some $u_0$ in $M^p_{1,H}(X,\mu)$ such that $$C = \| u_0 \|_{L^{p^{\ast}}(X,\mu)}\ /\ \| u_0 \|_{M^p_{1,H}(X,\mu)} .$$

Define the functional $\mathcal{I} :  M^p_{1,H}(X,\mu) \thicksim  \{ 0\} \to  \mathbb{R}$ by $$\mathcal{I}[u] := \|  u \|_{M^p_1(X,\mu)} ,$$ and set $$D := \inf \{\ \mathcal{I}[u]\ :\ u \in M^p_{1,H}(X,\mu)\ ,\   \| u \|_{L^{p^{\ast}}(X,\mu)} = 1   \ \} .$$ Let $\{ u_n \}$ be a minimizing sequence, i.e. such that $u_n \in M^p_{1,H}(X,\mu)$ and $ \| u_n \|_{L^{p^{\ast}}(X,\mu)} = 1$ for every $n$, with $\mathcal{I}[u_n] \to D$. Since $\{ u_n \}$ is bounded in $M^p_{1,H}(X,\mu)$, by Theorem \[main2\] there is a subsequence[^11] of $\{ u_n \}$ and some $w$ in $M^p_{1,H}(X,\mu)$ such that $$u_n \to w\ \text{weakly in}\ M^p_{1,H}(X,\mu),$$ $$\text{and}\ u_n \to w\ \text{strongly in}\ L^{p^{\ast}}(X,\mu).$$

But $\| w \|_{L^{p^{\ast}}(X,\mu)} = 1$ by strong convergence in $L^{p^{\ast}}(X,\mu)$, hence $$D = \lim_{n \to \infty} \mathcal{I}[u_n] = \liminf_{n \to \infty} \|  u_n \|_{M^p_1(X,\mu)} \geq \|  w \|_{M^p_1(X,\mu)} = \mathcal{I}[w] \geq D.$$

Therefore $\mathcal{I}[w] = D$, and it follows that $C = 1/D$, with $u_0 = w$.

Riemannian applications {#examples}
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The next result is not surprising and probably not new, however we could not find it in the literature. To satisfy the interested reader, and justify the 