Symmetry in Fourier Analysis: Heisenberg Group to Stein–Weiss Integrals

Abstract

Embedded symmetry within the Heisenberg group is used to couple geometric insight and analytic calculation to obtain a new sharp Stein–Weiss inequality with mixed homogeneity on the line of duality. SL(2,R) invariance and Riesz potentials define a natural bridge for encoded information that connects distinct geometric structures. The intrinsic character of the Heisenberg group makes it the natural playing field on which to explore the laws of symmetry and the interplay between analysis and geometry on a manifold.

Appendix

1.1 1. Stein–Weiss Theorem with Mixed Homogeneity

A short argument is given in the appendix for [9] of the general Stein–Weiss theorem on $\mathbb {R}^n$ for general integrability classes. This argument easily extends to show that Stein–Weiss inequalities for mixed homogeneity are controlled by the general result. This argument is implicit in Theorem 5 from that paper which addresses maps on the diagonal where sharp constants are obtained. The key step is to use Young’s inequality to integrate out the “free variable.”

Theorem 8

For $f \in \mathcal {S}(\mathbb {R}^{n+m}), 1< p \le q< \infty , v = (x,y) \in \mathbb {R}^n \times \mathbb {R}^m, 0< \lambda< n+m, \alpha + \beta > 0, \alpha< m/q, \beta < m/p' , \alpha + \beta + \lambda = (n+m)/r, 1/q = 1/r + 1/p - 1, 1/r = 1/p' + 1/q$

$$\begin{aligned} \parallel |y|^{-\alpha } (|v|^{-\lambda } * |y|^{-\beta } f(v) ) \parallel _{L^q(\mathbb {R}^{n+m})}\ \le A \parallel f \parallel _{L^p(\mathbb {R}^{n+m})} \end{aligned}$$

Proof

Use Minkowski’s inequality for integrals and Young’s inequality applied to the x-integration to obtain

$$\begin{aligned} \parallel \ |y|^{-\alpha }(|v|^{-\lambda }*|y|^{-\beta }f(v))\parallel _{L^q(\mathbb {R}^{n+m})} \le C\ \parallel |y|^{-\alpha }(|y|^{-\sigma }*|y|^{-\beta }F(y)\parallel _{L^q(\mathbb {R}^m)} \end{aligned}$$

where $\sigma = \lambda - n/r = m/r - \alpha - \beta$ with $0< \sigma < m/r$, and

$$\begin{aligned} F(y) = [\int _{\mathbb {R}^n} |f(x,y)|^P \ dx]^{1/p}, \ C = [\int _{\mathbb {R}^n} (1+|x|^2)^{\lambda r/2}\ dx]^{1/r} \end{aligned}$$

with $\lambda r -n >0$. Using the general result

$$\begin{aligned} \parallel |y|^{-\alpha }(|y|^{-\sigma }*|y|^{-\beta }F(y))\parallel _{L^q(\mathbb {R}^m)} \le A_m \ \parallel F\parallel _{L^p(\mathbb {R}^m)} \end{aligned}$$

one obtains

$$\begin{aligned} \parallel |y|^{-\alpha }(|v|^{-\lambda }*|y|^{-\beta }f(v))\parallel _ {L^q(\mathbb {R}^{n+m})} \le C A_m \ \parallel F\parallel _{L^p(\mathbb {R}^m)} = C A_m \ \parallel f\parallel _{L^p(\mathbb {R}^{n+m})} \end{aligned}$$

$\square$

1.2 2. Hardy–Littlewood–Sobolev Inequality on Hyperbolic Space

Inverse powers of the Poincaré distance determine a natural Hardy–Littlewood–Sobolev inequality on $\mathbb {H}^n$ which is sharp with no extremals by direct correspondence with the Hardy–Littlewood–Sobolev inequality on $\mathbb {R}^n$.

Theorem 9

For non-negative $F,G \in L^p(\mathbb {H}^n)$, $1<p<2$

$$\begin{aligned} \int _{\mathbb {H}^n\times \mathbb {H}^n} F(w) d(w,w'')^{-2n/p'} G(w') \ d\nu d\nu ' \le A_p \ \parallel F\parallel _{L^p(\mathbb {H}^n)} \parallel G\parallel _{L^p(\mathbb {H}^n)} \end{aligned}$$

This inequality is sharp with no extremals.

Proof

Write out the functional

$$\begin{aligned}&\int F(w) \ d(w,w')^{-2n/p'}G(w')\ d\nu d\nu '\nonumber \\&\quad =\int F(w)\ |w-w'|^{-2n/p'}\ (4yy')^{n/p'} G(w') \ d\nu d\nu '; \end{aligned}$$

let $f = Fy^{-n/p}$, $g = Gy'^{-n/p}$, and the equation becomes

$$\begin{aligned}&\int f(x,y) \ \left[ |x-x'|^2 + |y-y'|^2 \right] ^{-n/p'} g(x',y') \ dmdm'\\&\quad \le 4^{-n/p'} A_p \parallel f\parallel _{L^p(\mathbb {R}^n)} \ \parallel g\parallel _ {L^p(\mathbb {R}^n)} \end{aligned}$$

where the integration is for $y,y'$ positive. By translating the $y, y' = 0$ boundary, this inequality is equivalent to the full Hardy–Littlewood–Sobolev inequality on $\mathbb {R}^n$, but because of the sliding boundary no extremals exist prior to filling the full space.The lack of an extremal corresponds to the realization of the equivalent $\mathbb {R}^n$ inequality on the two-sheeted hyperboloid while the upper-half-space matches only one sheet. The constant is obtained by matching the sharp constant for the Euclidean inequality:

$$\begin{aligned} A_p = (4\pi )^{n/p'} \frac{\Gamma (n(1/p - 1/2)}{\Gamma (n/p)}\ \left[ \frac{ \Gamma (n/2)}{\Gamma (n)}\right] ^{1-2/p} \end{aligned}$$

$\square$

This calculation exhibits by its close correspondence with Euclidean analysis the usefulness of the upper-half-space model for calculations on hyperbolic space. Similar features are at play in obtaining the Riesz–Sobolev rearrangement inequality on hyperbolic space. Overall, the estimate reflects the structural nature of analysis on manifolds:

$$\begin{aligned} F \in L^p(\mathbb {H}^n) \rightsquigarrow F*u^{-n/p'} \in L^{p'} (\mathbb {H}^n ) , 1\le p < 2 \end{aligned}$$

where $u=d^2(w,\hat{0})$. The more general Hardy–Littlewood–Sobolev Inequality on $\mathbb {H}^n$

$$\begin{aligned} F \in L^p(\mathbb {H}^n) \rightsquigarrow F*u^{-n/2q} \in L^r(\mathbb {H}^n), 1/r = 1/p + 1/q - 1 \end{aligned}$$

follows naturally from an application of the Stein–Weiss inequality combined with Young’s inequality. To demonstrate this map, one needs to show that

$$\begin{aligned} \int _{\mathbb {H}^n\times \mathbb {H}^n} F(w) \left[ d(w,w')\right] ^{-n/q} G(w')\ dwdw' \le C\ \parallel F\parallel _{L^p(\mathbb {H}^n)}\ \parallel G\parallel _{L^{r'}(\mathbb {H}^n)} \end{aligned}$$

Set $\tilde{f} = F\ |y|^{-n/p}$, $\tilde{g} = G\ |y|^{-n/ri}$; then the functional becomes

$$\begin{aligned} \int \tilde{f}(w)\ |y|^{-\alpha } |w-w'|^{-n/q}|y'|^{-\beta }\ \tilde{g}(w')\ dwdw' \end{aligned}$$

where $w = (x,y) \in \mathbb {R}^{n-1}\times \mathbb {R}_+$, $\alpha = n/p' - n/2q$, $\beta =n/r -n/2q$. Observe that $\alpha +\beta =0$ as $n/p' +n/r - n/q =0$. Apply Young’s inequality in the $x,x'$ variables in $\mathbb {R}^{n-1}$ and set

$$\begin{aligned} f(y = \left[ \int _{\mathbb {R}^{n-1}} |\tilde{f}(x,y)|^p dx\right] ^{1/p}, \ g(y') = \left[ \int _{\mathbb {R}^{n-1}} |\tilde{g}(x,y')|^{r'} dx\right] ^{1/r'} \end{aligned}$$

Then using the formula

$$\begin{aligned} \left[ \int _{\mathbb {R}^{n-1}} |w-w'|^{-n} dx\right] ^{1/q} = |y-y'|^{1/q} \left[ \int _{\mathbb {R}^{n-1}} [1+|x|^2]^{-n/2} dx\right] ^{1/q} \end{aligned}$$

the equation above is obtained by applying the Stein–Weiss theorem as given in the Appendix in [9] and letting f, g take the value zero for negative values of y and $y'$ for the functional

$$\begin{aligned} \int _{\mathbb {R}\times \mathbb {R}}f(y)\ |y|^{-\alpha } |y-y'|^{-1/q}|y'|^{-\beta } \ g(y') \ dydy' \end{aligned}$$

so that the Stein–Weiss inequality completes the proof of the general Hardy–Littlewood–Sobolev inequality on the hyperbolic space $\mathbb {H}^n$. Since the weights satisfy $\alpha + \beta = 0$, this calculation demonstrates the importance of allowing weights of opposite sign in the Stein–Weiss theorem. But for Stein–Weiss integrals with mixed homogeneity, estimates for Sobolev embedding for functions with radial symmetry lead to examples with positive power weights. This calculation extending Young’s inequality is not so much a digression as it illustrates the intrinsic connection between analysis on hyperbolic space and Stein–Weiss integrals.

Conformal geometry connects three representations for hyperbolic space by relatively simple coordinate transformations: (1) the Poincaré–Beltrami upper-half-space as used in this paper, (2) the Poincaré unit ball, and (3) one sheet of the hyperboloid invariant under the action of the group SO(n.1). To explicitly show the conformal equivalence between the two standard models for hyperbolic space, half-space and ball, as manifest in the Hardy–Littlewood–Sobolev inequality, consider the conformal metric relations as expressed by the following:

$$\begin{aligned} d^2(w,w') = \frac{|w-w'|^2}{4yy'} = \frac{|v-v'|^2}{4(1-|v|^2)(1-|v'|^2)} = d^2(v,v') \end{aligned}$$

where $w,w' \in \mathbb {H}^n$ and $v,v' \in B$ with B being the n-dimensional unit ball centered at the origin. Observe the clear similarity with the corresponding metric relations for the classical geometries representing the plane, the sphere, and the two-sheeted hyperboloid (see [5]). Direct transference using this metric relation allows first the realization of the Hardy–Littlewood–Sobolev inequality in terms of the Poincaré ball model, and then a simple return to the inequality on $\mathbb {R}^n$ with a second explicit demonstration that extremal functions do not exist for the hyperbolic inequality. The first discussion of an equivalent form of the Hardy–Littlewood–Sobolev inequality on the hyperbolic plane $\mathbb {H}^2$ with an existing extremal appears in my written lecture from the 1991 Princeton conference honoring Eli Stein (see Theorem 18 in [5]) and my lecture at El Escorial [7]. Another approach was taken by Lu and Yang [42, 43] with their program to develop Fourier analysis on n-dimensional hyperbolic space using the ball model with explicit formulas for the Fourier transform, Harish-Chandra’s c-function, and the Plancherl measure though using a different group structure and a different definition of convolution on the manifold.

A more fundamental realization of the Hardy–Littlewood–Sobolev inequality with respect to hyperbolic geometry appears in the context of the conformal equivalence of the classical geometries—the plane, the sphere, the hyperboloid, and the equivalent forms for functionals and metric relations for Riesz potentials acting on the line of duality where the operator is positive-definite and self-adjoint. The domain is now the n-dimensional two-sheeted hyperboloid in $\mathbb {R}^{n+1}$:

$$\begin{aligned} \hat{\mathbb {H}}^n = \{p = (p_o, p_1) \in \mathbb {R}\times \mathbb {R}^n : p_o^2 - p_1^2 = 1\} \end{aligned}$$

$\hat{\mathbb {H}}^n$ is a homogeneous space under the action of the Lorentz group O(1, n). Consider the map from $\mathbb {R}^n - \{|x|=1\}$ to $\hat{\mathbb {H}}^n$ given by

$$\begin{aligned} p = \left( \frac{1+|x|^2}{1-|x|^2},\frac{2x}{1-|x|^2}\right) \end{aligned}$$

with

$$\begin{aligned} |pq-1| = 2\frac{|x-y|^2}{|1-|x|^2||1-|y|^2|} \end{aligned}$$

where $pq = p_oq_o-p_1q_1$ and

$$\begin{aligned} d\nu (p) = 2 \delta (1-p^2)\ dp = 2\delta (1+p_1^2 - p_o^2)\ dp_odp_1 = 2^n |1-|x|^2|^{-n} \ dx \end{aligned}$$

is an O(1, n) invariant measure on $\hat{\mathbb {H}}^n$. The corresponding Hardy–Littlewood–Sobolev inequality on $\hat{\mathbb {H}}^n$ is given by:

Theorem 10

For $F\in L^p(\mathbb {H}^n)$ with $1<p<2$

$$\begin{aligned}&\int _{\hat{\mathbb {H}}^n\times \hat{\mathbb {H}}^n}F(q)\ |qq' -1|^{-n/p'} \ F(q')\ d\nu (q)d\nu (q') \le A_p\ \left[ \parallel _{L^p(\hat{\mathbb {H}}^n)}\right] ^2\\&A_p = (4\pi )^{n/p'}\ \frac{\Gamma (n(1/p - 1/2))}{\Gamma (n/p)}\left[ \Gamma (n/2) /\Gamma (n)\right] ^{1-2/p} \end{aligned}$$

This inequality is sharp and the constant is attained for the extremal function

$$\begin{aligned} F(q) = A\ (1+|q_1|^2)^{-n/p} \end{aligned}$$

The proof follows directly from a change of variables with the Hardy–Littlewood–Sobolev inequality on $\mathbb {R}^n$. Potentials that depend on decreasing functions of distance do allow rearrangement inequalities through application of “two-point symmetrization.”

Lemma

For non-negative functions F, G, and K a non-negative decreasing function

$$\begin{aligned}&\int _{\hat{\mathbb {H}}^n\times \hat{\mathbb {H}}^n} F(p)K(|pq-1|)G(q)\ d\nu (p)d\nu (q) \\&\quad \le \int _{\hat{\mathbb {H}}^n\times \hat{\mathbb {H}}^n} F^{*}(p)K(|pq-1|) G^{*}(q) \ d\nu (p)d\nu (q) \end{aligned}$$

where $F^{*}, G^{*}$ are equimeasurable radial decreasing rearrangements of the respective component functions on each component of $\hat{\mathbb {H}}^n$.

The domain consists of two separated components so it’s not possible to “mix points” between components using the action of the group SO(1, n). This domain can be viewed as $\mathcal {S}^o\times \mathbb {H}^n$ where $\mathcal {S}^o = \{+1,-1\}$ so that more generally domains corresponding to the indefinite orthogonal group SO(m, n) could be viewed as the cylindrical surface $\mathcal {S}^{m-1}\times \mathbb {H}^n$. For $m\ge 2$, the surface has only one component which can be represented as the rotation of a hyperboloid and perhaps best visualized in the case $m=2$.

Observe that the reverse Hardy–Littlewood–Sobolev inequality for hyperbolic space $\mathbb {H}^n$ follows naturally in the same way from the Euclidean reverse inequality [11]. Here the difference is reflected only by the index $p'$ being negative and the positive powers of the metric which can be related to lower bounds for the Wasserstein distance. These inequalities characterize the continuity of the Hardy–Littlewood–Sobolev functional in terms of the integrability classes relative to the index $0<p<2$ of non-negative functions which results from conformal geometry.

1.3 3. Gamma Function and Beta Function Estimates

In order to show that the Hardy–Littlewood–Sobolev inequality restricted to radial functions on the Heisenberg group $\mathcal {H}_n$ is controlled by Euclidean inequalities on the sphere, the following estimate for gamma functions is used:

$$\begin{aligned} \left[ \frac{\Gamma \left( \frac{n+1}{p}\right) \ \Gamma \left( \frac{n+1}{p'} + l\right) }{\Gamma \left( \frac{n+1}{p'}\right) \ \Gamma \left( \frac{n+1}{p} + l\right) } \right] ^2 \le \frac{\Gamma \left( \frac{(2n+1)}{p}\right) \ \Gamma [\frac{(2n+1)}{p'} +2l]}{\Gamma \left( \frac{(2n+1)}{p'}\right) \ \Gamma [\frac{(2n+1)}{p}+ 2l]} \end{aligned}$$

Here $p<p'$ and equality occurs for $p=2$. Equivalently it suffices to show

$$\begin{aligned} \left[ \frac{\Gamma \left( \frac{n+1}{p}\right) }{\Gamma \left( \frac{n+1}{p} +l\right) }\right] ^2 \frac{\Gamma \left( \frac{(2n+1)}{p}+2l\right) }{\Gamma \left( \frac{(2n+1)}{p}\right) } \le \left[ \frac{\Gamma \left( \frac{n+1}{p'}\right) }{\Gamma \left( \frac{n+1}{p'} + l\right) }\right] ^2 \frac{\Gamma \left( \frac{(2n+1)}{p'} +2l\right) }{\Gamma \left( \frac{(2n+1)}{p'}\right) } \end{aligned}$$

Set $\alpha = 1/p$ with $\alpha \ge 1/2$, and consider

$$\begin{aligned} F(\alpha )= & {} \left[ \frac{\Gamma ((n+)\alpha )}{\Gamma ((n+1)\alpha +l)}\right] ^2 \frac{\Gamma ((2n+1)\alpha +2l)}{\Gamma ((2n+1)\alpha )} \\= & {} \frac{(2n+1)\alpha \cdots [(2n+1)\alpha +2l-1]}{[(n+1)\alpha \cdots [(n+1)\alpha + l-1]]^2} \end{aligned}$$

Our objective is to show that $F(\alpha )$ is decreasing for $\alpha > 0$. Now compute $F'(\alpha )$ and observe $F'(\alpha ) < 0$

$$\begin{aligned} F'(\alpha )= & {} \left[ (2n+1)\ \sum _{k=0}^{2l -1} \frac{1}{((2n+1)\alpha + k} - 2(n+1)\ \sum _{k=0}^{l-1} \frac{1}{(n+1)\alpha + k}\right] \ F(\alpha )\\= & {} (2n+1)\left[ \sum _{k=0}^{2l-1} \frac{1}{(2n+1)\alpha + k} - 2 \ \sum _{k=0}^{l-1} \frac{1}{(2n+1)\alpha +\frac{2n+1}{n+1}k}\right] \ F(\alpha ) \end{aligned}$$

Write the first sum in terms of even and odd indices

$$\begin{aligned}&\sum _{k=0}^{2l-1} \ \frac{1}{(2n+1)\alpha +k} = \sum _{k=0}^{l-1}\frac{1}{(2n+1)\alpha +2k} +\sum _{k=0}^{l-1}\frac{1}{(2n+1)\alpha +2k+1}\\&\quad< 2\sum _{k=0}^{l-1}\frac{1}{(2n+1)\alpha +2k} < 2 \sum _{k=0}^{l-1} \frac{1}{(2n+1)\alpha +\frac{2n+1}{n+1}k} \end{aligned}$$

This argument shows that $F'(\alpha ) <0$ and $F(\alpha )$ is decreasing for $\alpha >0$.The estimate proves the majorization for Euclidean spherical harmonics and that conformal estimates for the Hardy–Littlewood–Sobolev inequality on the sphere control Heisenberg estimates for radial functions. More generally, this shows for the beta function $B(\alpha ,\beta )$ with $\alpha ,\beta >0$

$$\begin{aligned} B(\alpha .\beta ) = \frac{\Gamma (\alpha )\ \Gamma (\beta )}{\Gamma (\alpha +\beta )}, \ G(\alpha ,l) = \frac{[B((n+1)\alpha ,l)]^2}{B((2n+1)\alpha ,2l)} \end{aligned}$$

then $G(\alpha ,l)$ is decreasing for $\alpha >0$ and l is a positive integer.

Using series representations for

$$\begin{aligned} \psi (x) = \dfrac{d}{dx} \ln \Gamma (x) \end{aligned}$$

one can extend the result above to show that for $\alpha , \beta >0$

$$\begin{aligned} F(\alpha ,\beta ) = \left[ \frac{\Gamma (d\alpha )}{\Gamma (d\alpha + \beta )}\right] ^2 \frac{\Gamma (c\alpha + 2\beta )}{\Gamma (c\alpha )} \end{aligned}$$

is decreasing as a function of $\alpha >0$ where $0<d$, $0<c<2d$. This result can be equivalently stated in terms of beta functions so that the function $G(\alpha ,\beta )$

$$\begin{aligned} G(\alpha ,\beta ) = \frac{\left[ B(d\alpha ,\beta )\right] ^2}{B(c\alpha ,2\beta )} \end{aligned}$$

is decreasing as a function of $\alpha >0$. It is natural to think in terms of integral representations, but it is more direct to use the series representations for $\psi$ with splittings over even and odd indices as above. Set

$$\begin{aligned} \Lambda (\alpha ) = \ln F(\alpha ,\beta ) = 2\ln \Gamma (d\alpha ) - 2\ln \Gamma (d\alpha +\beta ) +\ln \Gamma (c\alpha + 2\beta ) -\ln \Gamma (c\alpha ), \end{aligned}$$

then by differentiating in $\alpha$

$$\begin{aligned} \Lambda '(\alpha )= & {} 2d[\psi (d\alpha ) -\psi (d\alpha +\beta )] + c[\psi (c\alpha +2\beta ) -\psi (c\alpha )]\\ \Lambda ''(\alpha )= & {} 2d^2[\psi '(d\alpha ) -\psi '(d\alpha +\beta )]+ c^2[\psi '(c\alpha +2\beta ) -\psi '(c\alpha )] \end{aligned}$$

Using formulas due to Gauss and Legendre

$$\begin{aligned} \psi '(\alpha )= & {} \sum _{k=0}^{\infty } \frac{1}{(k+\alpha )^2}\\ \psi (\alpha ) - \psi (\alpha +\beta )= & {} \beta \ \sum _{k=0}^{\infty } \ \frac{1}{(k+\alpha )(k+\alpha +\beta )}\ \rightarrowtail 0,\ \alpha \rightarrowtail \infty \end{aligned}$$

(note second equation implies $\psi$ is decreasing), we show that $\Lambda ''(\alpha ) >0$ which implies that $\Lambda '(\alpha )$ is increasing and with $\lim \Lambda '(\alpha ) =0$ as $\alpha \rightarrowtail \infty$, then $\Lambda '(\alpha ) <0$ which implies $\Lambda (\alpha )$ is decreasing so that $F(\alpha ,\beta )$ is decreasing as a function of $\alpha$, and similarly the corresponding function $G(\alpha ,\beta )$, expressed in terms of beta functions, is decreasing as a function of $\alpha >0$.

To fully complete these arguments, use the series representation

$$\begin{aligned} \psi '(\alpha ) = \sum _{k=0}^{\infty }\ \frac{1}{(k+\alpha )^2} \end{aligned}$$

and group the terms for $\Lambda ''(\alpha )$

$$\begin{aligned} \Lambda ''(\alpha ) = 2d^2\left\{ \left[ \psi '(d\alpha ) -\frac{c^2}{2d^2} \psi '(c\alpha )\right] -\left[ \psi '(d\alpha +\beta ) - \frac{c^2}{2d^2} \psi '(c\alpha +2\beta )\right] \right\} \end{aligned}$$

To show that $\Lambda ''(\alpha )>0$ for $\alpha >0$, it suffices to show that

$$\begin{aligned} W(\alpha ) = \psi '(\alpha ) -(b^2/2)\psi '(b\alpha ) \end{aligned}$$

is positive and decreasing for $0\le b\le 2$

$$\begin{aligned} W(\alpha ) = \sum _{k=0}^{\infty }\ \frac{1}{(k+\alpha )^2} - \frac{1}{2}\sum _{k=0}^{\infty }\ \frac{1}{(k/b+\alpha )^2} \end{aligned}$$

Split the second sum with respect to even and odd indices to obtain

$$\begin{aligned} W(\alpha ) = \sum _{k=0}^{\infty }\ \frac{1}{(k+\alpha )^2} -\frac{1}{2}\left[ \sum _{k=0}^{\infty }\ \frac{1}{(2k/b +\alpha )^2} +\sum _{k=0}^{\infty }\ \frac{1}{((2k+1)/b +\alpha )^2}\right] \ > 0 \end{aligned}$$

Observe that $W'(\alpha ) < 0$ so that $W(\alpha )$ is decreasing This implies that $\Lambda '(\alpha )$ is increasing with the limit being zero at infinity and hence $\Lambda '(\alpha )$ is negative which means that $\Lambda (\alpha )$ is decreasing and completes the argument to show that $F(\alpha ,\beta )$ is decreasing as a function of $\alpha >0$. (For background on these estimates, see [31, 57]).