In this section we discuss the relationship between pointwise definitions (B), (I), (Î), (Ǐ), (D), (S) and (H) for a given *x*. Some of these results depend on explicit expressions for *Lf*(*x*) and thus extensions to more general operators are problematic.

Our first two results are standard.

*If f ∈ 𝒟(L*_{Î}, x) for some x ∈ **R**^{d}, then f ∈ 𝒟(*L*_{I}, x)*, and L*_{Î}f(x) = L_{I}f(x). Similarly, if f ∈ 𝒟(*L*_{Î}, x), then f ∈ 𝒟(*L*_{I}, x)*, and L*_{Î}f(*x*) *= L*_{I}f(*x*).

For the first statement, it suffices to observe that
$$\begin{array}{rl}& {\int}_{{\mathbf{R}}^{d}\setminus {B}_{r}}(f(x+z)-f(x)-\mathrm{\nabla}f(x)\cdot z{1}_{B}(z))\nu (z)dz\\ & \phantom{\rule{2em}{0ex}}={\int}_{{\mathbf{R}}^{d}\setminus {B}_{r}}(f(x+z)-f(x))\nu (z)dz\end{array}$$

and take a limit as *r →* 0^{+}. The other one is proved in a similar way. □

Note that any bounded *f* such that *f(x* + *z) = -f(x - z*) belongs to 𝒟(*L*_{I}, x), but not every such function has a gradient at *x*, so 𝒟(*L*_{Î}, x) is a proper subset of 𝒟(*L*_{I}, x). In a similar way, it is easy to construct a bounded function *f* such that the integral of *φ*(*z*) *=* (*f*(*x* + *z*)+ *f*(*x - z*) *-*2*f*(*x*))/|*z*|^{d+α} over **R**^{d} \ B_{r} has a finite limit as *r →* 0^{+}, but a similar limit for the integral of |*φ*(*z*)| is infinite. This shows that inclusions in the above lemma are proper.

*If f has second order partial derivatives at x* ∈ **R**^{d} *and* (1 + |*z|)*^{−d̄-α} f(z) is integrable, then f is in 𝒟(*L*_{D}, x)*, 𝒟*(*L*_{I}, x)*, ∈ 𝒟*(*L*_{Î}, x)*, 𝒟*(*L*_{Ǐ}, x)*, 𝒟*(*L*_{S}, x)*, 𝒟*(*L*_{h}, x) *and 𝒟*(*L*_{b}, x)*, and all corresponding definitions of Lf*(*x*) *agree. If f is of class 𝒞*^{2} in B(*x, r*)*, then the rates of convergence in each of the definitions* (D), (I), (S) *and* (H) *of Lf(x) depend only on r*, sup{max(|*f*(*y*)|*, |∇f*(*y*)|*, |∇*^{2}*f*(*y*)|) : *y ∈ B*(*x, r*)} *and ${\int}_{{\mathbf{R}}^{d}}(1+|z|{)}^{-d-\alpha}|f(x+z)|dz.$*

The result follows easily from Taylor’s expansion of *f* at *x*,
$$f(x+z)=f(x)+z\cdot \mathrm{\nabla}f(x)+O(|z{|}^{2}),$$

as well as the symmetry and upper bounds for the appropriate convolution kernels. We omit the details. □

The following result seems to be new.

*If f ∈ 𝒟*(*L*_{D}, x) *for some x ∈* **R**^{d}, then f ∈ 𝒟(*L*_{I}, x)*, and L*_{D}f(x) = L_{I}f(*x*).

Proof. Let 0 *< s < t*. Substituting *r*^{2} = s^{2}+v(*t*^{2} – s^{2}) and using [29, formula 3.197.4], we obtain
$$\begin{array}{rl}& {\int}_{s}^{t}\frac{1}{{t}^{d}({t}^{2}-{r}^{2}{)}^{\alpha /2}}\frac{1}{r({r}^{2}-{s}^{2}{)}^{1-\alpha /2}}dr\\ & \phantom{\rule{2em}{0ex}}=\frac{1}{2{s}^{2}{t}^{d}}{\int}_{0}^{1}\frac{{v}^{\alpha /2-1}(1-v{)}^{-\alpha /2}}{(1+v({t}^{2}/{s}^{2}-1))}dv\\ & \phantom{\rule{2em}{0ex}}=\frac{\mathrm{\Gamma}(\frac{\alpha}{2})\mathrm{\Gamma}(1-\frac{\alpha}{2})}{2{s}^{2}{t}^{d}({t}^{2}/{s}^{2}{)}^{\alpha /2}}=\frac{\alpha \mathrm{\Gamma}(\frac{\alpha}{2})|\mathrm{\Gamma}(-\frac{\alpha}{2})|}{4{s}^{2-\alpha}{t}^{d+\alpha}}.\end{array}$$

Taking *s = R*, *t =* | *z* |and multiplying both sides by *c*_{d,α} leads to
$${\nu}_{R}(z)=\frac{4{R}^{2-\alpha}}{\alpha \mathrm{\Gamma}(\frac{\alpha}{2})|\mathrm{\Gamma}(-\frac{\alpha}{2})|}{\int}_{R}^{\mathrm{\infty}}{\stackrel{~}{v}}_{r}(z)\frac{1}{r({r}^{2}-{R}^{2}{)}^{1-\alpha /2}}dr.$$(3.1)

Note that both sides of the above equality are zero in ${\overline{B}}_{R},$ and so it extends to all *R >* 0 and all *z*.

The remaining part of the proof is standard. Let *f* ∈ 𝒟(*L*_{d}, x) and
$$\phi (r)={\int}_{{\mathrm{R}}^{d}}(f(x+z)-f(x)){\stackrel{~}{\nu}}_{R}(z)dz$$

for *r >* 0. Then *φ* converges to a limit *φ*(0^{+}) = *L*_{D} f (*x*) as *r →* 0^{+}. In particular, *ϕ* is bounded on some interval (0*, r*_{0}). Since ${v}_{r}(z)\le {\stackrel{~}{v}}_{r}(z),$ the definition (D) requires that (*f* (*x* + *z*) *– f*(*x*))*ν*_{r}(*z*) is absolutely integrable for all *r >* 0. Hence, by (3.1) and Fubini,
$$\begin{array}{rl}& {\int}_{{\mathrm{R}}^{d}}(f(x+z)-f(x)){\nu}_{R}(z)dz\\ & \phantom{\rule{2em}{0ex}}=\frac{4{R}^{2-\alpha}}{\alpha \mathrm{\Gamma}(\frac{\alpha}{2})|\mathrm{\Gamma}(-\frac{\alpha}{2})|}{\int}_{R}^{\mathrm{\infty}}\phi (r)\frac{1}{r({r}^{2}-{R}^{2}{)}^{1-\alpha /2}}dr\\ & \phantom{\rule{2em}{0ex}}=\frac{4}{\alpha \mathrm{\Gamma}(\frac{\alpha}{2})|\mathrm{\Gamma}(-\frac{\alpha}{2})|}{\int}_{1}^{{r}_{0}/R}\phi (Rs)\frac{1}{s({s}^{2}-1{)}^{1-\alpha /2}}ds\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+\frac{4}{\alpha \mathrm{\Gamma}(\frac{\alpha}{2})|\mathrm{\Gamma}(-\frac{\alpha}{2})|}{\int}_{{r}_{0}}^{\mathrm{\infty}}\phi (r)\frac{1}{r((\frac{r}{R}{)}^{2}-1{)}^{1-\alpha /2}}dr,\end{array}$$(3.2)

and all integrals above are absolutely convergent. We consider the two integrals in the right-hand side of (3.2) separately. By dominated convergence, as *R* →0^{+}, the former one converges to (see [29, formula 3.191.2])
$$-\frac{4\phi ({0}^{+})}{\alpha \mathrm{\Gamma}(\frac{\alpha}{2})\mathrm{\Gamma}(-\frac{\alpha}{2})}{\int}_{1}^{\mathrm{\infty}}\frac{1}{s({s}^{2}-1{)}^{1-\alpha /2}}ds=\phi ({0}^{+}).$$

In the latter integral, the absolute value of the integrand decreases to 0 as *R* →0^{+}, and the integral is absolutely integrable for all *R ∈* (0*, r*_{0}).

Again by dominated convergence, the latter integral in the right-hand side of (3.2) converges to 0 as *R →* 0^{+}.

It is easy to construct *f* ∈𝒟(*L*_{I}, *x*) which is not in 𝒟(*L*_{d}, x).

Let
$$f(x+z)=\sum _{n=1}^{\mathrm{\infty}}\frac{{\epsilon}_{n}}{(|z{|}^{2}-{r}_{n}^{2}{)}^{1-\alpha /2}}{1}_{[{r}_{n},2{r}_{n}]}(|z|),$$

where, for example, *r*_{n} = 2^{–n}and *ε*_{n} = 5^{–n}. Then
$${\int}_{{\mathbf{R}}^{d}}f(x+z)\nu (z)dz=\sum _{n=1}^{\mathrm{\infty}}\frac{{c}_{d,\alpha}{\epsilon}_{n}}{{r}_{n}^{2}}{\int}_{{B}_{2}\mathrm{\setminus}{B}_{1}}\frac{1}{|y{|}^{d+\alpha}(|y{|}^{2}-1{)}^{1-\alpha /2}}dy<\mathrm{\infty},$$

so *f ∈𝒟(LI, x*), but
$${\int}_{{\mathbf{R}}^{d}}(f(x+z){\stackrel{~}{\nu}}_{{r}_{n}}(z)dz=\mathrm{\infty},$$

so that *f ∉ 𝒟(L*_{D}, x).

The next result is likely well-known, although the author could not find it in the literature. Clearly, if *f*(*x* + *z) = –f*(*x – z*) for *z* ∈ R^{d} and *f*(*x+z*)|*z*|^{d+α} is integrable *in* **R**^{d}\*B*(*x, r*) for every *r >* 0, then *f ∈ 𝒟*(*L*_{I}, *x*) and *L*_{I}f(*x*) *=* 0. Nevertheless, *f* may fail to be locally integrable near *x*, and so *f* ∉ *𝒟*(*L*_{S}, *x*). For locally integrable *f*, however, the pointwise definition of *L*_{S} is indeed an extension of the pointwise definition of *L*_{I}.

*If f ∈ 𝒟*(*L*_{I}, x) *for some x ∈* **R**^{d} and f is locally integrable near x, then f ∈𝒟(*L*_{S}, x)*, and L*_{I} f(*x*) *= L*_{S}f(*x*).

Let *m*(*r*) be the profile function of ${c}_{d,\alpha}^{-1}|z{|}^{d+\alpha}{p}_{1}(z).$ Later in this proof we show that |*m′*(*r*)| is integrable on (0, ∞). Once this is done, the argument is very similar to the one used in the proof of Lemma 3.3. Observe that *m*(0) = 0 and *m(r) →* 1 as *r → ∞* , so that the integral of *m′*(*r*) is equal to 1. Since
$$\begin{array}{ll}{c}_{d,\alpha}^{-1}|z{|}^{d+\alpha}{p}_{t}(z)& ={c}_{d,\alpha}^{-1}{t}^{-d/\alpha}|z{|}^{d+\alpha}{p}_{1}({t}^{-1/\alpha}z)=tm({t}^{-1/\alpha}|z|)\\ & =t{\int}_{0}^{{t}^{-1/\alpha}|z|}{m}^{\prime}(r)dr=t{\int}_{0}^{\mathrm{\infty}}{\mathbf{1}}_{{\mathbf{R}}^{d}\mathrm{\setminus}{B}_{r}}({t}^{-1/\alpha}z){m}^{\prime}(r)dr,\end{array}$$

we have, by Fubini,
$$\begin{array}{ll}{p}_{t}(z)& =t{\int}_{0}^{\mathrm{\infty}}{c}_{d,\alpha}|z{|}^{-d-\alpha}{\mathbf{1}}_{{\mathbf{R}}^{d}\mathrm{\setminus}{B}_{r}}({t}^{-1/\alpha}z){m}^{\prime}(r)dr\\ & =t{\int}_{0}^{\mathrm{\infty}}{\nu}_{{t}^{1/\alpha}r}(z){m}^{\prime}(r)dr.\end{array}$$

Let *f ∈ 𝒟*(*L*_{I}, x) and
$$\phi (r)={\int}_{{\mathbf{R}}^{d}}(f(x+z)-f(x)){\nu}_{r}(z)dz$$

for *r >* 0. Then *φ* is bounded and it converges to a limit *φ*(0^{+}) = *L*_{I}f(*x*) as *r →* 0^{+}. Since
$${\int}_{{\mathbf{R}}^{d}}(f(x+z)-f(x))\frac{{p}_{t}(z)}{t}dz={\int}_{0}^{\mathrm{\infty}}\phi ({t}^{1/\alpha}r){m}^{\prime}(r)dr,$$

the desired result follows by dominated convergence.

It remains to prove integrability of |*m*′(*r*)|. Observe that
$${m}^{\prime}(|z|)={c}_{d,\alpha}^{-1}|z{|}^{d+\alpha -1}((d+\alpha ){p}_{1}(z)+z\cdot \mathrm{\nabla}{p}_{1}(z)),$$

so that the Fourier transform of *g*(*z*) *= c*_{d,α}|*z*|^{1–d–α} *m′*(|*z*|) is equal to
$$\begin{array}{ll}\mathcal{F}g(\xi )& =(d+\alpha ){e}^{-|\xi {|}^{\alpha}}-\mathrm{\nabla}\cdot (\xi {e}^{-|\xi {|}^{\alpha}})\\ & =(d+\alpha ){e}^{-|\xi {|}^{\alpha}}-(d-\alpha |\xi {|}^{\alpha}){e}^{-|\xi {|}^{\alpha}}\\ & =\alpha {e}^{-|\xi {|}^{\alpha}}(1+|\xi {|}^{\alpha}).\end{array}$$

The right-hand side is smooth in **R**^{d} \{0}, and its series expansion at 0 is $\mathcal{F}g(\xi )=\alpha (1-\frac{1}{2}|\xi {|}^{2\alpha})+O(|\xi {|}^{3\alpha}).$ By [40, Theorem 4],
$$\underset{r\to \mathrm{\infty}}{lim}{c}_{d,\alpha}{r}^{1+\alpha}{m}^{\prime}(r)=\underset{|z|\to \mathrm{\infty}}{lim}|z{|}^{d+2\alpha}g(z)=\frac{{2}^{2\alpha -d/2-1}\mathrm{\Gamma}(\frac{d}{2}+\alpha )}{\mathrm{\Gamma}(-\alpha )},$$

where the right-hand side is understood to be equal to 0 if *α =* 1. It follows that for *α ∈* (0,1), *m′*(*r*) is ultimately positive, while for *α ∈* (1,2), *m’*(*r*) is ultimately negative. By (2.7), for *α =* 1, *m′*(*r*) is everywhere positive (ultimate positivity also follows from [40, Theorem 4] by considering the next term in the series expansion of *Fg*(*ξ*)). Since *m’*(*r*) is smooth on [0*, ∞*) and
$$\underset{R\to \mathrm{\infty}}{lim}{\int}_{0}^{R}{m}^{\prime}(r)dr=\underset{R\to \mathrm{\infty}}{lim}m(R)-m(\mathrm{O})=\underset{|z|\to \mathrm{\infty}}{lim}{c}_{d,\alpha}^{-1}|z{|}^{d+\alpha}{p}_{1}(z)=1,$$

we conclude that |*m*′(*r*)| is integrable, as desired.

In a similar way, we obtain the following result.

*If f* ∈ 𝒟(*L*_{I}, x) *for some x ∈* **R**^{d} *and f is locally integrable near x, then f ∈* 𝒟(*L*_{h},*x*)*, and L*_{I} f(*x*) *= L*_{H} f(*x*).

The argument is exactly the same as the proof of Lemma 3.4, once we note that by (2.19), the profile function of ${c}_{d,\alpha}^{-1}|z{|}^{d+\alpha}{q}_{1}(z),$namely $m(r)=({r}^{2}/(c{\alpha}^{2/\alpha}+{r}^{2}){)}^{(d+\alpha )/2},$is increasing.

In order to prove a similar result connecting (H) and (S), one would need some relationship between *q*_{y}(*z*) and *p*_{t}(*z*). For *α* = 1, *q*_{y}(*z*) *= p*_{y}(*z*) and equivalence of (H) and (S) is trivial. For *α* ∈ (1,2), we conjecture that $\phi (r)=\sqrt{r}{K}_{\alpha /2}({r}^{1/\alpha})$ is completely monotone. If this is the case, then $\mathcal{F}{q}_{y}(\xi )$ can be expressed as the integral average of $\mathcal{F}{p}_{t}(\xi )={e}^{-t|\xi {|}^{\alpha}},$ and so *f* ∈ *𝒟*(*L*_{S}, *x*) implies *f ∈ 𝒟*(*L*_{H}, *x*). For *α* ∈ (0,1) one might expect the converse; however, it is unclear whether $\mathcal{F}{p}_{t}(\xi )$ can be expressed as an integral average of $\mathcal{F}{q}_{y}(\xi ).$

*For α* ∈ (1,2) *the function $\sqrt{r}{K}_{\alpha /2}({r}^{1/\alpha})$ is completely monotone in r* ∈ (0, ∞).

As in Lemma 3.3, the inclusions in Lemmas 3.4 and 3.5 are proper. An example of *f* ∈ *𝒟*(*L*_{S}, x) which is not in 𝒟(*L*_{I}, *x*) is, however, more complicated, and we only sketch the argument.

Let
$$f(x+z)=\sum _{n=1}^{\mathrm{\infty}}{\epsilon}_{n}|z{|}^{1+\alpha}({1}_{[{r}_{n},(1+{\delta}_{n}){r}_{n}]}(|z|)-{1}_{[(1-{\delta}_{n}){r}_{n},{r}_{n}]}(|z|)),$$

where, for example, *r*_{n} = 2^{–n}, δ_{n} = 4^{–n} and *ε*_{n} = n8^{n}. First of all, *f* is easily proved to be integrable. Due to cancelations, the integral of *f*(*x* + *z*)|*z*|^{–d–α} over $z\in {B}_{(1+{\delta}_{n}){r}_{n}}\mathrm{\setminus}{B}_{(1-{\delta}_{n}){r}_{n}}$is zero, and therefore
$$\begin{array}{ll}{\int}_{{\mathbf{R}}^{d}}f(x+z){\nu}_{{r}_{n}}(z)dz& ={c}_{d,\alpha}{\int}_{{B}_{(1+{\delta}_{n}){r}_{n}}\mathrm{\setminus}{B}_{{r}_{n}}}f(x+z)|z{|}^{-d-\alpha}dz\\ & ={c}_{d,\alpha}d|B|{\epsilon}_{n}{\delta}_{n}{r}_{n}\to \mathrm{\infty}\end{array}$$

as *n* →∞. It follows that *f* ∉ *𝒟(L*_{I}, *x*). However, if *m*(*r*) denotes the profile function of |*z*|^{d}^{+α}*p*_{1}(*z*) and *M* is the supremum of |*m*′(*r*)|(1 + *r*^{1+α}) (which was shown to be finite in the proof of Lemma 3.4), then

$$\begin{array}{rl}& |{\int}_{{\mathrm{R}}^{d}}f(x+z)\frac{{p}_{t}(z)}{t}dz|\\ & \begin{array}{ll}& \le \sum _{n=1}^{\mathrm{\infty}}{\epsilon}_{n}\left|\left({\int}_{{B}_{(1+{\delta}_{n}){r}_{n}}\mathrm{\setminus}{B}_{{r}_{n}}}-{\int}_{{B}_{{r}_{n}}\mathrm{\setminus}{B}_{(1-{\delta}_{n}){r}_{n}}}\right)|z{|}^{1+\alpha}\frac{{p}_{t}(z)}{t}dz\right|\\ & =\sum _{n=1}^{\mathrm{\infty}}d|B|{\epsilon}_{n}\left|\left({\int}_{{r}_{n}}^{(1+{\delta}_{n}){r}_{n}}-{\int}_{(1-{\delta}_{n}){r}_{n}}^{{r}_{n}}\right)m({t}^{-1/\alpha}r)dr\right|\\ & =\sum _{n=1}^{\mathrm{\infty}}d|B|{\epsilon}_{n}{r}_{n}\left|{\int}_{0}^{{\delta}_{n}}(m({t}^{-1/\alpha}(1+s){r}_{n})-m({t}^{-1/\alpha}(1-s){r}_{n}))ds\right|\\ & \le \sum _{n=1}^{\mathrm{\infty}}d|B|{\epsilon}_{n}{r}_{n}\frac{{t}^{-1/\alpha}{r}_{n}M{\delta}_{n}^{2}}{1+{\left(\frac{1}{2}{{t}^{-1}}^{/\alpha}{r}_{n}\right)}^{1+\alpha}}\\ & =\sum _{n=1}^{\mathrm{\infty}}{2}^{1+\alpha}d|B|M{\epsilon}_{n}{\delta}_{n}^{2}{r}_{n}^{1-\alpha}\frac{t}{(2{t}^{1/\alpha}{r}_{n}^{-1}{)}^{1+\alpha}+1}\to 0\end{array}\end{array}$$

as *t* →0^{+}. This proves that *f* ∈ 𝒟(*L*_{S}, *x*), and a similar argument shows that *f* ∈ *𝒟*(*L*_{h}, *x*). We omit the details.

Apparently an example can be given to prove that *f* ∈ *𝒟(L*_{S}, *x*) does not imply *f* ∈ *𝒟*(*L*_{b}, *x*), but the author could not work out the technical details. On the other hand, it is not true that *f* ∈ *𝒟*(*L*_{b}, *x*) implies *f* ∈ 𝒟(*L*_{S}, *x*): pointwise convergence in (B) does not require integrability of (1 + |*z*|)^{–d–α}*f*(*z*) at infinity. For example, when *d* > 2, the function $f(x)={x}_{1}^{2}-{x}_{2}^{2}$ can be proved to belong to 𝒟(*L*_{b}, *x*)(and in fact *L*_{b} f is everywhere zero according to (B)), but due to fast growth of *f* at infinity, *f* does not belong to 𝒟(*L*_{S}, x). This is, however, the only obstacle in the proof of the following result.

*If f* ∈ 𝒟(*L*_{b}, *x*) *for some x* ∈ **R**^{d} *and* (1 + |*z*|)^{–d–α} *f*(*z*) *is integrable, then f* ∈ *𝒟*(*L*_{S},*x*) *and f* ∈ *𝒟*(*L*_{h}, *x*)*, and L*_{b} f(*x*) *= L*_{S}f(*x*) *= L*_{H} f(*x*).

Recall that by the Bochner’s subordination formula (2.9),
$${p}_{t}(z)={t}^{-2/\alpha}{\int}_{0}^{\mathrm{\infty}}{k}_{r}(z)\eta ({t}^{-2/\alpha}r)dr,$$

where *k*_{r}(*z*) is the Gauss-Weierstrass kernel and *η*(*s*) is a smooth function such that 0 *< η*(*s*)< *C*_{α} min(1*, s*^{–}^{1–α}^{/2}) for *s >* 0 and $\underset{s\to \mathrm{\infty}}{lim}{s}^{1+\alpha /2}\eta (s)=1/\left|\mathrm{\Gamma}\left(-\frac{\alpha}{2}\right)\right|$ (see [51, Remark 14.18]). Therefore, by Fubini,

Figure 1 Relationship between pointwise definitions of *Lf*(*x*) for *f* such that (1 + |*z*|)^{–d–α} *f*(*x* + *z*) is integrable. An arrow indicates inclusion of appropriate domains, and an arrow with a tail indicates proper inclusion.

$${\int}_{{\mathbf{R}}^{d}}(f(x+z)-f(x))\frac{{p}_{t}(z)}{t}dz={\int}_{0}^{\mathrm{\infty}}({k}_{r}\ast f(x)-f(x)){t}^{-1-2/\alpha}\eta ({t}^{-2/\alpha}r)dr.$$

Suppose that *f* ∈ *𝒟*(*L*_{b}, *x*). Then, by dominated convergence,
$$\begin{array}{rl}& \underset{t\to {0}^{+}}{lim}{\int}_{{\mathbf{R}}^{d}}(f(x+z)-f(x))\frac{{p}_{t}(z)}{t}dz\\ & \phantom{\rule{2em}{0ex}}=\frac{1}{|\mathrm{\Gamma}(-\frac{\alpha}{2})|}{\int}_{0}^{\mathrm{\infty}}({k}_{r}\ast f(x)-f(x)){r}^{-1-\alpha /2}dr,\end{array}$$

as desired.

A similar argument involving the identity
$${q}_{y}(z)={\displaystyle \frac{{y}^{-2/\alpha}}{|\mathrm{\Gamma}(-\frac{\alpha}{2})|}{\int}_{0}^{\mathrm{\infty}}{k}_{r}(z)\frac{1}{(y-2/\alpha r{)}^{1+\alpha /2}}\mathrm{exp}\left(-\frac{1}{4({c}_{\alpha}{)}^{2/\alpha -2/\alpha}yr}\right)dr,}$$

which follows easily from the gamma integral, shows that *f ∈ 𝒟*(*L*_{H},*x*) and *L*_{H} f(*x*) *= L*_{B}f(*x*)(this is a variant of a result proved in [56]).

The proofs of pointwise results can be re-used for the corresponding statements for norm convergence in any of the spaces ℒ^{p}, 𝒞_{0}*, 𝒞*_{bu} and 𝒞_{b}.

*Let 𝒳* *be any of the spaces ℒ*^{p}, p ∈ [1,∞]*, 𝒞*_{0}*,𝒞*_{bu} and 𝒞_{b}. If f ∈ *𝒟*(*L*_{d}, 𝒳)*, then f ∈ 𝒟*(*L*_{I}, 𝒳)*, which in turn implies f* ∈ 𝒟 (*L*_{S}, 𝒳) ⋂ (*L*_{H}, 𝒳)*. Also, if f ∈* *𝒟*(*L*_{b}, *𝒳*)*, then f ∈* *3>*(*L*_{S}, 𝒳) ⋂𝒟(*L*_{h}, 𝒳)*. Furthermore, all definitions of Lf agree on appropriate domains*.

We only prove that *f ∈ 𝒟* (*L*_{I}, L^{p}) implies *f ∈* 𝒟(*L*_{S}, *L*^{p}), the other statements being very similar. Recall that
$${p}_{t}(z)=t{\int}_{0}^{\mathrm{\infty}}{\nu}_{{t}^{1/\alpha}r}(z){m}^{\prime}(r)dr,$$

where *m*′(*r*) is an absolutely integrable function, with integral 1. Denote
$${\phi}_{r}(x)={\int}_{{\mathbf{R}}^{d}}(f(x+z)-f(x)){\nu}_{r}(z)dz.$$

If *f ∈ 𝒟(L*_{I}, ℒ^{p}), then *φ*_{r} converges to *φ*_{0+} = *L*_{I} f in ℒ^{p}. By Fubini,
$$\begin{array}{r}{\int}_{{\mathbf{R}}^{d}}(f(x+z)-f(x))\frac{{p}_{t}(z)}{z}dz-{\phi}_{0+}(x)\\ ={\int}_{0}^{\mathrm{\infty}}({\phi}_{{t}^{1/\alpha}}r(x)-{\phi}_{0+}(x)){m}^{\prime}(r)dr,\end{array}$$

and so, by dominated convergence,
$$\begin{array}{rl}\underset{t\to {0}^{+}}{lim}& \parallel \frac{1}{t}({P}_{t}f-f)-{\phi}_{0+}{\parallel}_{p}\\ & \le \underset{t\to {0}^{+}}{lim}{\int}_{0}^{\mathrm{\infty}}\parallel {\phi}_{{t}^{1/\alpha}}r(x)-{\phi}_{0+}(x){\parallel}_{p}|{m}^{\prime}(r)|dr=0,\end{array}$$

as desired.

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