Appendix
1.1 A Stein like identity
The following lemma is a rewrite of Lemma 2.1 in Fourdrinier and Strawderman (2015).
Lemma A.1
Let (X, U) have a spherically symmetric distribution about \((\varvec{\theta },{\textbf{0}}) \in {\mathbb {R}}^p \times {\mathbb {R}}^k\) as in (1.1). For any function h(x, t) from \({{\mathbb {R}}}^p\times {{\mathbb {R}}}_{+}\) into \({{\mathbb {R}}}^p\) differentiable in x for any t, we have
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ (X-\varvec{\theta })^{\mathrm{\scriptscriptstyle T}}h(X,\Vert U\Vert ^2) \right] = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{ \Vert U\Vert ^2}{2} \int _0^1 \textrm{div}_X h(X,\Vert U\Vert ^2\, t) \, t^{k/2-1} \, dt \right] , \nonumber \\ \end{aligned}$$
(A.1)
provided either of these expectations exists. Here, \(\textrm{div}_X\) denotes the weak divergence operator with respect to X.
Thanks to Lemma A.1 we have the following result.
Corollary A.1
Let (X, U) have a spherically symmetric distribution about \((\varvec{\theta },{\textbf{0}}) \in {\mathbb {R}}^p \times {\mathbb {R}}^k\) as in (1.1). For any function H(x, t) from \({{\mathbb {R}}}^p\times {{\mathbb {R}}}_{+}\) into \({{\mathbb {R}}}\) twice weakly differentiable in x for any t, we have
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert X - \varvec{\theta }\Vert ^2 \, H(X,\Vert U\Vert ^2) \right]= & {} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ p \, \frac{\Vert U\Vert ^2}{2} \displaystyle \int _0^1 H(X,\Vert U\Vert ^2 \, t) \, t^{k/2-1} \, dt \right] \nonumber \\{} & {} + \, {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^4}{4} \displaystyle \int _0^1 {\varDelta }_X H(X,\Vert U\Vert ^2\,t) \,(1-t)\, t^{k/2-1} \, dt \right] \! \!, \nonumber \\ \end{aligned}$$
(A.2)
provided either of these expectations exists and, for any \(w \in {\mathbb {R}}_+\) and any \(x \in {\mathbb {R}}^p\), the following differentiation under the integral sign
$$\begin{aligned} \textrm{div}_x \int _0^1 \nabla _x H(x, w \, t) \, t^{k/2-1} \, dt = \int _0^1 {\varDelta }_x H(x, w \, t) \, t^{k/2-1} \, dt \end{aligned}$$
(A.3)
is valid. Here, \({{\varDelta }}_X\) denotes the weak Laplacian operator with respect to X.
Proof
Applying Lemma A.1 with
$$\begin{aligned} h(X,\Vert U\Vert ^2) = (X - \varvec{\theta }) \, H(X,\Vert U\Vert ^2), \end{aligned}$$
we have
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert X - \varvec{\theta }\Vert ^2 \, H(X,\Vert U\Vert ^2) \right]= & {} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^2}{2} \int _0^1 \textrm{div}_X \bigl \{(X - \varvec{\theta }) \, H(X,\Vert U\Vert ^2\, t)\bigr \} \, t^{k/2-1} \, dt \right] \nonumber \\= & {} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^2}{2} \int _0^1 (X - \varvec{\theta })^{\mathrm{\scriptscriptstyle T}}\, \nabla _X H(X,\Vert U\Vert ^2 \, t) \bigr \} \, t^{k/2-1} \, dt \right] \nonumber \\{} & {} + \, {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^2}{2} \int _0^1 p \, H(X,\Vert U\Vert ^2\, t) \, t^{k/2-1} \, dt \right] , \end{aligned}$$
(A.4)
according to the product rule for the divergence and since \(\textrm{div}_X (X - \varvec{\theta }) = p\). Now, the elimination of the parameter \(\varvec{\theta }\) in the first term of the right-hand side of (A.4) can be done again through the application of Lemma A.1 with
$$\begin{aligned} h(X,\Vert U\Vert ^2) = \frac{\Vert U\Vert ^2}{2} \int _0^1 \nabla _X H(X,\Vert U\Vert ^2 \, t) \, t^{k/2-1} \, dt. \end{aligned}$$
This leads to
$$\begin{aligned}{} & {} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^2}{2} \int _0^1 (X - \varvec{\theta })^{\mathrm{\scriptscriptstyle T}}\nabla _X H(X,\Vert U\Vert ^2 \, t) \bigr \} \, t^{k/2-1} \, dt \right] \\{} & {} \quad = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^2}{2} \int _0^1 \textrm{div}_X h(X,\Vert U\Vert ^2 s) \, s^{k/2-1} \, ds \right] \\{} & {} \quad = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^4}{4} \int _0^1 \textrm{div}_X \! \left\{ \int _0^1 \nabla _X H(X,\Vert U\Vert ^2 \,s \, t) \, t^{k/2-1} \, dt \right\} \, s^{k/2} \, ds \right] \\{} & {} \quad = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^4}{4} \int _0^1 \int _0^1 {\varDelta }_X H(X,\Vert U\Vert ^2 \,s \, t) \, t^{k/2-1} \, dt \, s^{k/2} \, ds \right] , \end{aligned}$$
thanks to (A.3) and the definition of the Laplacian. Then, setting \(\tau = st\) and applying Fubini’s theorem, we have
$$\begin{aligned}{} & {} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^2}{2} \int _0^1 (X - \varvec{\theta })^{\mathrm{\scriptscriptstyle T}}\nabla _X H(X,\Vert U\Vert ^2 \, t) \bigr \} \, t^{k/2-1} \, dt \right] \nonumber \\{} & {} \quad = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^4}{4} \int _0^1 \int _0^s {\varDelta }_X H(X,\Vert U\Vert ^2 \,\tau ) \, \tau ^{k/2-1} \, d\tau \, ds \right] \nonumber \\{} & {} \quad = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^4}{4} \int _0^1 {\varDelta }_X H(X,\Vert U\Vert ^2 \,\tau ) \, \tau ^{k/2-1} \int _\tau ^1\, ds \, d\tau \right] \nonumber \\{} & {} \quad = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^4}{4} \int _0^1 {\varDelta }_X H(X,\Vert U\Vert ^2 \,\tau ) \, \tau ^{k/2-1} \, (1-\tau ) \, d\tau \right] . \end{aligned}$$
(A.5)
Finally, the result follows from (A.4) and (A.5). \(\square \)
As an application of Corollary A.1, we have the following.
Proposition A.1
Under the density in (1.2), we have, as soon as \(k > q\),
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2}\Bigl [ \Vert X - \varvec{\theta }\Vert ^2 \, \Vert U\Vert ^{-q} \Bigr ] = \frac{p}{k-q} \, {\mathbb {E}}_{\varvec{\theta },\sigma ^2}\bigl [ \Vert U\Vert ^{2-q} \bigr ] \,, \end{aligned}$$
(A.6)
and, as soon as \(k > 2 \, q\),
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert X - \varvec{\theta }\Vert ^4 \, \Vert U\Vert ^{-2q} \right] = \frac{p \, (p + 2)}{(k - 2 \, q) \, (k - 2 \, q + 2)} \, {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{4 - 2 \, q} \right] , \end{aligned}$$
(A.7)
provided either of these expectations is finite.
Proof
First, assume that \({\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert X - \varvec{\theta }\Vert ^2 \, \Vert U\Vert ^{-q} \right] < \infty \). We can apply Corollary A.1 with \( H(X,\Vert U\Vert ^2) = \big ( \Vert U\Vert ^2 \big )^{\! -q} \), noticing that Condition (2.8) is satisfied since \( H(X,\Vert U\Vert ^2) \) does not depend on X. Then
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \left[ \Vert X - \varvec{\theta }\Vert ^ 2 \, \Vert U\Vert ^{-q} \right]= & {} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ p \, \frac{\Vert U\Vert ^2}{2} \int _0^1 \big ( \Vert U\Vert ^2 \, t \big )^{\! -q/2} \, t^{k/2-1} \, dt \right] \\= & {} \frac{p}{2} \, \int _0^1 t^{(k-q)/2-1} \, dt \, {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \big ( \Vert U\Vert ^2 \big )^{1-q/2} \right] \\= & {} \frac{p}{k - q} \, {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{2-q} \right] . \end{aligned}$$
Now, assume that \({\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert X - \varvec{\theta }\Vert ^4 \, \Vert U\Vert ^{-2q} \, \right] < \infty \). We can apply Corollary A.1 with \( H(X,\Vert U\Vert ^2) = \Vert X - \varvec{\theta }\Vert ^2 \, \big ( \Vert U\Vert ^2 \big )^{\! -q} \), noticing that Condition (2.8) is satisfied since \( H(X,\Vert U\Vert ^2) \) is the product of a function of X and a function of \( \Vert U\Vert ^2 \). Then
$$\begin{aligned}{} & {} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \left[ \Vert X - \varvec{\theta }\Vert ^4 \, \Vert U\Vert ^{-2q} \right] = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert X - \varvec{\theta }\Vert ^2 \, \big \{ \Vert X - \varvec{\theta }\Vert ^2 \, \Vert U\Vert ^{-2q} \big \} \right] \\{} & {} \quad = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ p \, \frac{\Vert U\Vert ^2}{2} \int _0^1 \Vert X - \varvec{\theta }\Vert ^2 \, \big ( \Vert U\Vert ^2 \, t \big )^{\! -q} \, t^{k/2-1} \, dt \right] \\{} & {} \qquad + {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^4}{4} \int _0^1 {\varDelta }_X \! \Big ( \Vert X - \varvec{\theta }\Vert ^2 \big ( \Vert U\Vert ^2 \, t \big )^{\! -q} \Big ) (1 - t) \, t^{k/2-1} \, dt \right] . \end{aligned}$$
Factorizing and expressing \({\varDelta }_X \! \Big (\Vert X - \varvec{\theta }\Vert ^2 \Big ) = 2 \, p\), it follows that
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert X - \varvec{\theta }\Vert ^4 \, \Vert U\Vert ^{-2q} \right]= & {} \frac{p}{k - 2 \, q} \, {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert X - \varvec{\theta }\Vert ^2 \, \Vert U\Vert ^{2-2q} \right] \nonumber \\{} & {} + \frac{2 \, p}{(k - 2 \, q) \, (k - 2 \, q + 2)} \, {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{4 - 2 \, q} \right] \nonumber \\ \end{aligned}$$
(A.8)
since
$$\begin{aligned} \int _0^1 t^{k/2-q-1} \, dt = \frac{2}{k - 2 \, q} \quad \text {and} \quad \int _0^1 (1 - t) \, t^{k/2-1} \, dt = \frac{4}{(k - 2 \, q) \, (k - 2 \, q + 2)} \,. \end{aligned}$$
Now, applying again Corollary A.1 to the first expectation on the right-hand side of (A.8) with \( H(X,\Vert U\Vert ^2) = \big ( \Vert U\Vert ^2 \big )^{\! 1-q} \), we obtain
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert X - \varvec{\theta }\Vert ^2 \, \Vert U\Vert ^{2-2q} \right]= & {} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ p \, \frac{\Vert U\Vert ^2}{2} \int _0^1 \big ( \Vert U\Vert ^2 \, t \big )^{\! 1-q} \, t^{k/2-1} \, dt \right] \nonumber \\= & {} \frac{p}{k - 2 \, q + 2} \, {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{4 - 2 \, q} \right] \end{aligned}$$
(A.9)
since
$$\begin{aligned} {\varDelta }_X \! \Big ( \Vert U\Vert ^2 \big )^{\! 1-q} \Big ) = 0 \quad \text {and} \quad \int _0^1 t^{k/2-q} \, dt = \frac{2}{k - 2 \, q + 2} \,. \end{aligned}$$
Hence it follows from (A.8) and (A.9) that
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert X - \varvec{\theta }\Vert ^4 \, \Vert U\Vert ^{-2q} \right] = \frac{p^2 + 2 \, p}{(k - 2 \, q) \, (k - 2 \, q + 2)} \, {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{4 - 2 \, q} \right] , \end{aligned}$$
which is is the desired result in (A.7).
To show (A.6) assuming \({\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{2-q} \right] < \infty \) (respectively (A.7) assuming \({\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{4-2\,q} \right] < \infty \)) it suffices essentially to reverse the steps in the above argument. \(\square \)
Remark 1
The finiteness conditions of the expectations in Proposition A.1 are weaker for \(q = 2\) than for \(q = 0\). Indeed, for \(q = 2\), the expectation in (A.6) is finite as soon as \(k > 3\) since it reduces to
$$\begin{aligned} \frac{p}{k - 2} \,, \end{aligned}$$
while, for \(q = 0\), it equals
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2}\Bigl [ \Vert X - \varvec{\theta }\Vert ^2 \Bigr ] = \frac{p}{k} \, {\mathbb {E}}_{\varvec{\theta },\sigma ^2}\bigl [ \Vert U\Vert ^{2} \bigr ] \,. \end{aligned}$$
Similarly, for \(q = 2\), the expectation in (A.7) is finite as soon as \(k > 4\) since it reduces to
$$\begin{aligned} \frac{p \, (p + 2)}{(k - 4) \, (k - 2)} \,, \end{aligned}$$
while, for \(q = 0\), it equals
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert X - \varvec{\theta }\Vert ^4 \right] = \frac{p \, (p + 2)}{k \, (k + 2)} \, {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{2} \right] . \end{aligned}$$
1.2 Proof of Theorem 2.1
We use repeatedly the Cauchy–Schwarz inequality to prove the risk finiteness of \( \delta _{q,0} \bigl (X,\Vert U\Vert ^2\bigr ) \) in (2.2). Clearly, it is equivalent to
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \left[ \ell _q^2(\varphi _{a,g},\varvec{\theta }) \right]< \infty \quad \text {and} \quad {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \delta _{q,0}^2 \bigl (X,\Vert U\Vert ^2\bigr ) \right] < \infty \end{aligned}$$
which is guaranteed if, respectively, one has
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert X - \varvec{\theta }\Vert ^4 \, \Vert U\Vert ^{-2q} \right]< \infty \quad \text {and} \quad {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{8-2q} \, \Vert g\bigl (X,\Vert U\Vert ^2\bigr )\Vert ^4 \right] < \infty \nonumber \\ \end{aligned}$$
(A.10)
and
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{4-2q} \right]< \infty \; \text {and} \; {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{8-2q} \, \left\{ \int _0^1 \!\! {\textrm{div}}_X g(X,\Vert U\Vert ^2 \, t) \, t^{(k-q)/2} \, dt \right\} ^{\! \! 2} \right] < \infty \,. \nonumber \\ \end{aligned}$$
(A.11)
According to (A.7) in Proposition A.1, the first condition in (A.10) is reduced to
$$\begin{aligned} k > 2 \, q \quad \text {and} \quad {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{4-2\,q} \right] < \infty \end{aligned}$$
which implies the first condition in (A.11). Note that the second condition in (A.10) implies the risk finiteness condition of \(\varphi _{a,g}\bigl (X,\Vert U\Vert ^2\bigr )\) in (2.4). Finally, under these above conditions which guarantee the finiteness of the risk of \( \delta _{q,0} \bigl (X,\Vert U\Vert ^2\bigr ) \) in (2.2), a risk finiteness condition for \( \delta _{q,\gamma } \bigl (X,\Vert U\Vert ^2\bigr ) \) in (2.5) is
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{8 - 2 \, q} \, \gamma ^2\bigl (X,\Vert U\Vert ^2\bigr ) \right] < \infty \,, \end{aligned}$$
also thanks to the Cauchy–Schwarz inequality.
Thus the finiteness of the risk difference between \( \delta _{q,\gamma } \bigl (X,\Vert U\Vert ^2\bigr ) \) and \( \delta _{q,0} \bigl (X,\Vert U\Vert ^2\bigr ) \) is guaranteed. It is expressed as
$$\begin{aligned} D_{\varvec{\theta },\sigma ^2}^q \!= & {} \! {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Bigl (\delta _{q,\gamma } \bigl (X,\Vert U\Vert ^2\bigr ) - \ell _q(\varphi _{a,g},\varvec{\theta }) \Bigr )^{\! 2} - \Bigl (\delta _{q,0} \bigl (X,\Vert U\Vert ^2\bigr ) - \ell _q(\varphi _{a,g},\varvec{\theta }) \Bigr )^{\! 2} \right] \\ \!= & {} \! {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \Vert U\Vert ^{8-2q} \, \gamma ^2\bigl (X,\Vert U\Vert ^2\bigr ) - 2 \, \Vert U\Vert ^{4-q} \, \gamma \bigl (X,\Vert U\Vert ^2\bigr ) \bigl ( \delta _{q,0} \bigl (X,\Vert U\Vert ^2\bigr ) - \ell _q(\varphi _{a,g},\varvec{\theta }) \bigr ) \right] , \nonumber \\ \end{aligned}$$
according to the expression of \(\delta _{q,\gamma }\bigl (X,\Vert U\Vert ^2\bigr ) \) in (2.5). Then, using \(\delta _{q,0} \bigl (X,\Vert U\Vert ^2\bigr )\) in (2.2) and \(\varphi _{a,g}\) in (2.1), and expressing the loss in (1.9), it follows that
$$\begin{aligned} D_{\varvec{\theta },\sigma ^2}^q= & {} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \bigl [ \bigr . \; \Vert U\Vert ^{8-2q} \, \gamma ^2\bigl (X,\Vert U\Vert ^2\bigr ) \nonumber \\{} & {} - 2 \, \tfrac{p}{k-q} \, \Vert U\Vert ^{6-2q} \, \gamma \bigl (X,\Vert U\Vert ^2\bigr ) \nonumber \\{} & {} - 2 \, a \, \Vert U\Vert ^{8-2q} \, \gamma \bigl (X,\Vert U\Vert ^2\bigr ) \int _0^1 {\textrm{div}}_X g(X,\Vert U\Vert ^2 \, t) \, t^{(k-q)/2} \, dt \nonumber \\{} & {} + 4 \, a \, \Vert U\Vert ^{6-2q} \, \gamma \bigl (X,\Vert U\Vert ^2\bigr ) \, (X - \varvec{\theta })^{{\mathrm{\scriptscriptstyle T}}} g\bigl (X,\Vert U\Vert ^2\bigr ) \nonumber \\{} & {} + 2 \, \Vert U\Vert ^{4-2q} \, \gamma \bigl (X,\Vert U\Vert ^2\bigr ) \, \Vert X - \varvec{\theta }\Vert ^2 \left. \right] , \end{aligned}$$
(A.12)
where the terms involving \(2 \, a^2 \, \Vert U\Vert ^{8-2q} \, \gamma \bigl (X,\Vert U\Vert ^2\bigr ) \bigl \Vert g\bigl (X,\Vert U\Vert ^2\bigr ) \bigr \Vert ^2\) have been simplified.
Note that, not only \(D_{\varvec{\theta },\sigma ^2}^q\) in (A.12) is finite, but also it can be shown, using again the Cauchy–Schwarz inequality (through appropriate factorizations), that the expectation of each of the four last integrand terms in (A.12) exists. We first replace the two last integrand terms in (A.12) by terms independent of \(\varvec{\theta }\). As for the penultimate term in (A.12), applying Lemma A.1 with
$$\begin{aligned} h(X,\Vert U\Vert ^2) = 4 \, a \, \Vert U\Vert ^{6-2q} \, \gamma \bigl (X,\Vert U\Vert ^2\bigr ) \, g\bigl (X,\Vert U\Vert ^2\bigr ), \end{aligned}$$
we have
$$\begin{aligned}{} & {} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ 4 \, a \, \Vert U\Vert ^{6-2q} \, \gamma \bigl (X,\Vert U\Vert ^2\bigr ) \, (X - \varvec{\theta })^{{\mathrm{\scriptscriptstyle T}}} g\bigl (X,\Vert U\Vert ^2\bigr ) \right] \nonumber \\{} & {} \quad = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ 2 \, a \, \Vert U\Vert ^{8-2q} \, \int _0^1 \textrm{div}_X \! \left\{ \gamma \bigl (X,\Vert U\Vert ^2 t\bigr ) \, g\bigl (X,\Vert U\Vert ^2 t\bigr ) \right\} \, t^{k/2-q+2} \, dt \right] . \nonumber \\ \end{aligned}$$
(A.13)
Also, to deal with the last term on the right-hand side of (A.12), we apply Corollary A.1 with
$$\begin{aligned} H(X,\Vert U\Vert ^2) = 2 \, \Vert U\Vert ^{4-2q} \, \gamma \bigl (X,\Vert U\Vert ^2\bigr )\,, \end{aligned}$$
thanks to Condition (2.8). This gives
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \left[ 2 \, \Vert U\Vert ^{4-2q} \, \gamma \bigl (X,\Vert U\Vert ^2\bigr ) \, \Vert X - \varvec{\theta }\Vert ^2 \right] = A_1 + A_2, \end{aligned}$$
(A.14)
with
$$\begin{aligned} A_1 = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ p \, \Vert U\Vert ^{6-2q} \, \int _0^1 \gamma \bigl (X,\Vert U\Vert ^2 t\bigr ) \, \, t^{k/2-q+1} \, dt \right] \end{aligned}$$
and
$$\begin{aligned} A_2 = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \frac{\Vert U\Vert ^{8-2q}}{2} \, \int _0^1 \Delta _X \! \left\{ \gamma \bigl (X,\Vert U\Vert ^2 t\bigr ) \right\} \, (1-t)\, t^{k/2-q+1} \, dt \right] . \end{aligned}$$
Consequently, according to (A.13) and (A.14), the risk difference in (A.12) can be written as
$$\begin{aligned} D_{\varvec{\theta },\sigma ^2}^q = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \Bigl [ \Vert U\Vert ^{8-2q}\left( \frac{A_3}{\Vert U\Vert ^2} + A_4\right) \Bigr ] \end{aligned}$$
(A.15)
with
$$\begin{aligned} A_3 = p \left( \int _0^1 \gamma \bigl (X,\Vert U\Vert ^2 t\bigr ) \, t^{k/2-q+1} \, dt - \frac{2}{k-q} \, \gamma \bigl (X,\Vert U\Vert ^2 \bigr ) \right) , \end{aligned}$$
(A.16)
and
$$\begin{aligned} A_4= & {} \gamma ^2\bigl (X,\Vert U\Vert ^2\bigr ) \; -\; 2 \, a \, \gamma \bigl (X,\Vert U\Vert ^2\bigr ) \int _0^1 {\textrm{div}}_X g(X,\Vert U\Vert ^2 \, t) \, t^{(k-q)/2} \, dt \nonumber \\{} & {} + \, 2 \, a \int _0^1 \textrm{div}_X \! \left\{ \! \gamma \bigl (X,\Vert U\Vert ^2 \, t \bigr ) \, g\bigl (X,\Vert U\Vert ^2 \, t\bigr ) \! \right\} t^{k/2-q+2} \, dt \nonumber \\{} & {} +\; \frac{1}{2} \, \int _0^1 {\Delta }_X \! \left\{ \! \gamma \bigl (X,\Vert U\Vert ^2 \, t\bigr ) \! \right\} (1-t) \, t^{k/2-q+1} \, dt. \end{aligned}$$
(A.17)
Replacing (A.16) and (A.17) in (A.15) gives that the risk difference between \( \delta _{q,\gamma } \bigl (X,\Vert U\Vert ^2\bigr ) \) and \( \delta _{q,0} \bigl (X,\Vert U\Vert ^2\bigr ) \) is
$$\begin{aligned} D_{\varvec{\theta },\sigma ^2}^q = {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \Bigl [ \Vert U\Vert ^{8-2q} \, {{{\mathscr {O}}}}\bigl (X,\Vert U\Vert ^2\bigr ) \Bigr ] \end{aligned}$$
with \({{{\mathscr {O}}}}\bigl (X,\Vert U\Vert ^2\bigr ) \) in (2.9), which is the first part of the theorem.
The second part is immediate.
1.3 Proof of Proposition 3.1
1.3.1 Checking Condition (3.8)
Note that, according to (3.8), the moment assumption \( {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \bigl [ \Vert U\Vert ^{4-q} \bigr ] < \infty \) in Theorem 2.1 is satisfied. Now we check the moment conditions involving g given at the beginning of Theorem 2.1. As according to (3.2), for any \(({{\textbf{x}}},{{\textbf{u}}}) \in {\mathbb {R}}^p \times {\mathbb {R}}^k\),
$$\begin{aligned} g({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2) = - \frac{2 \, (p-2)}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c} \, {{\textbf{x}}}\,, \end{aligned}$$
the moment condition \({\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \bigl [ \Vert U\Vert ^{8-2q} \, \Vert g\bigl (X,\Vert U\Vert ^2\bigr )\Vert ^4 \bigr ] < \infty \) is satisfied when \(q = 2\) since it reduces to
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \! \left( \frac{\Vert U\Vert ^{2}}{\Vert X\Vert ^2 + \Vert U\Vert ^2 + c} \right) ^{\! \! 2} \left( \frac{\Vert X\Vert ^2}{\Vert X\Vert ^2 + \Vert U\Vert ^2 + c} \right) ^{\! \!2} \right] < \infty \,. \end{aligned}$$
Also, when \(q = 0\), we have
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \left[ \! \left( \! \frac{\Vert U\Vert ^{2}}{\Vert X\Vert ^2 + \Vert U\Vert ^2 + c} \! \right) ^{\! \! \! 4} \Vert X\Vert ^4 \! \right] < \infty \,, \end{aligned}$$
which is satisfied as soon as \({\mathbb {E}}_{\varvec{\theta },\sigma ^2} [\Vert X\Vert ^4] < \infty \). This is the case since this last condition is equivalent to \({\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! [\Vert X - \varvec{\theta }\Vert ^4] < \infty \), and hence by sphericity of the distribution of (X, U), equivalent to \({\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \bigl [ \Vert U\Vert ^{4} \bigr ] < \infty \) (also according to (A.7)).
At the beginning of Sect. 3, we have addressed the issue of the weak differentiability of \(g({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2)\) in \({{\textbf{x}}}\in {\mathbb {R}}^p\); it weakly differentiable for \(p \ge 3\). Now, thanks to the product rule for the divergence, we can express
$$\begin{aligned} \textrm{div}_{{\textbf{x}}}g({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2)= & {} \textrm{div}_{{\textbf{x}}}\left( \frac{- 2 \, (p - 2)}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c} \, {{\textbf{x}}}\right) \nonumber \\= & {} - p \, \frac{2 \, (p - 2)}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2+ c} - {{\textbf{x}}}^{{\mathrm{\scriptscriptstyle T}}}\, \nabla _{{\textbf{x}}}\frac{1}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2+ c} \nonumber \\= & {} - p \, \frac{2 \, (p - 2)}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2+ c} - \, {{\textbf{x}}}^{{\mathrm{\scriptscriptstyle T}}} \, \frac{-2 \, {{\textbf{x}}}}{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2+ c\bigr )^2} \nonumber \\= & {} - \frac{2 \, (p - 2)}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2+ c} \left( \! p - 2 \, \frac{\Vert {{\textbf{x}}}\Vert ^2}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2+ c} \right) \end{aligned}$$
(A.18)
Then, according to (A.18), the last moment condition reduces to show that
$$\begin{aligned}{} & {} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \! \left\{ \Vert U\Vert ^{8 - 2 \, q} \int _0^1 \!\! {\textrm{div}}_X g(X,\Vert U\Vert ^2 \, t) \, t^{(k-q)/2} \, dt \right\} ^{\! \! 2} \right] \! \! \nonumber \\{} & {} \quad ={\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \! \left\{ \Vert U\Vert ^{8 - 2 \, q} \int _0^1 \frac{2 \, (p - 2)}{\Vert X\Vert ^2+ \Vert U\Vert ^2 \, t+ c} \left( \! p - 2 \, \frac{\Vert X\Vert ^2}{\Vert X\Vert ^2+ \Vert U\Vert ^2 \, t + c} \right) \, t^{(k-q)/2} \, dt \right\} ^{\! \! 2} \right] \! \! < \infty \,. \nonumber \\ \end{aligned}$$
(A.19)
First, note that the integrand term in (A.19) is non increasing in \(\Vert {{\textbf{u}}}\Vert ^2 \, t + c\) since, setting
$$\begin{aligned} m(s) = \frac{1}{\Vert {{\textbf{x}}}\Vert ^2+ s} \left( \! p - 2 \, \frac{\Vert {{\textbf{x}}}\Vert ^2}{\Vert {{\textbf{x}}}\Vert ^2+ s } \right) , \end{aligned}$$
it can be seen that its derivative satisfies
$$\begin{aligned} m^\prime (s) = - \frac{1}{(\Vert {{\textbf{x}}}\Vert ^2+ s )^2} \left( \! p - 4 \, \frac{\Vert {{\textbf{x}}}\Vert ^2}{\Vert {{\textbf{x}}}\Vert ^2+ s } \right) , \end{aligned}$$
which is non positive for \(p \ge 4\). Hence (A.19) is bounded above by
$$\begin{aligned} \left( \frac{4 \,(p-2) \, p}{k - q + 2} \right) ^{\! \! 2} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \! \left\{ \frac{\Vert U\Vert ^{8 - 2 \, q}}{\Vert X\Vert ^2} \right\} ^{\!\!2} \right] . \end{aligned}$$
(A.20)
It is pointed out in Fourdrinier and Strawderman (2015) that, by superharmocity of the function \({{\textbf{x}}}\mapsto {1}/{\Vert {{\textbf{x}}}\Vert ^2}\) for \(p \ge 4\), we have
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^{2}} \! \!\left[ \! \left\{ \frac{\Vert U\Vert ^{8 - 2 \, q}}{\Vert X\Vert ^2} \right\} ^{\!\!2} \right] \le {\mathbb {E}}_{0,\sigma ^2} \! \! \left[ \! \left\{ \frac{\Vert U\Vert ^{8 - 2 \, q}}{\Vert X\Vert ^2} \right\} ^{\!\!2} \right] \,, \end{aligned}$$
for any \(\varvec{\theta }\) in \({\mathbb {R}}^p\). Now we have
$$\begin{aligned} {\mathbb {E}}_{0,\sigma ^2} \! \! \left[ \! \left\{ \frac{\Vert U\Vert ^{8 - 2 \, q}}{\Vert X\Vert ^2} \right\} ^{\!\!2} \right] = {\mathbb {E}}\! \! \left[ \! \left( \frac{Z^{8-2q}}{(1 - Z)^2} \right) \big ( \Vert X\Vert ^2 + \Vert U\Vert ^2 \big )^{6-2q} \right] \,, \end{aligned}$$
where \(Z = \Vert U\Vert ^2 / \big ( \Vert X\Vert ^2 + \Vert U\Vert ^2 \big )\) has the beta distribution \({{{\mathscr {B}}}}(k/2,p/2)\) and is independent of \( \Vert X\Vert ^2 + \Vert U\Vert ^2\). Then
$$\begin{aligned} {\mathbb {E}}_{0,\sigma ^2} \! \! \left[ \! \left\{ \frac{\Vert U\Vert ^{8 - 2 \, q}}{\Vert X\Vert ^2} \right\} ^{\!\!2} \right] = \frac{B(k/2 - 8 + 2q,p/2-2) }{B(k/2,p/2)} \; {\mathbb {E}}_{0,\sigma ^2} \! \! \left[ \! \big ( \Vert X\Vert ^2 + \Vert U\Vert ^2 \big )^{6-2q} \right] \,, \end{aligned}$$
where, for any \(a > 0\) and \(b>0\), \(B(a,b) = {{\varGamma }(a) \, {\varGamma }(b)}/{{\varGamma }(a+b)}\). This expectation is finite if \(p > 4\), \(k > 16 - 4\, q\) and if the radius of the underlying distribution has a moment of order \(12-4\, q\) which are conditions in (3.8). Thus, under these conditions, the expectation in (A.20) is finite and hence, the expectations in (A.19) are finite as well.
The alternative estimator \( \delta _{q,\gamma }\bigl (X,\Vert U\Vert ^2\bigr ) \) is expressed through the loss correction \(\gamma \) in (3.7), that is,
$$\begin{aligned} \gamma ({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2 ) = b \, \frac{ 1 }{\Vert {{\textbf{x}}}\Vert ^2 + \Vert {{\textbf{u}}}\Vert ^2 + c} \,. \end{aligned}$$
(A.21)
The moment condition \({\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \big [ \! \Vert U\Vert ^{8 - 2 \, q} \, \gamma ^2\bigl (X,\Vert U\Vert ^2\bigr ) \! \big ]\) \(< \infty \) in Theorem 2.1 is satisfied when \(q = 2\) since it is expressed as
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \! \left[ \! \left( \frac{\Vert U\Vert ^{2}}{\Vert X\Vert ^2 + \Vert U\Vert ^2 + c} \right) ^{\! \! 2} \right] < \infty \,. \end{aligned}$$
Also, when \(q = 0\), we have
$$\begin{aligned} {\mathbb {E}}_{\varvec{\theta },\sigma ^2} \! \left[ \! \left( \! \frac{\Vert U\Vert ^{2}}{\Vert X\Vert ^2 + \Vert U\Vert ^2 + c} \! \right) ^{\! \! \! 2} \Vert U\Vert ^4 \! \right] < \infty \,, \end{aligned}$$
which is satisfied as soon as \({\mathbb {E}}_{\varvec{\theta },\sigma ^2} [\Vert U\Vert ^4] < \infty \), which is implied by (3.8).
We have mentioned, at the beginning of Sect. 3, that \(\gamma ({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2 )\) is twice weakly differentiable as a function of \({{\textbf{x}}}\) for \(p \ge 5\). In that case, its weak Laplacian can be calculated as follows. Indeed its weak gradient with respect to x is
$$\begin{aligned} \nabla _{{\textbf{x}}}\gamma ({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2 ) = - b \, \frac{ 2 }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2+ c\bigr )^2} \, {{\textbf{x}}}\end{aligned}$$
(A.22)
which, taking the divergence of this gradient, gives rise to the Laplacian
$$\begin{aligned}{} & {} {{\varDelta }}_{{\textbf{x}}}\gamma ({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2 )\nonumber \\ {}{} & {} = - p \, \frac{2 \, b }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2+ c\bigr )^2} + {{\textbf{x}}}^{{\mathrm{\scriptscriptstyle T}}} \nabla _{{\textbf{x}}}\frac{- 2 \, b }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2+ c\bigr )^2} \nonumber \\= & {} - p \, \frac{2 \, b }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2+ c\bigr )^2} + {{\textbf{x}}}^{{\mathrm{\scriptscriptstyle T}}} \frac{8 \, b \, {{\textbf{x}}}}{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c\bigr )^3} \nonumber \\= & {} -\frac{2 \, b }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c\bigr )^2} \left( p - 4 \, \frac{ \Vert {{\textbf{x}}}\Vert ^2 }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c}\right) . \end{aligned}$$
(A.23)
Also the divergence of the product of \(\gamma ({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2 )\) and \(g ({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2 )\) is
$$\begin{aligned}{} & {} \textrm{div}_{{\textbf{x}}}\bigl (\gamma ({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2 ) \, g({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2) \bigr ) \nonumber \\ {}{} & {} = \frac{b }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c} \left( \! \frac{- 2 \, (p - 2)}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c} \left( \! p - \, \frac{2 \, \Vert {{\textbf{x}}}\Vert ^2}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c} \! \right) \! \! \right) \nonumber \\{} & {} \quad + \frac{- 2 \, b }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c\bigr )^2} \, {{\textbf{x}}}^{\mathrm{\scriptscriptstyle T}}\frac{ - 2 \, (p - 2)}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c} \, {{\textbf{x}}}\nonumber \\= & {} - 2 \, (p - 2) \, \frac{ b }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c\bigr )^2} \left( \! p - 4 \, \frac{ \Vert {{\textbf{x}}}\Vert ^2}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c} \! \right) . \end{aligned}$$
(A.24)
We may note in passing that, for this example,
$$\begin{aligned} (p - 2) \, {{\varDelta }}_{{\textbf{x}}}\gamma ({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2 ) = \textrm{div}_{{\textbf{x}}}\bigl (\gamma ({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2 ) \, g({{\textbf{x}}}, \Vert {{\textbf{u}}}\Vert ^2) \bigr ) \,. \end{aligned}$$
It remains to check Condition (2.8). According to (A.22), this reduces to verify, for any \(i =1, \ldots , p\) and for any \(w \in {\mathbb {R}}_+\), the following derivation under the integral sign:
$$\begin{aligned} \frac{\partial }{\partial x_i} \int _0^1 \frac{ x_i }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ w \, t + c\bigr )^{\! 2}} \, t^{k/2-q+1} \, dt = \int _0^1 \frac{\partial }{\partial x_i} \, \frac{ x_i }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ w \, t + c\bigr )^{\! 2}} \, t^{k/2-q+1} \, dt \,. \nonumber \\ \end{aligned}$$
(A.25)
The left-hand side of (A.25) equals
$$\begin{aligned}{} & {} \frac{\partial }{\partial x_i} \! \left( \! x_i \int _0^1 \frac{ t^{k/2-q+1} }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ w \, t + c\bigr )^{\! 2}} \, dt \! \right) = \int _0^1 \frac{ t^{k/2-q+1} }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ w \, t + c\bigr )^{\! 2}} \, dt \nonumber \\ {}{} & {} \qquad \qquad \quad + x_i \, \frac{\partial }{\partial x_i} \! \int _0^1 \frac{ t^{k/2-q+1} }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ w \, t + c\bigr )^{\! 2}} \, dt \,, \end{aligned}$$
(A.26)
since the Lebesgue dominated convergence theorem applies for \(k > 2 \, q\), in so far as
$$\begin{aligned} \forall {{\textbf{x}}}\in {\mathbb {R}}^p \quad \forall t \in [0,1] \quad \frac{ t^{k/2-q+1} }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ w \, t + c\bigr )^{\! 2}} \le \frac{ t^{k/2-q-1} }{w^2} \,. \end{aligned}$$
Thus the conditions of Theorem 2.1 are satisfied.
1.3.2 Checking (3.11)
We can see that, using (A.18), (A.21), (A.23) and (A.24), the term \({{{\mathscr {O}}}}\bigl ({{\textbf{x}}},\Vert {{\textbf{u}}}\Vert ^2\bigr )\) in (2.9) equals
$$\begin{aligned}{} & {} \left( \! \frac{ 1 }{\Vert {{\textbf{x}}}\Vert ^2 + \Vert {{\textbf{u}}}\Vert ^2 + c } \! \right) ^{\! \! \! \! ^2} b^2 \nonumber \\{} & {} \quad + \frac{p}{\Vert {{\textbf{u}}}\Vert ^2} \, b \left( \int _0^1 \frac{ 1 }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \, t^{k/2-q+1} \, dt - \frac{2}{k - q} \, \frac{ 1 }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c } \right) \nonumber \\{} & {} \quad + 2 \, \alpha \, \frac{ 1 }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c } \, b \int _0^1 \frac{1}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \left( \! p - 2 \, \frac{\Vert {{\textbf{x}}}\Vert ^2}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \! \right) t^{(k-q)/2} \, dt \nonumber \\{} & {} \quad - \, b \int _0^1 \frac{ 1 }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c\bigr )^{\!2}} \left( \! p - 4 \, \frac{ \Vert {{\textbf{x}}}\Vert ^2}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \! \right) (2 \, \alpha \, t + 1 - t) \, t^{k/2-q+1} \, dt, \nonumber \\ \end{aligned}$$
(A.27)
which corresponds to the expression in (3.10).
We will now show that the polynomial in (A.27) is non positive. For any \(t \in [0,1]\), as for the third term in (A.27), we can write
$$\begin{aligned} \frac{1}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \left( p - 2 \, \frac{\Vert {{\textbf{x}}}\Vert ^2}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \right) \le \frac{p}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \,, \end{aligned}$$
and, as for the fourth term,
$$\begin{aligned} \begin{array}{lcl} &{} - \displaystyle \frac{ 1 }{\bigl (\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c\bigr )^2} \left( p - 4 \, \frac{ \Vert {{\textbf{x}}}\Vert ^2}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \right) \nonumber \\ &{} \le \displaystyle \frac{ - ( p - 4 ) }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c} \, \frac{ 1 }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \end{array} \end{aligned}$$
Then the polynomial in (A.27) is bounded from above by
$$\begin{aligned}{} & {} \left( \frac{ 1 }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c } \right) ^{\! \! \! \! ^2} b^2 \\ {}{} & {} + \frac{p}{\Vert {{\textbf{u}}}\Vert ^2} \, b \left( \int _0^1 \frac{ 1 }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \, t^{k/2-q+1} \, dt - \frac{2}{k - q} \, \frac{ 1 }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c } \right) \\{} & {} \quad + 2 \, \alpha \, p \, \frac{ 1 }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c } \, b \int _0^1 \frac{t^{(k-q)/2}}{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \, dt \\{} & {} \quad - \, \frac{ p - 4 }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c} {\tiny {\tiny }} \, b \int _0^1 \frac{ 1 }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \, (2 \, \alpha \, t+1-t)\, t^{k/2-q+1} \, dt, \end{aligned}$$
so that the expression in (A.27) will be non positive as soon as
$$\begin{aligned}{} & {} 0 \le b \le \frac{p}{\Vert {{\textbf{u}}}\Vert ^2} \left( \! \frac{2}{k - q} \, \big ( \Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c \big ) - \big (\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c\big )^{\! 2}\right. \nonumber \\ {}{} & {} \left. \int _0^1 \frac{ 1 }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \, t^{k/2-q+1} \, dt \! \right) \nonumber \\{} & {} \quad - 2 \, \alpha \, p \, \big ( \Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c \big ) \int _0^1 \frac{t^{(k-q)/2}}{ \Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c } \, dt \nonumber \\{} & {} \quad + \, ( p - 4 ) \, \big ( \Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c \big ) \int _0^1 \frac{ 1 }{\Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c} \, (2 \, \alpha \, t + 1 - t) \, t^{k/2-q+1} \, dt \,. \nonumber \\ \end{aligned}$$
(A.28)
Using the fact that, for any \(t \in (0,1]\),
$$\begin{aligned} \frac{ \Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c }{ \Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c } \le \frac{1}{t} \,, \end{aligned}$$
(A.29)
the second term in parentheses on the right-hand side of (A.28) is bounded below by
$$\begin{aligned} - \big ( \Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c \big ) \int _0^1 \, t^{k/2-q} \, dt = - \frac{2}{k - 2 \, q + 2} \, \big ( \Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c \big ) \,. \end{aligned}$$
Hence the sum of the two first terms in (A.28) is bounded below by
$$\begin{aligned} 2 \, p \, \frac{ \Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c }{ \Vert {{\textbf{u}}}\Vert ^2 } \left( \! \frac{1}{k-q} - \frac{1}{k-2 q+2} \! \right) \ge \frac{2 \, p \, (2-q)}{(k - q) (k - 2 q + 2)} \,. \end{aligned}$$
(A.30)
Now, using again (A.29), the third term in the right-hand side of (A.28) is bounded from below by
$$\begin{aligned} - 2 \,p \, \alpha \, \int _0^1 t^{(k-q)/2 - 1} \, dt = - \frac{ 4 \, p }{ k - q} \, \alpha \,. \end{aligned}$$
(A.31)
Also, as
$$\begin{aligned} \frac{ \Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 + c }{ \Vert {{\textbf{x}}}\Vert ^2+ \Vert {{\textbf{u}}}\Vert ^2 \, t + c } \ge 1 \,, \end{aligned}$$
the fourth term on the right-hand side of (A.28) is bounded from below by
$$\begin{aligned}{} & {} (p - 4) \int _0^1 (2 \, \alpha \, t + 1-t)\, t^{ k/2 -q + 1} \, dt \nonumber \\{} & {} = (p - 4) \left[ \frac{ 4 }{ k - 2 \, q + 6 } \, \alpha + \frac{ 4 }{ (k - 2 \, q + 4) \, (k - 2 \, q + 6) } \right] \nonumber \\{} & {} = \frac{4 \, ( p - 4 )}{ k - 2 \, q + 6 } \left[ \alpha + \frac{ 1 }{ k - 2 \, q + 4 } \right] . \end{aligned}$$
(A.32)
Therefore, according to (A.30), (A.31) and (A.32), the double inequality in (A.28) is satisfied as soon as
$$\begin{aligned} 0\le & {} b \le \displaystyle 4 \left( \! \frac{p\, (2-q)}{ 2 (k - q) \, (k - 2\, q + 2)} - \frac{ p }{ k - q } \, \alpha + \frac{ p - 4 }{ k - 2 \, q + 6 } \left[ \! \alpha + \frac{ 1 }{ k - 2 \, q + 4 } \! \right] \! \right) \nonumber \\= & {} \displaystyle 4 \left( \! \left[ \! \frac{ p - 4 }{k - 2 \, q + 6 } - \frac{ p }{ k-q } \! \right] \! \alpha + \frac{ p \, (2 - q) }{2 \, (k - q ) \, (k - 2 \, q + 2) } + \frac{ p - 4 }{(k - 2 \, q + 4) \, (k - 2 \, q + 6) } \! \right) \!, \nonumber \\ \end{aligned}$$
(A.33)
which is (3.11). Thus Proposition 3.1 is proved.
