Proof By (3.4) and a change of variable, we find

(3.7) $$ \begin{align} \iint\limits_{M\times M}\frac{\mathrm{d}\rho(x) \mathrm{d}\rho(y)}{d(x, y)^\lambda} & \leq \iint\limits_{M\times M}\frac{\mathrm{d}\rho(x) \mathrm{d}\rho(y)}{|f(x)-f(y)|^\lambda} \nonumber \\ & =\iint\limits_{T_o M\times T_o M}\frac{\mathrm{d} f_\#\rho(u)\, \mathrm{d} f_\#\rho(v)}{|u-v|^\lambda}, \end{align} $$

where $f_\#\rho \in \mathcal {P}_{ac}(T_o M)$ denotes the pushforward (as measures) of $\rho $ by f. Note that

(3.8) $$ \begin{align} f_\#\rho(u)=\rho(f^{-1}(u))|J(f^{-1})(u)|, \qquad \text{ for all } u\in T_o M, \end{align} $$

where $J(f^{-1})$ denotes the Jacobian of the map $f^{-1}$ . The inverse map $f^{-1}:T_o M\to M$ is the Riemannian exponential map

$$ \begin{align*} f^{-1}(u)=\exp_o u, \qquad \text{ for all } u\in T_o M, \end{align*} $$

whose Jacobian was discussed in Section 2.2.

Since $T_o M$ can be identified with $\mathbb R^n $ , we can apply (3.2) to get

(3.9) $$ \begin{align} \iint\limits_{T_o M\times T_o M}\frac{\mathrm{d} f_\#\rho(u)\, \mathrm{d} f_\#\rho(v)}{|u-v|^\lambda}&\leq C(\lambda, n) \int_{T_o M}(f_\#\rho(u))^{m_c}\mathrm{d} u \nonumber \\ &=C(\lambda, n)\int_M |J(f^{-1})(f(x))|^{m_c-1}\rho(x)^{m_c} \mathrm{d} x, \end{align} $$

where and for the second line we used (3.8) and the change of variable $u = f(x)$ . We also note here since $\rho $ is compactly supported and in $L^{m_c}(M)$ , $f_\#\rho $ is in $L^{m_c}(T_o M)$ . This follows from the equality in (3.9), given that $|J(f^{-1})|$ is bounded on compact sets.

In (3.9), we use the upper bound on the Jacobian of the exponential map in Theorem 2.3 (note that $\|\log _o x\| = \theta _x$ ) to get

(3.10) $$ \begin{align} \iint\limits_{T_o M\times T_o M}\frac{\mathrm{d} f_\#\rho(u)\, \mathrm{d} f_\#\rho(v)}{|u-v|^\lambda}\leq C(\lambda, n)\int_M\left(\frac{\sinh(\sqrt{c_m} \theta_x)}{\sqrt{c_m} \theta_x}\right)^{(n-1)(m_c-1)} \rho(x)^{m_c} \mathrm{d} x. \end{align} $$

Finally, combine (3.7) and (3.10) to obtain (3.6).

Proof If $y\in \mathcal {S}_x$ , then we have $d(x, y)<6r$ , which implies $y\in B_{6r}(x)$ . Hence,

$$\begin{align*}\mathcal{S}_x\subset B_{6r}(x). \end{align*}$$

We also infer from here that for any $y \in \mathcal {S}_x\subset B_{6r}(x)$ , $B_r(y)$ is included in $B_{7r}(x)$ . Also note that by construction, $\{B_r(y)\}_{y\in \mathcal {S}_x}$ are disjoint. Hence, conclude that $\{B_r(y)\}_{y\in \mathcal {S}_x}$ are disjoint sets in $B_{7r}(x)$ . From this observation, we get

$$\begin{align*}\#(\mathcal{S}_x)\inf_{z\in M}|B_r(z)|\leq\sum_{y\in\mathcal{S}_x}|B_r(y)|\leq |B_{7r}(x)|\leq \sup_{z\in M}|B_{7r}(z)|. \end{align*}$$

We then find

(3.15) $$ \begin{align} \#(\mathcal{S}_x)\leq \frac{\sup_{z\in M}|B_{7r}(z)|}{\inf_{z\in M}|B_r(z)|}. \end{align} $$

From the comparison with volumes of balls in $\mathbb R^n$ (see (2.5)), we have

$$\begin{align*}\inf_{z\in M}|B_r(z)|\geq w(n)r^n, \end{align*}$$

while from the lower bound of curvatures and Theorem 2.4, we get

$$\begin{align*}\sup_{z\in M}|B_{7r}(z)|\leq n w(n)\int_0^{7r}\left(\frac{\sinh(\sqrt{c_m}s)}{\sqrt{c_m}}\right)^{n-1}\mathrm{d} s. \end{align*}$$

Using these bounds in (3.15), we find (3.13).

Proof Take $x, y$ to be two points in M which satisfy $d(x, y)<r$ . Since $x\in M$ and $M=\bigcup _{z\in \mathcal {C}}B_{5r}(z)$ , there exists $z\in \mathcal {C}$ such that $x\in B_{5r}(z)$ . Then,

$$\begin{align*}d(z, y)\leq d(z, x)+d(x, y)<5r+r=6r, \end{align*}$$

which implies $y\in B_{6r}(z)$ . Since $x\in B_{5r}(z)\subset B_{6r}(z)$ , we have

$$\begin{align*}(x, y)\in B_{6r}(z)\times B_{6r}(z). \end{align*}$$

This implies

(3.17) $$ \begin{align} \{(x, y)\in M\times M: d(x, y)<r\}\subset \bigcup_{z\in \mathcal{C}}B_{6r}(z)\times B_{6r}(z). \end{align} $$

Use (3.17) to estimate

(3.18) $$ \begin{align} \iint\limits_{M\times M}\frac{\mathrm{d}\rho(x)\mathrm{d}\rho(y)}{d(x, y)^\lambda}& = \iint\limits_{d(x, y)\geq r}\frac{\mathrm{d}\rho(x)\mathrm{d}\rho(y)}{d(x, y)^\lambda}+\iint\limits_{d(x, y)<r}\frac{\mathrm{d}\rho(x)\mathrm{d}\rho(y)}{d(x, y)^\lambda} \nonumber \\ &\leq r^{-\lambda}+\sum_{z\in\mathcal{C}}\iint\limits_{B_{6r}(z)\times B_{6r}(z)}\frac{\mathrm{d}\rho(x)\mathrm{d}\rho(y)}{d(x, y)^\lambda}. \end{align} $$

We remove all elements $z\in \mathcal {C}$ which satisfy

$$\begin{align*}\int_{B_{6r}(z)}\mathrm{d} \rho(x)=0, \end{align*}$$

and define $\mathcal {C}' \subset \mathcal {C}$ by

$$\begin{align*}\mathcal{C}'=\mathcal{C}\cap\left\{z\in M:\int_{B_{6r}(z)}\mathrm{d} \rho(x)>0\right\}. \end{align*}$$

Then, by (3.18) we have

(3.19) $$ \begin{align} \iint\limits_{M\times M}\frac{\mathrm{d}\rho(x)\mathrm{d}\rho(y)}{d(x, y)^\lambda}&\leq r^{-\lambda}+\sum_{z\in\mathcal{C}'}\iint\limits_{B_{6r}(z)\times B_{6r}(z)}\frac{\mathrm{d}\rho(x)\mathrm{d}\rho(y)}{d(x, y)^\lambda}. \end{align} $$

Now, for all $z\in \mathcal {C}'$ we consider the normalized restrictions of measure $\rho $ on $B_{6r}(z)$ , given by

$$\begin{align*}\rho_z(x)=\begin{cases} \displaystyle\frac{\rho(x)}{\|\rho\|_{L^1(B_{6r}(z))}}\qquad&\text{ when }x\in B_{6r}(z),\\[10pt] \displaystyle 0&\text{ otherwise}. \end{cases} \end{align*}$$

Note that $\rho _z$ is compactly supported, for all z. Fix $z \in \mathcal {C}'$ and substitute $\rho _z$ into (3.6) with z as pole (use $\rho _z$ in place of $\rho $ , and z in place of o in Proposition 3.1 – see also Remark 3.1), to get

(3.20) $$ \begin{align} & \frac{1}{\|\rho\|_{L^1(B_{6r}(z))}^2}\iint\limits_{B_{6r}(z)\times B_{6r}(z)}\frac{\mathrm{d}\rho(x) \mathrm{d}\rho(y)}{d(x, y)^\lambda}= \iint\limits_{M\times M}\frac{\mathrm{d}\rho_z(x) \mathrm{d}\rho_z(y)}{d(x, y)^\lambda} \nonumber \\ &\hspace{2cm} \leq C(\lambda, n)\int_M \left(\frac{\sinh(\sqrt{c_m} d(x, z))}{\sqrt{c_m} d(x, z)}\right)^{(n-1)(m_c-1)}\rho_z(x)^{m_c} \mathrm{d} x \nonumber \\ &\hspace{2cm} =\frac{C(\lambda, n)}{\|\rho\|_{L^1(B_{6r}(z))}^{m_c}}\int_{B_{6r}(z)}\left(\frac{\sinh(\sqrt{c_m}d(x, z))}{\sqrt{c_m}d(x, z)}\right)^{(n-1)(m_c-1)}\rho(x)^{m_c}\mathrm{d} x. \end{align} $$

Since

$$\begin{align*}\frac{\mathrm{d}}{\mathrm{d}\theta}\left(\frac{\sinh\theta}{\theta}\right)= \frac{\cosh\theta}{\theta^2}(\theta-\tanh\theta), \end{align*}$$

and $\theta \geq \tanh \theta $ for all $\theta \geq 0$ , we know that $\frac {\sinh \theta }{\theta }$ is non-decreasing for all $\theta \geq 0$ . From this property, we get

$$\begin{align*}\frac{\sinh(\sqrt{c_m}d(x, z))}{\sqrt{c_m}d(x, z)}\leq \frac{\sinh(6\sqrt{c_m}r)}{6\sqrt{c_m}r}, \qquad \text{ for all } x\in B_{6r}(z). \end{align*}$$

Then, multiply (3.20) by $\|\rho \|_{L^1(B_{6r}(z))}^2$ and use the inequality above to find

$$\begin{align*}\iint\limits_{B_{6r}(z)\times B_{6r}(z)}\frac{\mathrm{d}\rho(x) \mathrm{d}\rho(y)}{d(x, y)^\lambda}\leq \frac{C(\lambda, n)}{\|\rho\|_{L^1(B_{6r}(z))}^{m_c-2}}\left(\frac{\sinh(6\sqrt{c_m}r)}{6\sqrt{c_m}r}\right)^{(n-1)(m_c-1)}\int_{B_{6r}(z)}\rho(x)^{m_c}\mathrm{d} x. \end{align*}$$

Furthermore, use $\|\rho \|_{L^1(B_{6r}(z))}\leq \|\rho \|_{L^1(M)}=1$ and $m_c-2=\frac {\lambda }{n}-1<0$ , to get

(3.21) $$ \begin{align} \iint\limits_{B_{6r}(z)\times B_{6r}(z)}\frac{\mathrm{d}\rho(x) \mathrm{d}\rho(y)}{d(x, y)^\lambda}\leq C(\lambda, n)\left(\frac{\sinh(6\sqrt{c_m}r)}{6\sqrt{c_m}r}\right)^{(n-1)(m_c-1)}\int_{B_{6r}(z)}\rho(x)^{m_c}\mathrm{d} x. \end{align} $$

Finally, use (3.21) in (3.19) to get

(3.22) $$ \begin{align} \iint\limits_{M\times M}\frac{\mathrm{d}\rho(x)\mathrm{d}\rho(y)}{d(x, y)^\lambda} \leq r^{-\lambda}+C(\lambda, n)\left(\frac{\sinh(6\sqrt{c_m}r)}{6\sqrt{c_m}r}\right)^{(n-1)(m_c-1)}\sum_{z\in \mathcal{C}'}\int_{B_{6r}(z)}\rho(x)^{m_c}\mathrm{d} x. \end{align} $$

By Lemma 3.2, for any $x\in M$ there are at most $N(n,c_m, r)$ balls $\{B_{6r}(z)\}_{z\in \mathcal {C}'}$ which contain x. This implies

$$\begin{align*}\sum_{z\in\mathcal{C}'}\chi_{B_{6r}(z)}\leq N(n,c_m, r)\chi_{M}, \end{align*}$$

where $\chi _A$ is the characteristic function of the set A, defined as

$$\begin{align*}\chi_A(x)=\begin{cases} 1&\quad\text{when }x\in A,\\ 0&\quad\text{when }x\not\in A. \end{cases} \end{align*}$$

Then, we have

(3.23) $$ \begin{align} \sum_{z\in\mathcal{C}'}\int_{B_{6r}(z)}\rho(x)^{m_c}\mathrm{d} x&=\sum_{z\in\mathcal{C}'}\int_{M}\chi_{B_{6r}(z)}(x)\rho(x)^{m_c}\mathrm{d} x=\int_M\left(\sum_{z\in\mathcal{C}'}\chi_{B_{6r}(z)}(x)\right)\rho(x)^{m_c}\mathrm{d} x \nonumber \\ &\leq \int_M N(n,c_m, r)\rho(x)^{m_c}\mathrm{d} x=N(n,c_m, r)\int_M \rho(x)^{m_c}\mathrm{d} x. \end{align} $$

Finally, using (3.23) in (3.22), we arrive at

(3.24) $$ \begin{align} \iint\limits_{M\times M}\frac{\mathrm{d}\rho(x)\mathrm{d}\rho(y)}{d(x, y)^\lambda}\leq r^{-\lambda}+C(\lambda, n)\left(\frac{\sinh(6\sqrt{c_m}r)}{6\sqrt{c_m}r}\right)^{(n-1)(m_c-1)}N(n,c_m, r)\int_M \rho(x)^{m_c}\mathrm{d} x. \end{align} $$

For $m>m_c$ , the Hölder inequality with $p=\frac {m-1}{m-m_c}$ and $q=\frac {m-1}{m_c-1}$ , yields

$$\begin{align*}\int_M\rho(x)^{m_c}\mathrm{d} x\leq \left(\int_M\rho(x)\mathrm{d} x\right)^{\frac{m-m_c}{m-1}} \left(\int_M\rho(x)^{m}\mathrm{d} x\right)^{\frac{m_c-1}{m-1}}. \end{align*}$$

Since $\rho $ integrates to $1$ , this inequality together with (3.24) lead to the conclusion of the theorem.

Proof Fix $0\leq \beta <\min (n(m-1), n)$ , and define

$$\begin{align*}\tilde{\alpha}=\max\left(\frac{1}{2}(\alpha+\min(n(m-1), n)), \beta\right), \end{align*}$$

where $\alpha $ is the constant from (4.1). Then, $0<\alpha <\tilde {\alpha }<\min (n(m-1), n)$ , and hence

$$\begin{align*}\lim_{\theta\to0+}h(\theta)\theta^{\tilde{\alpha}}=\lim_{\theta\to0+}\left(h(\theta)\theta^\alpha\times \theta^{\tilde{\alpha}-\alpha}\right)=0. \end{align*}$$

From the assumptions on h (see (H), (2.1) and (4.1)) and Lemma 2.1, there exist ${\gamma _1,\gamma _2\geq 0}$ such that

$$\begin{align*}h(\theta)\geq -\gamma_1\theta^{-\tilde{\alpha}}-\gamma_2,\qquad\forall \theta>0. \end{align*}$$

Then, given that $\{\rho _k\}_{k\in \mathbb {N}}$ is a minimizing sequence, we have

(4.3) $$ \begin{align} \begin{aligned} E[\rho_1] &\geq E[\rho_k] \\ & = \frac{1}{m-1}\int_{{\mathbb H}^n}\rho_k(x)^m\mathrm{d} x+\frac{1}{2}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y \\ & \geq\frac{1}{m-1}\int_{{\mathbb H}^n}\rho_k(x)^m\mathrm{d} x-\frac{\gamma_1}{2}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}\frac{\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y}{d(x, y)^{\tilde{\alpha}}} - \frac{\gamma_2}{2}. \end{aligned} \end{align} $$

Fix $r>0$ arbitrary. By using the HLS inequality on Cartan–Hadamard manifolds (Theorem 3.2) for $M={\mathbb H}^n$ , $c_m=1$ and $\lambda =\tilde {\alpha ,}$ we find

(4.4) $$ \begin{align} \iint\limits_{{\mathbb H}^n\times {\mathbb H}^n}\frac{\mathrm{d} \rho_k(x)\mathrm{d} \rho_k(y)}{d(x, y)^{\tilde{\alpha}}}\leq r^{-\tilde{\alpha}}+\tilde{C}(\tilde{\alpha}, n, r, 1)\left(\int_{{\mathbb H}^n} \rho_k(x)^m\mathrm{d} x\right)^{\frac{\tilde{\alpha}}{n(m-1)}}. \end{align} $$

Combine (4.3) and (4.4) to get

$$\begin{align*}E[\rho_1]\geq \frac{1}{m-1}\int_{{\mathbb H}^n}\rho_k(x)^m\mathrm{d} x-\frac{\gamma_1\tilde{C}(\tilde{\alpha}, n, r, 1)}{2}\left(\int_{{\mathbb H}^n} \rho_k(x)^m\mathrm{d} x\right)^{\frac{\tilde{\alpha}}{n(m-1)}}-\frac{1}{2}(\gamma_1 r^{-\tilde{\alpha}}+\gamma_2). \end{align*}$$

Since $0<\frac {\tilde {\alpha }}{n(m-1)}<1$ , if $\int _{{\mathbb H}^n}\rho _k(x)^m\mathrm {d} x$ tends to $\infty $ as $k \to \infty $ , then the right hand side of the above inequality also tends to $\infty $ . Therefore, we conclude that $ \int _{{\mathbb H}^n}\rho _k(x)^m\mathrm {d} x $ is bounded from above uniformly in $k\in \mathbb {N}$ , which shows the statement for the first term in (4.2); note that a uniform lower bound is immediate, as the expressions are non-negative.

Given that $ \int _{{\mathbb H}^n}\rho _k(x)^m\mathrm {d} x $ are uniformly bounded from above, from (4.4), we then get the uniform boundedness in $k \in \mathbb {N}$ of $\iint\limits _{{\mathbb H}^n \times {\mathbb H}^n}\frac {\rho _k(x)\rho _k(y)\mathrm {d} x\mathrm {d} y}{d(x, y)^{\tilde {\alpha }}}$ . This allows us to define the uniform upper bound $0<U<\infty $ as

$$\begin{align*}U:=\sup_{k\in\mathbb{N}}\iint\limits_{{\mathbb H}^n\times {\mathbb H}^n}\frac{\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y}{d(x, y)^{\tilde{\alpha}}}. \end{align*}$$

Since $\tilde {\alpha }\geq \beta $ , we can apply the Hölder inequality to find

$$\begin{align*}&\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}\frac{\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y}{d(x, y)^{\beta}}\\ &\quad\leq \left(\iint\limits_{{\ \mathbb H}^n \times {\mathbb H}^n}\frac{\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y}{d(x, y)^{\tilde{\alpha}}}\right)^{\frac{\beta}{\tilde{\alpha}}}\left(\iint\limits_{\ {\mathbb H}^n \times {\mathbb H}^n}\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\right)^{\frac{\tilde{\alpha}-\beta}{\tilde{\alpha}}}. \end{align*}$$

In the above, use $\rho _k\in \mathcal {P}({\mathbb H}^n)$ and the definition of U to further get

$$\begin{align*}&\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}\frac{\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y}{d(x, y)^{\beta}}\leq U^{\frac{\beta}{\tilde{\alpha}}}. \end{align*}$$

Conclude then that $\iint\limits _{{\mathbb H}^n \times {\mathbb H}^n}\frac {\rho _k(x)\rho _k(y)\mathrm {d} x\mathrm {d} y}{d(x, y)^{\beta }}$ is uniformly bounded from above in $k\in \mathbb {N}$ , for arbitrary $0\leq \beta <\min (n(m-1), n)$ . This shows that the second term in (4.2) is uniformly bounded from above in $k \in \mathbb {N}$ , and since the expressions are non-negative, it is also bounded uniformly from below.

It remains to argue about $ \iint\limits _{{\mathbb H}^n \times {\mathbb H}^n} h(d(x, y))\rho _k(x)\rho _k(y)\mathrm {d} x\mathrm {d} y$ , the third term in (4.2). Its uniform boundedness from above follows immediately (see (4.3)) as

$$ \begin{align*} E[\rho_1]&\geq E[\rho_k]\geq \frac{1}{2}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n } h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y. \end{align*} $$

We will now show that it is also uniformly bounded from below.

From assumption (4.1) on h and Lemma 2.1, there exist $\gamma _1', \gamma _2'\geq 0$ such that

$$\begin{align*}h(\theta)\geq -\gamma_1'\theta^{-\alpha}-\gamma_2', \qquad \forall \theta>0. \end{align*}$$

If h is singular at $0+$ , then there exists $\delta _0>0$ such that $h(\theta )<0$ for any $0<\theta <\delta _0$ . Hence, for any $0<\delta <\delta _0$ , we have

$$ \begin{align*} &\Biggl|\ \iint\limits_{d(x, y)<\delta}h(d(x, y))\rho_k(x)\rho_k(y) \mathrm{d} x\mathrm{d} y\Biggr|=\iint\limits_{d(x, y)<\delta}|h(d(x, y))|\mathrm{d}\rho_k(x)\mathrm{d}\rho_k(y)\\ &\quad\leq \iint\limits_{d(x, y)<\delta}(\gamma_1'd(x, y)^{-\alpha}+\gamma_2')\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &\quad\leq \iint\limits_{d(x, y)<\delta}(\gamma_1'+\gamma_2'\delta^\alpha )d(x, y)^{-\alpha}\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &\quad\leq (\gamma_1'+\gamma_2'\delta^\alpha )\iint\limits_{d(x, y)<\delta}d(x, y)^{-\tilde{\alpha}}\delta^{\tilde{\alpha}-\alpha}\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &\quad\leq (\gamma_1'+\gamma_2'\delta^\alpha ) U\delta^{\tilde{\alpha}-\alpha}. \end{align*} $$

We conclude from the above (note that $\alpha <\tilde {\alpha }$ ) that

(4.5) $$ \begin{align} \lim_{\delta\to0+}\iint\limits_{d(x, y)<\delta}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y=0, \end{align} $$

uniformly in $k \in \mathbb {N}$ . On the other hand, if h is not singular at $0$ , showing (4.5) is immediate.

Now fix $\epsilon>0$ . By (4.5), there exists $\delta _1>0$ such that

$$ \begin{align*} \Biggl|\iint\limits_{d(x, y)<\delta_1}h(d(x, y))\rho_k(x)\rho_k(y) \mathrm{d} x \mathrm{d} y\Biggr|<\epsilon, \qquad\forall k\in\mathbb{N}. \end{align*} $$

Then, compute and estimate using the monotonicity of h:

$$ \begin{align*} & \iint\limits_{{\mathbb H}^n \times {\mathbb H}^n} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y \\ &\qquad =\iint\limits_{d(x, y)<\delta_1} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y+\iint\limits_{d(x, y)\geq\delta_1} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &\qquad \geq-\epsilon+\iint\limits_{d(x, y)\geq \delta_1}h(\delta_1)\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &\qquad \geq-\epsilon+\min(0, h(\delta_1)), \end{align*} $$

which yields that $ \iint\limits _{{\mathbb H}^n \times {\mathbb H}^n} h(d(x, y))\rho _k(x)\rho _k(y)\mathrm {d} x\mathrm {d} y$ is uniformly bounded from below in $k\in \mathbb {N}$ .

Proof We will investigate the entropy and the interaction energy separately.

Part 1: Entropy. In this part, we study the lower semicontinuity of the entropy functional

$$\begin{align*}\frac{1}{m-1}\int_{{\mathbb H}^n}\rho(x)^m\mathrm{d} x. \end{align*}$$

We use again $f(x)=\log _o x$ (see (3.3) and (3.8)) with o a fixed pole in ${\mathbb H}^n$ , to transform

$$\begin{align*}\int_{{\mathbb H}^n}\rho(x)^m\mathrm{d} x=\int_{T_o {\mathbb H}^n}\left(\frac{f_\#\rho(u)}{J(f^{-1}(u))}\right)^m J(f^{-1}(u))\mathrm{d} u. \end{align*}$$

Here, $\mathrm {d} u$ denotes the Lebegue measure on ${T_o {\mathbb H}^n}$ , which is identified to $\mathbb R^n $ . In the notations of [Reference Ambrosio, Fusco and Pallara1, Section 2.6], the r.h.s. above can be written as

$$\begin{align*}G (\nu, \mu) = \int_{T_o {\mathbb H}^n}U \left(\frac{\nu }{\mu}\right) \mathrm{d} \mu(u), \end{align*}$$

where

$$\begin{align*}U(s)=s^m,\quad \nu=f_\#\rho(u)\mathrm{d} u,\quad \text{ and } \mu=J(f^{-1}(u))\mathrm{d} u. \end{align*}$$

Hence, for the sequence $\rho _k$ we can write

$$\begin{align*}\int_{{\mathbb H}^n}\rho_k(x)^m\mathrm{d} x = G (\nu_k,\mu), \end{align*}$$

where $\nu _k=f_\#\rho _k(u)\mathrm {d} u$ . Note that since $\rho _k$ converges weakly- $^{*}$ to $\rho _0$ , $\nu _k$ converges weakly- $^{*}$ to $\nu _0:=f_\#\rho _0(u)\mathrm {d} u$ . By [Reference Ambrosio, Fusco and Pallara1, Theorem 2.34], identifying $T_o {\mathbb H}^n$ with $\mathbb R^n$ , and using that U is convex and lower semicontinuous, we have

$$\begin{align*}\liminf_{k\to \infty} G (\nu_k,\mu) \geq G (\nu_0,\mu), \end{align*}$$

which then implies the lower-semicontinuity of the entropy component:

(4.6) $$ \begin{align} \liminf_{k\to\infty}\frac{1}{m-1}\int_{{\mathbb H}^n}\rho_k(x)^m\mathrm{d} x\geq \frac{1}{m-1}\int_{{\mathbb H}^n}\rho_0(x)^m\mathrm{d} x. \end{align} $$

Part 2: Interaction energy. We now study the lower semicontinuity of the interaction energy functional

$$\begin{align*}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y. \end{align*}$$

Fix $\epsilon>0$ . For $0<\delta <1$ (to be specified later), and $k\geq 1$ fixed, we divide the domain of the integration as follows:

$$ \begin{align*} &\iint\limits_{{\mathbb H}^n\times {\mathbb H}^n} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ & \quad =\iint\limits_{d(x, y)<\delta} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y +\iint\limits_{d(x, y)>\delta^{-1}} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ & \qquad +\iint\limits_{\delta\leq d(x, y)\leq \delta^{-1}} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y \\ & \quad =:\mathcal{I}_1+\mathcal{I}_2+\mathcal{I}_3. \end{align*} $$

We will investigate the three terms $\mathcal {I}_1$ , $\mathcal {I}_2,$ and $\mathcal {I}_3$ separately.

Estimate of $\mathcal {I}_1$ . We have already investigated $\mathcal {I}_1$ in Proposition 4.1. By the uniform (in k) convergence in (4.5), we can choose $\delta _1>0$ such that

(4.7) $$ \begin{align} -\epsilon<\mathcal{I}_1<\epsilon, \qquad \text{ for any } \delta\in(0, \delta_1). \end{align} $$

Estimate of $\mathcal {I}_2$ . Recall that we only have two cases: $h_\infty =0$ and $h_\infty =\infty $ . Since h is increasing and converges to $h_\infty $ , there exists $\delta _2>0$ such that $-\epsilon <h(\theta )$ for any $\theta>\delta _2^{-1}$ . Hence, for any $\delta \in (0, \delta _2)$ , we can estimate $\mathcal {I}_2$ as

(4.8) $$ \begin{align} \mathcal{I}_2\geq \iint\limits_{d(x, y)>\delta^{-1}}(-\epsilon)\rho_k(x)\rho_k(y)\mathrm{d} x \mathrm{d} y\geq -\epsilon. \end{align} $$

Note that both (4.7) and (4.8) hold for arbitrary $k \geq 1$ fixed, while $\delta _1$ and $\delta _2$ depend on $\epsilon $ , but do not depend on k.

Estimate of $\mathcal {I}_3$ . Since $\rho _k$ is radially symmetric and decreasing, for any $x\in {\mathbb H}^n,$ we have

$$\begin{align*}1\geq \int_{B_{\theta_x}(o)}\rho_k(y)\mathrm{d} y\geq \int_{B_{\theta_x}(o)}\rho_k(x)\mathrm{d} y=\rho_k(x)|B_{\theta_x}(o)|. \end{align*}$$

This implies

$$\begin{align*}\rho_k(x)\leq |B_{\theta_x}(o)|^{-1},\qquad \text{ for all } x\in {\mathbb H}^n, \text{ and } k \geq 1, \end{align*}$$

which yields

(4.9) $$ \begin{align} \|\rho_k\|_{L^\infty(B_R(o)^c)}\leq |B_R(o)|^{-1},\qquad \text{ for all } R>0, \text{ and } k \geq 1. \end{align} $$

Note that in these considerations we used implicitly the Riesz rearrangement property on ${\mathbb H}^n$ (Theorem 2.5), which allows us to consider only radially symmetric and decreasing minimizing sequences $\rho _k$ ; this is in fact the only place in the article where we use Theorem 2.5.

Fix $R>0$ . We divide $\mathcal {I}_3$ as follows:

$$ \begin{align*} \mathcal{I}_3&=\iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_x,\theta_y\leq R}} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y+\iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_x> R}} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &\iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_y> R}} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y-\iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_x,\theta_y> R}} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &=\iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_x,\theta_y\leq R}} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y+2\!\iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_x> R}} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &-\iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_x,\theta_y> R}} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &=:\mathcal{I}_{31}+2\mathcal{I}_{32}-\mathcal{I}_{33}. \end{align*} $$

To estimate $\mathcal {I}_{32}$ , note that

(4.10) $$ \begin{align} \iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_x> R}} \rho_k(y)\mathrm{d} x\mathrm{d} y&\leq \iint\limits_{d(x, y)\leq \delta^{-1}}\rho_k(y)\mathrm{d} x\mathrm{d} y =\int_{{\mathbb H}^n} \rho_k(y)|B_{\delta^{-1}}(y)|\mathrm{d} y \nonumber \\ &=\int_{{\mathbb H}^n} \rho_k(y)|B_{\delta^{-1}}(o)|\mathrm{d} y= |B_{\delta^{-1}}(o)|, \end{align} $$

where we integrated in x for the first equal sign, and for the second equal sign we used that on ${\mathbb H}^n$ all geodesic balls of a certain radius have the same volume. Then, by the monotonicity of h, along with (4.9) and (4.10), we get

(4.11) $$ \begin{align} |\mathcal{I}_{32}|&\leq \|\rho_k\|_{L^\infty(B_R(o)^c)}\max \left(|h(\delta)|, |h(\delta^{-1})|\right)\iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_x> R}} \rho_k(y)\mathrm{d} x\mathrm{d} y \nonumber \\ &\leq|B_R(o)|^{-1}\max\left(|h(\delta)|, |h(\delta^{-1})|\right)|B_{\delta^{-1}}(o)|. \end{align} $$

For $\mathcal {I}_{33}$ , one can proceed exactly as in (4.10) and write

$$ \begin{align*} \iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_x,\theta_y> R}} \rho_k(y)\mathrm{d} x\mathrm{d} y&\leq \iint\limits_{d(x, y)\leq \delta^{-1}}\rho_k(y)\mathrm{d} x\mathrm{d} y =\int_M \rho_k(y)|B_{\delta^{-1}}(y)|\mathrm{d} y \\ &=\int_M \rho_k(y)|B_{\delta^{-1}}(o)|\mathrm{d} y= |B_{\delta^{-1}}(o)|, \end{align*} $$

This estimate, together with the monotonicity of h and (4.9), lead to

(4.12) $$ \begin{align} |\mathcal{I}_{33}|&\leq \|\rho_k\|_{L^\infty(B_R(o)^c)}\max \left(|h(\delta)|, |h(\delta^{-1})|\right)\iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_x,\theta_y> R}} \rho_k(y)\mathrm{d} x\mathrm{d} y \nonumber \\ &\leq|B_R(o)|^{-1}\max \left(|h(\delta)|, |h(\delta^{-1})|\right)|B_{\delta^{-1}}(o)|. \end{align} $$

Then, combining (4.11) and (4.12), we find

(4.13) $$ \begin{align} \begin{aligned} \mathcal{I}_3&\geq\mathcal{I}_{31}-2|\mathcal{I}_{32}|-|\mathcal{I}_{33}| \\ &\geq\mathcal{I}_{31}-3|B_R(o)|^{-1}\max \left(|h(\delta)|, |h(\delta^{-1})|\right)|B_{\delta^{-1}}(o)|. \end{aligned} \end{align} $$

Note that (4.13) holds for an arbitrary $k\geq 1$ , with $R>0$ and $\delta>0$ fixed, independent of k.

Overall, from (4.7), (4.8), and (4.13), we can conclude that

(4.14) $$ \begin{align} \begin{aligned} &\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &\quad \geq \iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_x,\theta_y\leq R}} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &\qquad-2\epsilon-3|B_R(o)|^{-1}\max \left(|h(\delta)|, |h(\delta^{-1})|\right)|B_{\delta^{-1}}(o)|. \end{aligned} \end{align} $$

In (4.14), $\epsilon>0$ and $R>0$ are fixed, independent of k, and $0<\delta \leq \min (\delta _1,\delta _2)$ with $\delta _1$ and $\delta _2$ depending on $\epsilon $ but not on k.

Since $\{(x, y)\in {\mathbb H}^n \times {\mathbb H}^n: \delta \leq d(x, y)\leq \delta ^{-1}, \theta _x\leq R, \theta _y\leq R\}$ is a compact set in ${\mathbb H}^n \times {\mathbb H}^n$ , we can send $k\to \infty $ in (4.14) to get

$$ \begin{align*} &\liminf_{k\to\infty}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ & \geq \liminf_{k\to\infty}\iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_x,\theta_y\leq R}} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &\quad-2\epsilon-3|B_R(o)|^{-1}\max(|h(\delta)|, |h(\delta^{-1})|)|B_{\delta^{-1}}(o)|\\ &=\iint\limits_{\substack{\delta\leq d(x, y)\leq \delta^{-1}\\\theta_x,\theta_y\leq R}} h(d(x, y))\rho_0(x)\rho_0(y)\mathrm{d} x\mathrm{d} y\\ &\quad-2\epsilon-3|B_R(o)|^{-1}\max(|h(\delta)|, |h(\delta^{-1})|)|B_{\delta^{-1}}(o)|. \end{align*} $$

By taking the limits $R\to \infty $ first and then $\delta \to 0+$ in the above, we get

$$\begin{align*}\liminf_{k\to\infty}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\geq \iint\limits_{{\mathbb H}^n\times {\mathbb H}^n} h(d(x, y))\rho_0(x)\rho_0(y)\mathrm{d} x\mathrm{d} y-2\epsilon. \end{align*}$$

Since $\epsilon>0$ was arbitrary, we get the lower-semicontinuity of the interaction energy:

(4.15) $$ \begin{align} \liminf_{k\to\infty}\iint\limits_{{\mathbb H}^n\times {\mathbb H}^n} h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\geq \iint\limits_{{\mathbb H}^n\times {\mathbb H}^n} h(d(x, y))\rho_0(x)\rho_0(y)\mathrm{d} x\mathrm{d} y. \end{align} $$

The conclusion of the theorem now follows from (4.6) and (4.15).

Proof Let o be an arbitrary pole in ${\mathbb H}^n$ and $\{\rho _k\}_{k \geq 1}\subset L^m({\mathbb H}^n) \cap \mathcal {P}_o^{sd}({\mathbb H}^n)$ be a minimizing sequence of $E[\rho ]$ . From Proposition 4.1, we know that $\iint\limits _{{\mathbb H}^n\times {\mathbb H}^n}h(d(x, y))\rho _k(x)\rho _k(y)\mathrm {d} x\mathrm {d} y$ is uniformly bounded (above and below) and also, that (4.5) holds uniformly in $k\geq 1$ . Therefore, we can fix $\delta _1>0$ such that

$$\begin{align*}\Biggl|\iint\limits_{d(x, y)<\delta_1}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\Biggr|<1, \qquad \text{ for all } k\geq 1. \end{align*}$$

This implies that for any $\delta>\delta _1$ ,

$$ \begin{align*} &\iint\limits_{d(x, y)<\delta}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &\quad = \iint\limits_{d(x, y)<\delta_1}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y+\iint\limits_{\delta_1\leq d(x, y)<\delta}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &\quad \geq -1+\min(h(\delta_1), 0), \end{align*} $$

for all $k \geq 1$ . Furthermore, for any $\delta>\delta _1$ , we then have

(5.1) $$ \begin{align} \begin{aligned} &\iint\limits_{d(x, y) \geq \delta}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y \\ &\quad =\iint\limits_{{\mathbb H}^n\times {\mathbb H}^n}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\kern1.6pt{-}\iint\limits_{d(x, y)<\delta}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\\ &\quad \leq \iint\limits_{{\mathbb H}^n\times {\mathbb H}^n}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y+1-\min(h(\delta_1), 0), \end{aligned} \end{align} $$

for all $k\geq 1$ .

Define

$$\begin{align*}V:=\sup_{k\in\mathbb{N}}\Biggl(\, \iint\limits_{{\mathbb H}^n\times {\mathbb H}^n}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y \Biggr) +1-\min(h(\delta_1), 0). \end{align*}$$

Recall that $\delta _1$ is a fixed number, so V is constant. Since $h_\infty =\infty $ , there exists $\delta _2>0$ such that $h(\delta )>0$ for all $\delta>\delta _2$ . Then, for any $\delta>\delta _2$ , we have

(5.2) $$ \begin{align} \iint\limits_{d(x, y)\geq \delta}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y\kern1.6pt{\geq} \iint\limits_{\substack{d(x, y)\geq \delta\\ \angle (xo y)\geq \pi/2}}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y, \end{align} $$

for all $k \geq 1$ . By combining (5.1) and (5.2), we infer that for any $\delta>\max (\delta _1, \delta _2)$ , we have

(5.3) $$ \begin{align} V\geq \iint\limits_{\substack{d(x, y)\geq \delta\\ \angle (xo y)\geq \pi/2}}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y, \qquad \text{ for all } k\geq 1. \end{align} $$

We will use the Rauch comparison theorem – see setup in Section 3 leading to (3.4). First, by the cosine law in the Euclidean space $T_o {\mathbb H}^n,$ we have

$$\begin{align*}|\log_o x - \log_o y|^2 = \theta_x^2+\theta_y^2-2\theta_x\theta_y\cos\angle(xo y), \qquad \text{ for all } x,y \in {\mathbb H}^n. \end{align*}$$

By combining the above with (3.4), we then get

$$ \begin{align*} d(x, y)^2 &\geq \theta_x^2+\theta_y^2-2\theta_x\theta_y\cos\angle(xo y) \\[2pt] &\geq \theta_x^2+\theta_y^2, \end{align*} $$

for all points $x,y$ such that $\angle (xo y)\geq \pi /2$ . Furthermore, if x and y satisfy $\theta _x, \theta _y\geq \delta $ and $\angle (xo y)\geq \pi /2$ , for some $\delta>0$ , then

$$\begin{align*}d(x, y)\geq\sqrt{2}\delta>\delta. \end{align*}$$

It implies

$$ \begin{align*} &\{(x, y)\in {\mathbb H}^n\times {\mathbb H}^n: d(x, y)\geq \delta,\angle(xo y)\geq \pi/2\}\supset \\ &\quad\{(x, y)\in {\mathbb H}^n\times {\mathbb H}^n: \theta_x\geq \delta, \theta_y\geq \delta,\angle(xo y)\geq \pi/2\}. \end{align*} $$

Using these considerations in (5.3), we then find for any $\delta>\max (\delta _1, \delta _2)$ ,

(5.4) $$ \begin{align} \begin{aligned} V &\geq \iint\limits_{\substack{\theta_x, \theta_y\geq \delta\\ \angle (xo y)\geq \pi/2}}h(d(x, y))\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y \\ &\geq \iint\limits_{\substack{\theta_x, \theta_y\geq \delta\\\angle (xo y)\geq \pi/2}}h(\delta)\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y, \end{aligned} \end{align} $$

for all $k \geq 1$ . Now, from the radial symmetry of $\rho _k$ with respect to o, for any fixed $x\neq o$ , we have

$$\begin{align*}\int_{\substack{\theta_y\geq \delta\\ \angle(xo y)\geq \pi/2}}\rho_k(y)\mathrm{d} y=\frac{1}{2}\int_{\theta_y\geq\delta}\rho_k(y)\mathrm{d} y. \end{align*}$$

Hence, we get for any $\delta>\max (\delta _1, \delta _2)$ ,

$$ \begin{align*} \iint\limits_{\substack{\theta_x, \theta_y\geq \delta\\ \angle (xo y)\geq \pi/2}}h(\delta)\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y&=\frac{h(\delta)}{2}\iint\limits_{\theta_x,\theta_y\geq\delta}\rho_k(x)\rho_k(y)\mathrm{d} x\mathrm{d} y \\ &=\frac{h(\delta)}{2}\left(\int_{\theta_x\geq \delta}\rho_k(x)\mathrm{d} x\right)^2, \end{align*} $$

which used in (5.4) yields

(5.5) $$ \begin{align} V\geq\frac{h(\delta)}{2}\left(\int_{\theta_x\geq \delta}\rho_k(x)\mathrm{d} x\right)^2, \qquad \text{ for all } k \geq 1. \end{align} $$

Write (5.5) as

$$\begin{align*}\int_{\theta_x\geq \delta}\rho_k(x)\mathrm{d} x\leq\left(\frac{2V}{h(\delta)}\right)^{1/2},\qquad\text{ for all } \delta>\max(\delta_1, \delta_2) \text{ and } k\geq 1. \end{align*}$$

Finally, use $\lim _{\delta \to \infty }h(\delta )=\infty $ to get

$$\begin{align*}\lim_{\delta\to\infty}\int_{\theta_x\geq \delta}\rho_k(x)\mathrm{d} x=0, \qquad \text{ uniformly in } k \geq 1. \end{align*}$$

We infer from here that $\rho _k$ is tight. Then, from Prokhorov’s theorem, there exists a subsequence $\{ \rho _{k_l}\}$ of $\{\rho _k\}$ such that $\{ \rho _{k_l}\}$ converges weakly- $^{*}$ to $\rho _0\in \mathcal {P}({\mathbb H}^n)$ as ${l\to \infty }$ . Define

$$\begin{align*}E_{\inf}:=\inf_{\rho\in L^m({\mathbb H}^n) \cap \mathcal{P}({\mathbb H}^n)}E[\rho]. \end{align*}$$

From the definition of the minimizing sequence and the weak lower semicontinuity of the energy functional (see Proposition 4.2), we have

(5.6) $$ \begin{align} E[\rho_0] \leq \liminf_{l\to\infty}E[\rho_{k_l}] = E_{\inf}. \end{align} $$

If $\rho _0\not \in L^m({\mathbb H}^n)$ , then by Proposition 2.3 $E[\rho _0]=\infty $ , and the above inequality cannot hold. Hence, $\rho _0\in L^m({\mathbb H}^n) \cap \mathcal {P}({\mathbb H}^n)$ , which yields

(5.7) $$ \begin{align} E_{\inf}\leq E[\rho_0]. \end{align} $$

From (5.6) and (5.7), we conclude that $E[\rho _0]=E_{\inf }$ and hence $\rho _0$ is a global energy minimizer over $\mathcal {P}({\mathbb H}^n)$ .

Proof Since $\rho $ is a global minimizer, in particular, it is a critical point of the energy. The derivation of the Euler-Lagrange equation for the Euclidean space (see, for instance, [Reference Carrillo, Delgadino and Patacchini10, Section 2.3] or [Reference Kaib37]) extends immediately to the manifold setup. For the energy (2.2), the Euler-Lagrange equation is given by

(5.8) $$ \begin{align} \frac{m}{m-1}\rho(x)^{m-1}+\int_{{\mathbb H}^n} h(d(x, y))\rho(y)\mathrm{d} y=C,\qquad \forall x\in\mathrm{supp}(\rho). \end{align} $$

Assume that the support of $\rho $ is the entire ${\mathbb H}^n$ , so (5.8) holds for all $x \in {\mathbb H}^n$ . Since h is non-decreasing and $h_\infty =\infty $ , there exists $s\geq 0$ such that

$$\begin{align*}s:=\inf\{\theta:h(\theta)\geq 0\}. \end{align*}$$

If $s=0$ , then $h(\theta )\geq 0$ for all $\theta>0$ . If $s>0$ then $h(\theta )<0$ for $0<\theta < s$ , and $h(\theta )\geq 0$ for $\theta \geq s$ . Also, take $R>0$ fixed such that

(5.9) $$ \begin{align} \int_{\theta_y\geq R}\rho(y) \mathrm{d} y \geq \frac{1}{2}. \end{align} $$

For $x\in \mathrm {supp}(\rho ) = {\mathbb H}^n$ fixed, but arbitrary, that satisfies $\theta _x\geq s+R$ , write

(5.10) $$ \begin{align} \int_{{\mathbb H}^n} h(d(x, y))\rho(y)\mathrm{d} y\kern1.6pt{=}\int_{d(x, y) < s} h(d(x, y))\rho(y)\mathrm{d} y\kern1.6pt{+}\int_{d(x, y)\geq s} h(d(x, y))\rho(y)\mathrm{d} y. \end{align} $$

We will inspect separately the two integrals in the r.h.s. of (5.10). For the first integral, note that for $y \in \{y: d(x,y)<s \}$ , we have

$$\begin{align*}\theta_y\geq \theta_x-d(x, y)\geq s+R-s=R, \end{align*}$$

which implies

$$ \begin{align*} \{y: d(x,y) < s\}\subset B_R(o)^c. \end{align*} $$

We, then infer

(5.11) $$ \begin{align} \|\rho\|_{L^\infty(\{y:d(x, y)\leq s\})}&\leq \|\rho\|_{L^\infty(B_R(o)^c)} \nonumber \\[2pt] & \leq |B_R(o)|^{-1}, \end{align} $$

where for the second inequality sign we used that $\rho $ is radially symmetric and decreasing (see (4.9) and the argument leading to it).

From Lemma 2.1, there exist $\gamma _1, \gamma _2\geq 0$ such that

$$ \begin{align*} h(\theta)\geq -\gamma_1\theta^{-\alpha}-\gamma_2,\qquad\forall \theta>0. \end{align*} $$

If we set $\beta =\gamma _1+\gamma _2s^\alpha $ , then we have

(5.12) $$ \begin{align} h(\theta)\geq -\beta\theta^{-\alpha},\qquad\forall \theta\in(0, s]. \end{align} $$

Using (5.12) followed by (5.11), we then find

(5.13) $$ \begin{align} \int_{d(x, y) < s} h(d(x, y))\rho(y)\mathrm{d} y &\geq-\beta\int_{d(x, y) < s}\frac{\rho(y)\mathrm{d} y}{d(x, y)^\alpha} \nonumber \\ &\geq-\beta|B_R(o)|^{-1}\int_{d(x, y)\leq s}\frac{\mathrm{d} y}{d(x, y)^\alpha} \nonumber \\ & = - \beta|B_R(o)|^{-1} \int_0^s\frac{| \partial B_\theta(o)|}{\theta^\alpha}\mathrm{d}\theta. \end{align} $$

Since $\lim _{\theta \to 0+}\frac {| \partial B_\theta (o)|}{\theta ^{n-1}} = n w(n)$ and $\alpha <n$ , the integral in the r.h.s. above is convergent, and we can define $U>0$ as

$$ \begin{align*} U:=\beta|B_R(o)|^{-1} \int_0^s\frac{|\partial B_\theta(o)|}{\theta^\alpha} \mathrm{d}\theta. \end{align*} $$

For the second integral in the r.h.s. of (5.10), note that if y is such that $\theta _y\leq R$ , then $d(x, y)\geq \theta _x-\theta _y\geq s+R-R=s$ . This implies

$$\begin{align*}\{y: \theta_y\leq R\}\subset \{y:d(x, y)\geq s\}. \end{align*}$$

and hence,

(5.14) $$ \begin{align} \int_{d(x, y)\geq s} h(d(x, y))\rho(y)\mathrm{d} y\geq \int_{B_{R}(o)}h(d(x, y))\rho(y)\mathrm{d} y. \end{align} $$

Finally, by using (5.10) in (5.8), along with (5.13) and (5.14), we get

$$ \begin{align*} C&=\frac{m}{m-1}\rho(x)^{m-1}+\int_{d(x, y)< s} h(d(x, y))\rho(y)\mathrm{d} y+\int_{d(x, y)\geq s} h(d(x, y))\rho(y)\mathrm{d} y\\ &\geq0-U+\int_{B_{R}(o)}h(d(x, y))\rho(y)\mathrm{d} y, \end{align*} $$

for all $x\in {\mathbb H}^n$ , with $\theta _x\geq s+R$ . Hence, for all such $x,$ we get

$$ \begin{align*} C+U &\geq \int_{B_{R}(o)}h(d(x, y))\rho(y)\mathrm{d} y \\ & \geq \int_{B_{R}(o)}h(\theta_x-R)\rho(y)\mathrm{d} y \\ & =h(\theta_x-R) \int_{B_{R}(o)}\rho(y)\mathrm{d} y, \end{align*} $$

where for the second inequality sign we used that $d(x, y)\geq \theta _x-\theta _y\geq \theta _x-R$ for $\theta _y \leq R$ , and that by the monotonicity of h and the definition of s, we have $h(\theta _x-R)\geq h(s)\geq 0$ . Now, use how R was defined (see (5.9)) to conclude

$$\begin{align*}C+U \geq \frac{1}{2} h(\theta_x-R), \end{align*}$$

for any $x \in {\mathbb H}^n$ which satisfies $\theta _x\geq R+s$ . Since x is arbitrary, we then find

$$\begin{align*}C+U\geq \frac{1}{2}\lim_{\theta_x\to\infty}h(\theta_x-R)=\infty \end{align*}$$

which yields a contradiction. The only remaining possibility is that the support of $\rho $ is a closed ball centred of o with radius $R<\infty $ .

Proof Let o be a fixed pole in ${\mathbb H}^n$ . Given that h is non-decreasing and $h_\infty = 0$ , there exists $\tilde {\theta }>0$ such that

$$\begin{align*}|B_{\tilde{\theta}}(o)|\leq1\qquad\text{and}\qquad h(2\tilde{\theta})<0. \end{align*}$$

Then, for any $x, y\in B_{\tilde {\theta }}(z)$ for some z, we have

(6.1) $$ \begin{align} h(d(x, y))\leq h(2\tilde{\theta}) <0. \end{align} $$

For any $\alpha>0$ , define $\rho _\alpha =\alpha \chi _{B_{\tilde {\theta }}(o)}$ . By a simple calculation,

$$\begin{align*}E[\rho_\alpha]=A \alpha^m+B \alpha^2, \end{align*}$$

where

$$\begin{align*}A:=\frac{1}{m-1}\int_{B_{\tilde{\theta}}(o)}1\mathrm{d} x>0,\quad \text{ and } \quad B:=\frac{1}{2}\iint\limits_{B_{\tilde{\theta}}(o)\times B_{\tilde{\theta}}(o)}h(d(x, y))\mathrm{d} x\mathrm{d} y<0. \end{align*}$$

Since $m>2$ , by comparing the two terms of orders $\alpha ^m$ and $\alpha ^2$ , we have

$$\begin{align*}\lim_{\alpha\to0+}E[\rho_\alpha]<0. \end{align*}$$

This implies that there exists $\alpha _0>0$ such that

$$\begin{align*}E[\rho_\alpha]<0, \qquad\forall\alpha\in(0,\alpha_0]. \end{align*}$$

Note however that $\rho _\alpha $ is not a probability measure, so it cannot be used to conclude the theorem.

Let N be an integer such that $ N \geq \frac {1}{\alpha _0|B_{\tilde {\theta }}(o)|}$ . Then,

$$\begin{align*}\alpha_*:=\frac{1}{N|B_{\tilde{\theta}}(o)|}\leq \alpha_0, \end{align*}$$

and hence,

(6.2) $$ \begin{align} E[\rho_{\alpha_*}]<0. \end{align} $$

Since N is finite, we can choose N points $x_1, x_2, \cdots , x_N \in {\mathbb H}^n$ which satisfy

$$\begin{align*}B_{\tilde{\theta}}(x_i)\cap B_{\tilde{\theta}}(x_j)=\emptyset,\qquad \text{ for all } 1\leq i<j\leq N. \end{align*}$$

Using these points, construct the measure $\rho $ by

$$\begin{align*}\rho=\frac{1}{N|B_{\tilde{\theta}}(o)|}\sum_{i=1}^N\chi_{B_{\tilde{\theta}}(x_i)}. \end{align*}$$

Note that $\rho $ is a probability measure, as

$$\begin{align*}\int_{{\mathbb H}^n}\rho(x)\mathrm{d} x=\alpha_*\sum_{i=1}^N|B_{\tilde{\theta}}(x_i)| =\alpha_* N|B_{\tilde{\theta}}(o)|=1, \end{align*}$$

where we used that all balls with radius $\tilde {\theta }$ have the same volume since ${\mathbb H}^n$ has constant sectional curvature.

The conclusion can then be inferred from the following calculation:

$$ \begin{align*} E[\rho]& =\alpha_*^m\sum_{i=1}^N\left(\frac{1}{m-1}\int_{B_{\tilde{\theta}}(x_i)}1\mathrm{d} x\right)+\frac{\alpha_*^2}{2}\sum_{i, j=1}^N\left(\iint\limits_{B_{\tilde{\theta}}(x_i)\times B_{\tilde{\theta}}(x_j)}h(d(x, y))\mathrm{d} x \mathrm{d} y\right)\\ &\leq\alpha_*^m\sum_{i=1}^N\left(\frac{1}{m-1}\int_{B_{\tilde{\theta}}(x_i)}1\mathrm{d} x\right)+\frac{\alpha_*^2}{2}\sum_{i=1}^N\left(\iint\limits_{B_{\tilde{\theta}}(x_i)\times B_{\tilde{\theta}}(x_i)}h(d(x, y))\mathrm{d} x \mathrm{d} y\right)\\ &=\alpha_*^m\sum_{i=1}^N\left(\frac{1}{m-1}\int_{B_{\tilde{\theta}}(o)}1\mathrm{d} x\right)+\frac{\alpha_*^2}{2}\sum_{i=1}^N\left(\iint\limits_{B_{\tilde{\theta}}(o)\times B_{\tilde{\theta}}(o)}h(d(x, y))\mathrm{d} x \mathrm{d} y\right)\\ &=NE[\rho_{\alpha_*}]<0, \end{align*} $$

where for the second line we only retained the diagonal part ( $1\leq i=j\leq N$ ) of the double sum ( $1\leq i, j\leq N$ ) and used (6.1). Also, for the third line, we used that ${\mathbb H}^n$ is a homogeneous manifold, and translated by isometries the balls $B_{\tilde {\theta }}(x_i)$ to $B_{\tilde {\theta }}(o)$ . For the last line, we used (6.2).

Proof Consider again densities in the form (A.1), for $R>0$ . Then, by the monotonicity of h, we have

$$ \begin{align*} E[\rho_R] &= \frac{1}{(m-1)|B_R(o)|^{m-1}} + \frac{1}{2} \frac{1}{|B_R(o)|^2} \iint\limits_{B_R(o) \times B_R(o)} h(d(x,y)) \mathrm{d} x \mathrm{d} y \\ &\leq\frac{1}{(m-1)|B_R(o)|^{m-1}}+\frac{h(2R)}{2}. \end{align*} $$

By the extra assumption (6.3) on h, by choosing $R = \frac {\theta _0}{2}$ , we find

$$\begin{align*}E[\rho_R]<0, \end{align*}$$

and this concludes the proof.

Proof Let $\{\rho _k\}_{k\geq 1}\subset L^m({\mathbb H}^n) \cap \mathcal {P}_o^{sd}({\mathbb H}^n)$ be a minimizing sequence of the energy functional. From Theorem 2.1, there exists a subsequence $\{\rho _{k_l}\}_{l\geq 1 }$ that converges weakly- $^{*}$ to $\rho _0\in \mathcal {M}_+({\mathbb H}^n)$ , which satisfies $\int _{{\mathbb H}^n}\rho _0(x)\mathrm {d} x\leq 1$ .

From Propositions 6.1 and 6.2, we know that there exists $\rho \in L^m({\mathbb H}^n) \cap \mathcal {P}({\mathbb H}^n)$ such that $E[\rho ]<0$ . Then, since $\rho _{k_l}$ is a minimizing sequence we infer that $\liminf _{l \to \infty } E[\rho _{k_l}]<0$ , and hence, by the weak lower semicontinuity of the energy (Proposition 4.2), we get

(6.4) $$ \begin{align} E[\rho_0]\leq \liminf_{l \to \infty} E[\rho_{k_l}]<0. \end{align} $$

We want to show that $\rho _0$ is the global energy minimizer, and for this purpose we need to show that it is a probability measure (i.e., $\int _{{\mathbb H}^n}\rho _0(x)\mathrm {d} x=1$ ). Since $E[0]=0$ , $\rho _0$ cannot be the zero measure, and if $\|\rho _0\|_{L^1({\mathbb H}^n)}=1$ , then the proof is done. Hence, we assume $0<\|\rho _0\|_{L^1({\mathbb H}^n)}<1$ , and follow a proof by contradiction argument.

We consider the two cases $1<m\leq 2$ and $m>2$ separately.

Case 1: $1<m\leq 2$ . For this case, the proof is simple. Define $\tilde {\rho }:=\frac {1}{\|\rho _0\|_{L^1({\mathbb H}^n)}}\rho _0\in \mathcal {P}({\mathbb H}^n)$ . Then, we have

$$ \begin{align*} E[\tilde{\rho}]&=\frac{1}{\|\rho_0\|_{L^1({\mathbb H}^n)}^m} \times \frac{1}{m-1}\int_{{\mathbb H}^n} \rho_0^m(x)\mathrm{d} x+\frac{1}{\|\rho_0\|_{L^1({\mathbb H}^n)}^2}\\&\quad\times \frac{1}{2}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y))\rho_0(x)\rho_0(y)\mathrm{d} x\mathrm{d} y\\ &\leq \frac{1}{\|\rho_0\|_{L^1({\mathbb H}^n)}^2} \times \frac{1}{m-1}\int_{{\mathbb H}^n} \rho_0^m(x)\mathrm{d} x+\frac{1}{\|\rho_0\|_{L^1({\mathbb H}^n)}^2}\\&\quad\times \frac{1}{2}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y))\rho_0(x)\rho_0(y)\mathrm{d} x\mathrm{d} y\\ &=\frac{1}{\|\rho_0\|_{L^1({\mathbb H}^n)}^2} E[\rho_0], \end{align*} $$

where for the second line we used that $\|\rho _0\|_{L^1({\mathbb H}^n)}^2 \leq \|\rho _0\|_{L^1({\mathbb H}^n)}^m$ (as $\|\rho _0\|_{L^1({\mathbb H}^n)}<1$ ).

Furthermore, since $E[\rho _0]<0$ and $\|\rho _0\|_{L^1({\mathbb H}^n)}<1$ , we infer from above that

$$\begin{align*}E[\tilde{\rho}]<E[\rho_0], \end{align*}$$

which combined with (6.4) leads to the following contradiction:

$$\begin{align*}E_{\inf} \leq E[\tilde{\rho}] < \liminf_{l\to\infty}E[\rho_{k_l}] = E_{\inf}. \end{align*}$$

Hence, necessarily $\rho _0 \in \mathcal {P}({\mathbb H}^n)$ .

Case 2: $m>2$ . From Proposition 6.1, there exists $\rho \in L^m({\mathbb H}^n) \cap \mathcal {P}({\mathbb H}^n)$ such that $E[\rho ]<0$ . We can assume that both $\rho _0$ and $\rho $ are radially symmetric and decreasing about the pole o.

As ${\mathbb H}^n$ is a homogeneous manifold, for any arbitrary point $o'\in {\mathbb H}^n$ , there exists an isometry $i_{o'}:{\mathbb H}^n \to {\mathbb H}^n$ which sends o into $o'$ , i.e., $i_{o'}(o)=o'$ . Set

$$\begin{align*}\rho_{o'}:={(i_{o'})}_\# \rho. \end{align*}$$

Since $\rho \in \mathcal {P} ({\mathbb H}^n) $ and $i_{o'}$ is an isometry, we also have $\rho _{o'} \in \mathcal {P} ({\mathbb H}^n) $ . Also, as $i_{o'}$ is an isometry, we have

$$\begin{align*}\int_{{\mathbb H}^n} \rho(x)^m\mathrm{d} x=\int_{{\mathbb H}^n} \rho_{o'}(x)^m\mathrm{d} x, \end{align*}$$

and

(6.5) $$ \begin{align} \quad \iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y))\rho_{o'}(x)\rho_{o'}(y)\mathrm{d} x\mathrm{d} y=\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n} h(d(x, y))\rho(x)\rho(y)\mathrm{d} x\mathrm{d} y, \end{align} $$

which put together yield

$$\begin{align*}E[\rho_{o'}]=E[\rho]<0. \end{align*}$$

By a similar argument as above, we also have

$$\begin{align*}E[(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}]=E[(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho]. \end{align*}$$

Since $0<1-\|\rho _0\|_{L^1({\mathbb H}^n)}<1$ and $m>2$ , we also get

(6.6) $$ \begin{align} \begin{aligned} &E[(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}] \\ &=(1-\|\rho_0\|_{L^1({\mathbb H}^n)})^m\times \frac{1}{m-1}\int_{{\mathbb H}^n} \rho_{o'}(x)^m\mathrm{d} x+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})^2\\&\quad\times \frac{1}{2}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y))\rho_{o'}(x)\rho_{o'}(y)\mathrm{d} x\mathrm{d} y \\ &\leq (1-\|\rho_0\|_{L^1({\mathbb H}^n)})^2\times \frac{1}{m-1}\int_{{\mathbb H}^n} \rho_{o'}(x)^m\mathrm{d} x+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})^2\\&\quad\times \frac{1}{2}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y))\rho_{o'}(x)\rho_{o'}(y)\mathrm{d} x\mathrm{d} y \\ &=(1-\|\rho_0\|_{L^1({\mathbb H}^n)})^2E[\rho_{o'}]<0. \end{aligned} \end{align} $$

The goal is to justify the following claim:

$\underline {\textit {Claim}\!:}$ There exists $o'\in {\mathbb H}^n$ such that

(6.7) $$ \begin{align} E[\rho_0+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}]<E[\rho_0]. \end{align} $$

Given that $\rho _0+(1-\|\rho _0\|_{L^1({\mathbb H}^n)})\rho _{o'} \in \mathcal {P}({\mathbb H}^n)$ , this will lead to the desired contradiction.

By a direct calculation, we get

$$ \begin{align*} &E[\rho_0+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}]\\ &\quad = \frac{1}{m-1}\int_{{\mathbb H}^n} \left(\rho_0(x)+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x\\ & \qquad + \frac{1}{2}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y))\left(\rho_0(x)+ (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)\left(\rho_0(y)\right.\\ &\qquad\left.+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(y)\right)\mathrm{d} x\mathrm{d} y\\ & \quad = \frac{1}{m-1}\int_{{\mathbb H}^n} \left(\rho_0(x)+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x\\ &\qquad + \frac{1}{2}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y))\rho_0(x)\rho_0(y)\mathrm{d} x\mathrm{d} y\\ &\qquad + (1-\|\rho_0\|_{L^1({\mathbb H}^n)}) \iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y))\rho_0(x)\rho_{o'}(y)\mathrm{d} x\mathrm{d} y\\ &\qquad + \frac{1}{2} (1-\|\rho_0\|_{L^1({\mathbb H}^n)})^2 \iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y)) \rho_{o'}(x) \rho_{o'}(y)\mathrm{d} x\mathrm{d} y\\ & \quad \leq \frac{1}{m-1}\int_{{\mathbb H}^n} \left(\rho_0(x)+ (1-\|\rho_0\|_{L^1({\mathbb H}^n)} )\rho_{o'}(x) \right)^m\mathrm{d} x\\ &\qquad + \frac{1}{2}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y))\rho_0(x)\rho_0(y)\mathrm{d} x\mathrm{d} y \\ &\qquad + \frac{1}{2} (1-\|\rho_0\|_{L^1({\mathbb H}^n)})^2 \iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y)) \rho_{o'}(x) \rho_{o'}(y)\mathrm{d} x\mathrm{d} y, \end{align*} $$

where for the last inequality we used $h\leq 0$ and dropped the mixed term. Furthermore, use (6.5) to write the result above as

(6.8) $$ \begin{align} \begin{aligned} &E[\rho_0+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}]\\[2pt]& \quad \leq \frac{1}{m-1}\int_{{\mathbb H}^n} \left(\rho_0(x)+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x\\ & \qquad + \frac{1}{2}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y))\rho_0(x)\rho_0(y)\mathrm{d} x\mathrm{d} y \\& \qquad + \frac{1}{2} (1-\|\rho_0\|_{L^1({\mathbb H}^n)})^2 \iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y)) \rho(x) \rho(y)\mathrm{d} x\mathrm{d} y. \end{aligned} \end{align} $$

Next, we will focus the entropy part, i.e., the first term in the r.h.s. of (6.8). From the mean value theorem, for any nonnegative numbers a and b, we have

$$\begin{align*}(a+b)^m-a^m= mb(a+b_*)^{m-1}, \end{align*}$$

where $b_*\in [0, b]$ . This implies

$$\begin{align*}(a+b)^m-a^m\leq mb(a+b)^{m-1}. \end{align*}$$

Then, for any non-negative functions $a(x)$ and $b(x)$ on a measurable set $S\subset {\mathbb H}^n$ , we get

(6.9) $$ \begin{align} \int_S \left((a(x)+b(x))^m-a(x)^m \right)\mathrm{d} x\leq\int_S mb(x)(a(x)+b(x))^{m-1}\mathrm{d} x. \end{align} $$

Also, by Hölder inequality, we can further estimate

(6.10) $$ \begin{align} \int_S b(x)(a(x)+b(x))^{m-1}\mathrm{d} x\leq \left(\int_S b(x)^m\mathrm{d} x\right)^{\frac{1}{m}}\left(\int_S(a(x)+b(x))^m \mathrm{d} x\right)^{\frac{m-1}{m}}. \end{align} $$

Given the following elementary inequality:

(6.11) $$ \begin{align} (a(x)+b(x))^m\leq (2\max(a(x), b(x)))^m\leq (2a(x))^m+(2b(x))^m=2^m(a(x)^m+b(x)^m), \end{align} $$

we can combine (6.9), (6.10), and (6.11) to get

(6.12) $$ \begin{align} \int_S \left((a(x)+b(x))^m-a(x)^m \right)\mathrm{d} x\leq m2^{m-1}\left(\int_S b(x)^m\mathrm{d} x\right)^{\frac{1}{m}}\left(\int_S a(x)^m\mathrm{d} x+\int_S b(x)^m \mathrm{d} x\right)^{\frac{m-1}{m}}, \end{align} $$

which holds for all non-negative functions $a(x)$ and $b(x)$ , and measurable subsets $S\subset {\mathbb H}^n$ .

From

$$\begin{align*}{\mathbb H}^n = \{x\in {\mathbb H}^n:d(x, o)\geq d(o, o')/2\}\cup \{x\in M: d(x, o')\geq d(o, o')/2\}, \end{align*}$$

we can separate the integration on ${\mathbb H}^n$ of the entropy part as follows:

(6.13) $$ \begin{align} \begin{aligned} &\int_{{\mathbb H}^n} \left(\rho_0(x)+ (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x \\&\quad \leq \int_{d(x, o)\geq d(o, o')/2} \left(\rho_0(x)+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x \\ & \qquad+\int_{d(x, o')\geq d(o, o')/2} \left(\rho_0(x)+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x. \end{aligned} \end{align} $$

Substitute

$$\begin{align*}a(x)=(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x), \quad b(x)=\rho_0(x), \quad \text{ and } \quad S=\{x:d(x, o)\geq d(o, o')/2\} \end{align*}$$

into (6.12), to get

(6.14) $$ \begin{align} \begin{aligned} &\int_{d(x, o)\geq d(o, o')/2} \left(\rho_0(x)+ (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x\\&\quad-\int_{d(x, o)\geq d(o, o')/2} \left((1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x) \right)^m\mathrm{d} x\\ & \quad \leq m2^{m-1}\left(\int_{d(x, o)\geq d(o, o')/2} \rho_0(x)^m\mathrm{d} x\right)^{\frac{1}{m}}\\ &\qquad\times\left(\int_{d(x, o)\geq d(o, o')/2} \rho_0(x)^m\mathrm{d} x+\int_{d(x, o)\geq d(o, o')/2} \left((1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x\right)^{\frac{m-1}{m}}\\ & \quad \leq m2^{m-1}\left(\int_{d(x, o)\geq d(o, o')/2} \rho_0(x)^m\mathrm{d} x\right)^{\frac{1}{m}} \\&\qquad\left(\int_{{\mathbb H}^n} \rho_0(x)^m\mathrm{d} x+\int_{{\mathbb H}^n} \left( (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x) \right)^m\mathrm{d} x\right)^{\frac{m-1}{m}}. \end{aligned} \end{align} $$

Similarly, substitute

$$\begin{align*}a(x)=\rho_0(x),\quad b(x)=(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x), \quad \text{ and } \quad S=\{x:d(x, o')\geq d(o, o')/2\} \end{align*}$$

into (6.12), to get

(6.15) $$ \begin{align} \begin{aligned} &\int_{d(x, o')\geq d(o, o')/2} \left(\rho_0(x)+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x-\int_{d(x, o')\geq d(o, o')/2} \rho_0(x)^m\mathrm{d} x\\ &\quad \leq m2^{m-1}\left(\int_{d(x, o')\geq d(o, o')/2} \left((1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x\right)^{\frac{1}{m}}\\ &\qquad\times \left(\int_{d(x, o')\geq d(o, o')/2} \rho_0(x)^m\mathrm{d} x+\int_{d(x, o')\geq d(o, o')/2} \left( (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x) \right)^m\mathrm{d} x\right)^{\frac{m-1}{m}}\\ & \quad \leq m2^{m-1}\left(\int_{d(x, o')\geq d(o, o')/2} \left( (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x) \right)^m\mathrm{d} x\right)^{\frac{1}{m}} \\ & \qquad \times \left(\int_{{\mathbb H}^n} \rho_0(x)^m\mathrm{d} x+\int_{{\mathbb H}^n} \left( (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x) \right)^m\mathrm{d} x\right)^{\frac{m-1}{m}}. \end{aligned} \end{align} $$

Now combine (6.13), (6.14), and (6.15) to get

(6.16) $$ \begin{align} \begin{aligned} &\int_{{\mathbb H}^n} \left(\rho_0(x)+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x\\ &\quad \leq \int_{d(x, o)\geq d(o, o')/2} \left( ( 1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x+\int_{d(x, o')\geq d(o, o')/2} \rho_0(x)^m\mathrm{d} x\\ &\qquad + m2^{m-1}\left(\int_{d(x, o)\geq d(o, o')/2} \rho_0(x)^m\mathrm{d} x\right)^{\frac{1}{m}}\\&\qquad\left(\int_{{\mathbb H}^n} \rho_0(x)^m\mathrm{d} x+\int_{{\mathbb H}^n} \left( (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x\right)^{\frac{m-1}{m}}\\ &\quad\quad + m2^{m-1}\left(\int_{d(x, o')\geq d(o, o')/2} \left( (1-\|\rho_0\|_{L^1({\mathbb H}^n)} )\rho_{o'}(x) \right)^m\mathrm{d} x\right)^{\frac{1}{m}} \\ &\qquad \times \left(\int_{{\mathbb H}^n} \rho_0(x)^m\mathrm{d} x+\int_{{\mathbb H}^n} \left( (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x) \right)^m\mathrm{d} x\right)^{\frac{m-1}{m}}. \end{aligned} \end{align} $$

Since $\int _{{\mathbb H}^n}\rho _k(x)^m\mathrm {d} x$ is uniformly bounded from above in $k\geq 1$ (see Proposition 4.1), from the lower semicontinuity of the entropy functional we infer that $\int _{{\mathbb H}^n}\rho _0(x)^m\mathrm {d} x $ is also bounded from above. Also, we know that $\int _{{\mathbb H}^n}\rho (x)^m\mathrm {d} x$ is bounded, as $\rho \in L^m({\mathbb H}^n)$ . Hence, set

(6.17) $$ \begin{align} U:=\max\left(\int_{{\mathbb H}^n}\rho_0(x)^m\mathrm{d} x, \int_{{\mathbb H}^n} \left( (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho(x) \right)^m\mathrm{d} x\right) < \infty. \end{align} $$

Then, we have

$$\begin{align*}\left(\int_{{\mathbb H}^n} \rho_0(x)^m\mathrm{d} x+\int_{{\mathbb H}^n} \left((1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x\right)^{\frac{m-1}{m}}\leq (2U)^{\frac{m-1}{m}}, \end{align*}$$

which used in (6.16) yields

(6.18) $$ \begin{align} \begin{aligned} &\int_{{\mathbb H}^n} \left(\rho_0(x)+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x\\ & \quad \leq \int_{d(x, o)\geq d(o, o')/2} \left((1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x) \right)^m\mathrm{d} x+\int_{d(x, o')\geq d(o, o')/2} \rho_0(x)^m\mathrm{d} x\\[2pt] &\qquad +m2^{m-1} \left(2U\right)^{\frac{m-1}{m}} \\ & \qquad \quad \times \left( \left(\int_{d(x, o)\geq d(o, o')/2} \rho_0(x)^m\mathrm{d} x\right)^{\frac{1}{m}}\right.\\&\quad \left.+\left(\int_{d(x, o')\geq d(o, o')/2} \left( (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x) \right)^m\mathrm{d} x\right)^{\frac{1}{m}} \right) \\ & \quad \leq \int_{{\mathbb H}^n} \left((1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x) \right)^m\mathrm{d} x+\int_{{\mathbb H}^n} \rho_0(x)^m\mathrm{d} x\\[2pt] &\qquad + m2^{m-\frac{1}{m}}U^{\frac{m-1}{m}} \\ &\qquad \quad \times \left( \left(\int_{d(x, o)\geq d(o, o')/2}\rho_0(x)^m\mathrm{d} x\right)^{\frac{1}{m}} \right.\\&\qquad\left.+\left(\int_{d(x, o')\geq d(o, o')/2} \left( (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x) \right)^m\mathrm{d} x\right)^{\frac{1}{m}} \right). \end{aligned} \end{align} $$

By the property of isometries, we have

$$\begin{align*}&\int_{d(x, o')\geq d(o, o')/2} \left((1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x) \right)^m\mathrm{d} x\\&\quad=\int_{d(x, o)\geq d(o, o')/2} \left( (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho(x) \right)^m\mathrm{d} x, \end{align*}$$

and

$$\begin{align*}\int_{{\mathbb H}^n} \left( (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x) \right)^m\mathrm{d} x=\int_{{\mathbb H}^n} \left( (1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho(x) \right)^m\mathrm{d} x, \end{align*}$$

and these used in (6.18) yield

(6.19) $$ \begin{align} \begin{aligned} &\int_{{\mathbb H}^n} \left(\rho_0(x)+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x\\[2pt] & \quad \leq \int_{{\mathbb H}^n} \left((1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho(x) \right)^m\mathrm{d} x+\int_{{\mathbb H}^n} \rho_0(x)^m\mathrm{d} x\\ & \qquad + m2^{m-\frac{1}{m}}U^{\frac{m-1}{m}} \\ & \qquad \quad \times \left( \left(\int_{d(x, o)\geq d(o, o')/2}\rho_0(x)^m\mathrm{d} x\right)^{\frac{1}{m}}\right.\\&\quad\left.+\left(\int_{d(x, o)\geq d(o, o')/2} ((1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho(x))^m\mathrm{d} x\right)^{\frac{1}{m}} \right). \end{aligned} \end{align} $$

Now, choose an arbitrary $\epsilon $ that satisfies

(6.20) $$ \begin{align} 0<\epsilon<-\frac{E[(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho]}{m2^{m+1-\frac{1}{m}}U^{\frac{m-1}{m}}}. \end{align} $$

Note that this is possible, since $E[(1-\|\rho _0\|_{L^1({\mathbb H}^n)})\rho ]<0$ by (6.6). From the boundedness of integrals (see (6.17)), there exists $L>0$ (depending on $\epsilon $ , $\rho _0$ and $\rho $ ) such that

$$\begin{align*}\int_{d(x, o)\geq L}\rho_0(x)^m\mathrm{d} x<\epsilon^m\qquad\text{and}\qquad \int_{d(x, o)\geq L}((1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho(x))^m\mathrm{d} x<\epsilon^m. \end{align*}$$

Hence, for any $o'$ which satisfies $d(o,o')\geq 2L$ , we get from (6.19) that

(6.21) $$ \begin{align} \begin{aligned} &\int_{{\mathbb H}^n} \left(\rho_0(x)+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}(x)\right)^m\mathrm{d} x\\[2pt] & \quad \leq \int_{{\mathbb H}^n} \left((1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho(x) \right)^m\mathrm{d} x+\int_{{\mathbb H}^n} \rho_0(x)^m\mathrm{d} x +m2^{m-\frac{1}{m}}U^{\frac{m-1}{m}}\left(\epsilon+\epsilon\right)\\[2pt] & \quad < \int_{{\mathbb H}^n} \left((1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho(x)\right)^m\mathrm{d} x+\int_{{\mathbb H}^n} \rho_0(x)^m\mathrm{d} x-E[(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho], \end{aligned} \end{align} $$

where in the last inequality we used the assumption (6.20) on $\epsilon $ .

Finally, we use (6.21) in (6.8) to get

$$ \begin{align*} &E[\rho_0+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}]\\[2pt] &\quad < \int_{{\mathbb H}^n} \left((1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho(x) \right)^m\mathrm{d} x+\int_{{\mathbb H}^n} \rho_0(x)^m\mathrm{d} x-E[(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho]\\[2pt] &\qquad + \frac{1}{2}\iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y))\rho_0(x)\rho_0(y)\mathrm{d} x\mathrm{d} y \\&\qquad+ \frac{1}{2} (1-\|\rho_0\|_{L^1({\mathbb H}^n)})^2 \iint\limits_{{\mathbb H}^n \times {\mathbb H}^n}h(d(x, y))\rho(x)\rho(y)\mathrm{d} x\mathrm{d} y\\ &\quad = E[\rho_0]. \end{align*} $$

This shows the claim - see (6.7). Similar to Case 1, (6.7) combined with (6.4), together with the fact that $\rho _0+(1-\|\rho _0\|_{L^1({\mathbb H}^n)})\rho _{o'} \in \mathcal {P}({\mathbb H}^n)$ , leads to the following contradiction:

$$\begin{align*}E_{\inf} \leq E[\rho_0+(1-\|\rho_0\|_{L^1({\mathbb H}^n)})\rho_{o'}] < \liminf_{l\to\infty}E[\rho_{k_l}] = E_{\inf}. \end{align*}$$

We conclude that $\rho _0\in \mathcal {P}({\mathbb H}^n)$ and hence, $\rho _0$ is a global minimizer of the energy functional.