube with \(r = b^{\alpha}\), \(l = b-a\), and \(c = (\frac{a}{b} )^{\alpha n}\).

Using the sequence \(a_{j} = \frac{1}{2^{j}}\), consider the essential tubes \(T_{j}\) defined as above using \(a = a_{j}\) and \(b = a_{j-1}\) Then,

$$\begin{aligned} E_{T_{j}} &= \sum_{j = 3}^{\infty} \biggl(\frac{a_{j}}{a_{j-1}} \biggr)^{\alpha n} (a_{j-1})^{\alpha n} \biggl( \frac{a_{j-1} - a_{j}}{ (a_{j-1})^{\alpha}} \biggr)^{s+1} \\ &= 2^{-\alpha (s+1)} \sum_{j = 3}^{\infty} \bigl(2^{(\alpha - 1)(s-n+1)-n} \bigr)^{j}. \end{aligned}$$

If \((\alpha -1)(s-n+1) \geq n\) then this series diverges, so by Corollary 3.2, \(\Omega _{\alpha}\) cannot be \(L^{s}\)-averaging. □

For the other direction, we first establish some structure and initial results. Because \(\Omega _{\alpha}\) is symmetric about the *x*-axis, it is beneficial to work in cylindrical coordinates \((x, r, \theta )\). In these coordinates,

$$ \Omega _{\alpha} = \bigl\{ (x, r, \theta ):0 < x < 1, r < x^{\alpha}, \theta \in \mathbb{S}^{n-2}\bigr\} , $$

and the volume element is

$$ dx \,dy_{1} \cdots dy_{n-1} = r^{n-2} \,dx \,dr \,d \theta , $$

where *dθ* is the volume element for the unit \((n-2)\)-sphere \(\mathbb{S}^{n-2}\).

Given \(j,m \in \mathbb{Z}^{+}\), let

$$ S_{j,m} = \biggl\{ (x, r, \theta ): \frac{m}{2^{j+\ell}} \leq x \leq \frac{m+1}{2^{j+\ell}}, \biggl(1 - \frac{1}{2^{\ell -1}} \biggr) x^{ \alpha} \leq r \leq \biggl(1 - \frac{1}{2^{\ell}} \biggr) x^{\alpha} \biggr\} , $$

where \(\ell = \lfloor \log _{2}(m)\rfloor + 1\). See Fig. 6.

These sets are created by first dividing the domain into disks indexed by *j*, then layers indexed by *ℓ*, and finally further subdividing into the sets described. Since, in the third step, each layer is subdivided into twice as many sets as the previous layer, the number of digits in the base-2 representation of *m* is the layer *ℓ* so that \(m < 2^{\ell} \leq 2m\).

This subdivision is not quite a generalized Whitney subdivision for two different reasons. First, the union of all of the \(S_{j,m}\) misses a significant portion of Ω. We define \(S_{0} = \Omega _{\alpha} \cap \{(x, r, \theta ): x > \frac{1}{4} \}\). This set is a John domain and we do not attempt to subdivide it.

Secondly, when \(m > 1\), the sets \(S_{j,m}\) are not star-shaped. When \(n = 2\), each \(S_{j,m}\) is the disjoint union of two sets, one lying above the *x*-axis and one lying below. To resolve this, for \(m > 1\) let \(S_{j,m}^{+} = S_{j,m} \cap \{y_{1} > 0\}\) and let \(S_{j,m}^{-} = S_{j,m} \cap \{y_{1} < 0\}\).

When \(n > 2\), each \(S_{j,m}\) has a solid ring shape, or disk-like shape if \(m = 1\). To formally make use of the observations about diameter in the previous section, we could further subdivide each \(S_{j,m}\) into star-shaped regions through some subdivision of \(\mathbb{S}^{n-2}\). However, we will find that because of the choice of the path, each such region would contribute the same, so we keep \(S_{j,m}\) as a single set.

To help with calculations later, we have the following:

### Lemma 5.2

*For each* *j*, *m*, *denote the radial thickness and horizontal width of* \(S_{j,m}\) *by* \(d_{r}\) *and* \(d_{x}\), *respectively*. *Then*,

### Proof

The horizontal width is

$$\begin{aligned} d_{x} &= \frac{m+1}{2^{j + \ell}} - \frac{m}{2^{j + \ell}} \\ &= \frac{1}{2^{j + \ell}}. \end{aligned}$$

The radial thickness for a given *x* is

$$\begin{aligned} d_{r}(x) &= \biggl(1 - \frac{1}{2^{\ell}} \biggr)x^{\alpha} - \biggl(1 - \frac{1}{2^{\ell -1}} \biggr)x^{\alpha} \\ &= \frac{x^{\alpha}}{2^{\ell}}, \end{aligned}$$

and this quantity is maximized at the right end of \(S_{j,m}\) at \(x = \frac{m+1}{2^{j+\ell}}\), hence

$$ d_{r} = \frac{ (\frac{m+1}{2^{j+\ell}} )^{\alpha}}{2^{\ell}}. $$

Using the fact that \(m < 2^{\ell}\), the ratio of these distances is

$$\begin{aligned} \frac{d_{r}}{d_{x}} &= \frac{\frac{ (\frac{m+1}{2^{j+\ell}} )^{\alpha}}{2^{\ell}}}{ \frac{1}{2^{j+\ell}}} \\ &< \biggl(\frac{2^{\ell}+1}{2^{\ell}} \biggr)^{\alpha} \frac{1}{2^{(\alpha -1)j}} \\ &< \frac{2^{\alpha}}{2^{(\alpha -1)j}} \\ &< \frac{2^{\alpha}}{2^{\alpha -1}} \\ &= 2. \end{aligned}$$

□

The curves used to estimate the quasihyperbolic distance for \(\Omega _{\alpha}\) will incorporate only horizontal and radial directions. Therefore, since in the previous section, the diameter of the set is used as a proxy for the length of a curve for a bound on \(k_{\Omega}\), we can restrict attention to \(d_{r}\) and \(d_{x}\), and in light of the previous lemma, we may use \(d_{x}(S_{j,m})\) in place of \(d(S_{j,m})\) and we have

$$ d_{x}(S_{j,m}) = \frac{1}{2^{j+\ell}}. $$

Next, we focus on distance to the boundary.

### Lemma 5.3

*For each set* \(S_{j,m}\),

$$ \delta (S_{j,m}) \geq C(\alpha ) \frac{1}{2^{\alpha j+\ell}}. $$

### Proof

We first consider the two-dimensional case. Let \(f(x) = x^{\alpha}\) define the boundary. Let \(z = (x,y) \in \Omega _{\alpha} \cap \{x \leq \frac{1}{2} \}\), and let \(\delta (z)\) be its distance to the boundary. Then, since *f* is convex and increasing, \(\delta (z)\) is at least the distance to the tangent line at \((x, f(x))\), and the distance to this tangent line is bounded below by a multiple of the vertical distance \(x^{\alpha} - y\). This multiple, \(C_{1}(\alpha )\) depends on *α* only, and is realized at \(x = \frac{1}{2}\). The general case is similar, due to rotational symmetry.

Focusing now on \(S_{j,m}\), the points closest to \(\partial (\Omega _{\alpha})\) are \((\frac{m}{2^{j+\ell}}, (1-\frac{1}{2^{\ell}} ) (\frac{m}{2^{j+\ell}} )^{\alpha},\theta )\). If we restrict our attention to just the radial distance we find

$$\begin{aligned} \delta (S_{j,m}) &\geq C_{1}(\alpha ) \biggl[ \biggl( \frac{m}{2^{j+\ell}} \biggr)^{\alpha} - \biggl(1-\frac{1}{2^{\ell}} \biggr) \biggl(\frac{m}{2^{j+\ell}} \biggr)^{\alpha} \biggr] \\ &= C_{1}(\alpha ) \frac{1}{2^{\ell}} \biggl(\frac{m}{2^{j+\ell}} \biggr)^{\alpha} \\ & \geq C_{1}(\alpha ) \frac{1}{2^{\ell}} \biggl( \frac{2^{\ell -1}}{ 2^{j+\ell}} \biggr)^{\alpha} \\ &= C(\alpha ) \frac{1}{2^{\alpha j+\ell}}. \end{aligned}$$

□

With these estimates in hand, we now have the following lemma.

### Lemma 5.4

*Using the basepoint* \(z_{0} = (\frac{1}{3},0,0 )\), *for any point* \(z \in S_{j,m}\),

$$ k(z, z_{0}; \Omega _{\alpha}) \leq C(\alpha ) (1+ \ell )2^{(\alpha -1)j}. $$

### Proof

For any point \(z = (x, r, \theta ) \in S_{j,m}\), \(j \geq 1\), define the L-shaped path \(\gamma :[0,2] \longrightarrow \Omega _{\alpha}\) by

$$ \gamma (t) = \textstyle\begin{cases} (x t + \frac{1}{3}(1-t), 0, 0 )&\text{if } t \in [0,1], \\ (x, r(t-1), \theta )& \text{if } t \in [1,2]. \end{cases} $$

For the first part we only need the horizontal width of each set, and for the second part, we only need the radial thickness.

Now, we can estimate \(k_{\Omega _{\alpha}}\). For the first step, we determine which sets intersect *γ*. Let \((x, r, \theta ) \in S_{j,m}\). For the initial leg from \(z_{0}\) to \((x, 0, 0)\), we use the sets \(S_{i, 1}\) for \(i = 1, \ldots , j\). For the second leg from \((x, 0, 0)\) to \((x, r, \theta )\) we need to determine which sets lie between \(S_{j,m}\) and \(S_{j, 1}\) There is one at each layer out to the layer containing \(S_{j,m}\), and the specific sets \(S_{j, \lambda}\) are determined as follows. Express *m* in binary. Then, the *λ* values are represented in binary by truncating the binary representation of *m* by successively removing the rightmost digit. For example, if \(m = 51\) then the *λ* to use are:

$$\begin{aligned} &51 = 110011_{2}, \\ &25 = 11001_{2}, \\ &12 = 1100_{2}, \\ &6 = 110_{2}, \\ &3 = 11_{2}, \\ &1 = 1_{2}. \end{aligned}$$

Let \(\Lambda (j,m)\) be the set of indices corresponding to these sets lying below \(S_{j,m}\) and note that \(|\Lambda (j,m)| = \ell \). With the specific sets through which *γ* passes known, Lemma 4.1, modified to account for the fact that only the radial or horizontal distances are needed, then Lemmas 5.2 and 5.3 are used to approximate \(k_{\Omega _{\alpha}}\) as follows:

$$\begin{aligned} k(z, z_{0}; \Omega _{\alpha}) &\leq 2\sum _{i=1}^{j} \frac{d_{x}(S_{i,1})}{\delta (S_{i,1})} + 2\sum _{\lambda \in \Lambda (k,m)} \frac{d_{r}(S_{j,\lambda})}{\delta (S_{j,\lambda})} \\ &\leq 2\sum_{i=1}^{j} \frac{d_{x}(S_{i,1})}{\delta (S_{i,1})} + 4 \sum_{\lambda \in \Lambda (k,m)} \frac{d_{x}(S_{j,\lambda})}{\delta (S_{j,\lambda})} \\ &\leq C_{1}(\alpha ) \Biggl(\sum_{i=1}^{j} \frac{2^{\alpha i+1}}{2^{i+1}} + \sum_{\lambda \in \Lambda (j,m)} \frac{2^{\alpha j+\ell}}{2^{j+\ell}} \Biggr) \\ &= C_{1}(\alpha ) \Biggl(\sum_{i=1}^{j} 2^{(\alpha -1) i} + \sum_{ \lambda \in \Lambda (j,m)} 2^{(\alpha -1) j} \Biggr) \\ &\leq C(\alpha ) \bigl( 2^{(\alpha -1)j} + \ell 2^{(\alpha -1) j} \bigr) \\ &= C(\alpha ) (1 + \ell )2^{(\alpha -1) j}, \end{aligned}$$

where the first sum on the third-to-last line is approximated by a constant times the largest term. □

With \(k(z, z_{0}; \Omega _{\alpha})\) approximated, the next step is to estimate the measure of \(S_{j,m}\).

### Lemma 5.5

*For each* \(S_{j,m}\),

$$ \vert S_{j,m} \vert \leq C(\alpha , n) \frac{1}{2^{j[\alpha (n-1)+1]}} \frac{1}{2^{2\ell}}. $$

### Proof

We have

$$ \vert S_{j,m} \vert = \int _{\mathbb{S}^{n-2}} \int _{\frac{m}{2^{j+\ell}}}^{ \frac{m+1}{2^{j+\ell}}} \int _{ (1-\frac{1}{2^{\ell -1}} )x^{ \alpha}}^{ (1-\frac{1}{2^{\ell}} )x^{\alpha}} r^{n-2} \,dr \,dx \,d\theta . $$

The integral over the sphere just produces a dimensional constant. For the other two integrals, the given functions are increasing, and so are approximated by \(\int _{a}^{b} f(x) \,dx \leq f(b)(b-a)\), resulting in

$$\begin{aligned} \vert S_{j,m} \vert &\leq C_{1}(n) \int _{\frac{m}{2^{j+\ell}}}^{ \frac{m+1}{2^{j+\ell}}} \biggl[ \biggl(1-\frac{1}{2^{\ell}} \biggr)x^{ \alpha} \biggr]^{n-2}x^{\alpha}\frac{1}{2^{\ell}} \,dx \\ &= C_{1}(n) \frac{1}{2^{\ell}} \biggl(1-\frac{1}{2^{\ell}} \biggr)^{n-2} \int _{\frac{m}{2^{j+\ell}}}^{\frac{m+1}{2^{j+\ell}}}x^{\alpha (n-1)} \,dx \\ &\leq C_{1}(n) \frac{1}{2^{\ell}} \biggl(1-\frac{1}{2^{\ell}} \biggr)^{n-2} \biggl(\frac{m+1}{2^{j+\ell}} \biggr)^{\alpha (n-1)} \frac{1}{2^{j+\ell}}. \end{aligned}$$

Since \(m+1 \leq 2m\) and \(1-\frac{1}{2^{\ell}} < 1\), this simplifies to

$$ \vert S_{j,m} \vert \leq C_{1}(n) \frac{1}{2^{j[\alpha (n-1)+1]}} \frac{(2m)^{\alpha (n-1)}}{2^{2\ell}2^{\ell \alpha (n-1)}}. $$

The final estimate then follows from the fact that \(m < 2^{\ell}\). □

We are now in position to complete the proof of Theorem 5.1.

### Proof (part two)

The domain \(\Omega _{\alpha}\) can be subdivided into a John domain and the family of \(S_{j,m}\) as follows

$$\begin{aligned} \int _{\Omega _{\alpha}} \bigl[k(z, z_{0}; \Omega _{\alpha})\bigr]^{s} \,dz &= \int _{\Omega _{\alpha} \cap \{x > \frac{1}{2}\}} \bigl[k(z, z_{0}, \Omega _{\alpha})\bigr]^{s} \,dz + \sum_{j = 1}^{\infty} \sum_{m = 1}^{ \infty} \int _{S_{j,m}} \bigl[k(z, z_{0}; \Omega _{\alpha})\bigr]^{s} \,dz \\ &< \int _{S_{0}} \bigl[k(z, z_{0}; \Omega _{\alpha})\bigr]^{s} \,dz + \sum_{j = 1}^{\infty} \sum_{m = 1}^{\infty} \int _{S_{j,m}} \bigl[k(z, z_{0}; \Omega _{\alpha})\bigr]^{s} \,dz \\ &< \int _{S_{0}} \bigl[k(z, z_{0}; S_{0}) \bigr]^{s} \,dz + \sum_{j = 1}^{ \infty} \sum_{m = 1}^{\infty} \int _{S_{j,m}} \bigl[k(z, z_{0}; \Omega _{ \alpha})\bigr]^{s} \,dz. \end{aligned}$$

The first integral on the right is finite since the domain is a John domain.

For the sum of integrals, the estimate for \(k(z, z_{0}; \Omega _{\alpha})\) and \(|\Omega _{\alpha}|\) are combined to estimate \(\int _{\Omega _{\alpha}} k(z, z_{0}; \Omega _{\alpha})^{s} \,dz\) as follows:

$$\begin{aligned} \sum_{j = 1}^{\infty} \sum _{m = 1}^{\infty} \int _{S_{j,m}} \bigl[k(z, z_{0}; \Omega _{\alpha})\bigr]^{s} \,dz &\leq C(\alpha ,n) \sum _{j = 1}^{\infty} \sum_{m = 1}^{\infty} \bigl[(1 + \ell )2^{(\alpha -1) j}\bigr]^{s} \frac{1}{2^{j(\alpha (n-1)+1)}} \frac{1}{2^{2\ell}} \\ &= C(\alpha ,n) \sum_{j = 1}^{\infty} \bigl[2^{(\alpha -1) s - (\alpha (n-1)+1)}\bigr]^{j} \sum_{m = 1}^{\infty} \frac{(1+\ell )^{s}}{2^{2\ell}}. \end{aligned}$$

For the sum over *m*, note that \(2^{2 \ell} > m^{2}\) and \((1+ \ell ) \leq 2 + \log _{2}(m)\), hence

$$ \sum_{m = 1}^{\infty} \frac{(1+\ell )^{s}}{2^{2\ell}} \leq \sum_{m = 1}^{\infty} \frac{[2+\log _{2}(m)]^{s}}{m^{2}}, $$

which converges for all *s*.

For the sum over *j*, this converges if and only if

$$ (\alpha -1) s - \bigl(\alpha (n-1)+1\bigr) < 0, $$

which can be rearranged as \((\alpha -1) (s - n + 1) < n\). □

We end with two comments. First, if \(n = 2\), the domain \(\Omega _{\alpha}\) is a finite intersection of John domains. Secondly, for all *n*, *α*, \(\Omega _{\alpha}\) is star-shaped and therefore *p*-Poincaré for all \(1 \leq p < \infty \).