sup x inf inequality on manifolds of dimension 5

Samy Skander Bahoura

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Samy Skander Bahoura. sup x inf inequality on manifolds of dimension 5. 2026. hal-05512167

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sup × inf inequality on manifolds of dimension 5.

Samy Skander Bahoura *

Department of Mathematics, Pierre and Marie Curie University, 75005, Paris, France.

Abstract

We give an inequality of type sup × inf in dimension 5 for a Yamabe type equation.

Keywords: sup × inf, a priori estimate, Riemannian manifolds, Yamabe type equation, dimension 5.

# 1 Introduction and Main Result

We consider a Riemannian manifold $(M, g)$ of dimension $n \ge 2$ and we denote by $\Delta = -\nabla^i(\nabla_i)$ the Laplace-Beltrami operator.

The Yamabe type equation (also called Schrodinger equation) in dimension $n \ge 3$ is:

$$ \Delta u + h u = n(n-2)u^{N-1}, \quad u > 0, \quad N = \frac{2n}{n-2}. $$

with $h$ a smooth function. When $h = \frac{n-2}{4(n-1)}R_g$ with $R_g$ the scalar curvature of $(M, g)$, the equation is the Yamabe equation. If the number $n(n-2)$ is replaced by a function $V$, the equation is called, the prescribed scalar curvature equation.

About the existence result for the Yamabe equation, see [1].

In this paper, we are interested by inequalities of type sup × inf relative to Yamabe type equation.

These type of estimates were studied for the prescribed scalar curvature equation and Yamabe equation in [1-20].

In dimension 2 and on domains of $\mathbb{R}^2$, Brezis-Li-Shafrir, see [8], proved that sup + inf is bounded when the prescribed scalar curvature are Lipschitz, in [11], Chen-Lin extend this result to Hölder prescribed scalar curvatures.

In dimensions $n \ge 3$, there are many results about sup × inf inequalities, see, [2-9], [11-12], [17-19].

Here, we are on a Riemannian manifold $(M, g)$ of dimension $n = 5$, and we consider a Yamabe type equation:

$$ \Delta u + h u = 15u^{7/3}, \quad u > 0, \qquad (1) $$

with $h$ a smooth function, $||h||_{C^2(M)} \le h_0, h_0 > 0$. (Here: $n(n-2) = 15$).

<footnote>*Adresses e-mail: samybahoura@gmail.com</footnote>
When $M$ is compact without boundary, this equation, where studied by many authors. See [1,13,15,17,19].

An important result in dimensions $n = 3, 4$ by Li-Zhang in [18], they proved the boundedness of $\sup \times \inf$ for the Yamabe equation.

Our main result is:

**Theorem 1.1** For all compact subset $K$ of $M$, there exists a positive constant $c = c(K, M, h_0, g) > 0$ such that:

$$ (\sup_K u)^{1/7} \times \inf_M u \le c. $$

for all $u > 0$ solution of (1).

In [19], Li-Zhu, gave an example of solution of a Yamabe type equation on domain of $\mathbb{R}^5$ with $\sup \times \inf \to +\infty$. Here, we obtain a weaker inequality.

In [3], we considered the Yamabe equation. In this paper we take $h \neq \frac{3}{16}R_g$, with $R_g$ the scalar curvature. Here we consider more general equation, of Yamabe type.

## 2 Proof of the Theorem:

### Blow-up analysis:

Let $u$ a function on $M$ such that,

$$ \Delta u + hu = 15u^{7/3}, \quad u > 0. $$

We argue by contradiction and we suppose that $\sup \times \inf$ is not bounded.

We suppose:

$\forall c, R > 0 \exists u_{c,R}$ solution of (E) such that:

$$ R^3 \left( \sup_{B(x_0, R)} u_{c,R} \right)^{1/7} \times \inf_M u_{c,R} \ge c, \quad (H) $$

**Proposition 2.1** There are, $(y_i)_i, y_i \to x_0$ and $(l_i)_i, (L_i)_i, l_i \to 0, L_i \to +\infty$, such that if we set, $v_i(y) = \frac{u_i o \exp_{y_i}(y)}{u_i(y_i)}$, we have:

$$ 0 < v_i(y) \le \beta_i \le 2^{3/2}, \quad \beta_i \to 1. $$

$$ v_i(y) \to \left( \frac{1}{1 + |y|^2} \right)^{3/2}, \quad \text{uniforment sur tout compact de } \mathbb{R}^5. $$

$$ l_i^{3/2} [u_i(y_i)]^{1/7} \times \inf_M u_i \to +\infty. $$
Proof proposition 2.1:

We use (H), we supps that, there are $R_i > 0, R_i \to 0$ et $c_i \to +\infty$, such that,

$$R_i^3(\sup_{B(x_0, R_i)} u_i)^{1/7} \inf_M u_i \geq c_i \to +\infty,$$

Let, $x_i \in B(x_0, R_i)$, such that $\sup_{B(x_0, R_i)} u_i = u_i(x_i)$ and $s_i(x) = [R_i - d(x, x_i)]^{3/2} u_i(x), x \in B(x_i, R_i)$. Then, $x_i \to x_0$.

We have,

$$\max_{B(x_i, R_i)} s_i(x) = s_i(y_i) \geq s_i(x_i) = R_i^{3/2} u_i(x_i) \geq \sqrt{c_i} \to +\infty.$$

We set :

$$l_i = R_i - d(y_i, x_i), \quad \bar{u}_i(y) = u_i \exp_{y_i}(y), \quad v_i(z) = \frac{u_i \exp_{y_i}(z/[u_i(y_i)]^{2/3})}{u_i(y_i)}.$$

Clearly, $y_i \to x_0$. We obtain:

$$L_i = \frac{l_i}{(c_i)^{1/6}} [u_i(y_i)]^{2/3} = \frac{[s_i(y_i)]^{2/3}}{c_i^{1/6}} \geq \frac{c_i^{1/3}}{c_i^{1/6}} = c_i^{1/6} \to +\infty.$$

If $|z| \leq L_i$, then $y = \exp_{y_i}[z/[u_i(y_i)]^{2/3}] \in B(y_i, \delta_i l_i)$ with $\delta_i = \frac{1}{(c_i)^{1/6}}$ and $d(y, y_i) < R_i - d(y_i, x_i)$, thus, $d(y, x_i) < R_i$ and thus, $s_i(y) \leq s_i(y_i)$, we can write,

$$u_i(y)[R_i - d(y, x_i)]^{3/2} \leq u_i(y_i)(l_i)^{3/2}.$$

but, $d(y, y_i) \leq \delta_i l_i$, $R_i > l_i$ et $R_i - d(y, x_i) \geq R_i - d(y_i, x_i) - \delta_i l_i > l_i - \delta_i l_i = l_i(1 - \delta_i)$, we obtain,

$$0 < v_i(z) = \frac{u_i(y)}{u_i(y_i)} \leq \left[ \frac{l_i}{l_i(1 - \delta_i)} \right]^{3/2} \leq 2^{3/2}.$$

We set, $\beta_i = \left( \frac{1}{1 - \delta_i} \right)^{3/2}$, clearly $\beta_i \to 1$.

The function $v_i$ satisfies:

$$-g^{jk}[\exp_{y_i}(y)] \partial_{jk} v_i - \partial_k [g^{jk} \sqrt{|g|}] [\exp_{y_i}(y)] \partial_j v_i + \frac{h[\exp_{y_i}(y)]}{[u_i(y_i)]^{4/(n-2)}} v_i = n(n-2)v_i^{N-1},$$

We use elliptic estimates to have a subsequence, $(v_i)_i$ which converge uniformly on compact subsets of the euclidean space to a function $v$ solution of,
$$ \Delta v = 15v^{7/3}, \quad v(0) = 1, \quad 0 \le v \le 1 \le 2^{3/2}, $$

By the maximum principle, $v > 0$ on $\mathbb{R}^5$. By a result of Caffarelli-Gidas-Spruck (see [10]), we have: $v(y) = \left( \frac{1}{1 + |y|^2} \right)^{3/2}$.

**Polar geodesic coordinates and moving-plane method**

We set:

$$ w_i(t, \theta) = e^{3t/2} \tilde{u}_i(e^t, \theta) = e^{3t/2} u_i o \exp_{y_i}(e^t \theta), \quad \text{and} \quad a(y_i, t, \theta) = \log J(y_i, e^t, \theta). $$

**Lemma 2.2** La fonction $w_i$ est solution de:

$$ -\partial_{tt} w_i - \partial_t a \partial_t w_i + \Delta_{\theta} w_i + cw_i = 15w_i^{7/3}, $$

with,

$$ c = c(y_i, t, \theta) = \left(\frac{3}{2}\right)^2 + \frac{3}{2}\partial_t a + he^{2t}, $$

**Proof of the Lemma.** See [3].

$$ \text{We write, } \partial_t a = \frac{\partial_t b_1}{b_1}, \quad b_1(y_i, t, \theta) = J(y_i, e^t, \theta) > 0, $$

and,

$$ -\frac{1}{\sqrt{b_1}} \partial_{tt}(\sqrt{b_1} w_i) + \Delta_{\theta} w_i + [c(t) + b_1^{-1/2} b_2(t, \theta)] w_i = 15 w_i^{7/3}, $$

$$ \text{où, } b_2(t, \theta) = \partial_{tt}(\sqrt{b_1}) = \frac{1}{2\sqrt{b_1}} \partial_{tt} b_1 - \frac{1}{4(b_1)^{3/2}} (\partial_t b_1)^2. $$

We set,

$$ \tilde{w}_i = \sqrt{b_1} w_i, $$

**Lemma 2.3** The function $\tilde{w}_i$ is solutions of:

$$ \begin{aligned} -\partial_{tt} \tilde{w}_i + \Delta_{\theta}(\tilde{w}_i) + 2\nabla_{\theta}(\tilde{w}_i) \cdot \nabla_{\theta} \log(\sqrt{b_1}) + (c + b_1^{-1/2} b_2 - c_2) \tilde{w}_i &= \\ &= 15 \left(\frac{1}{b_1}\right)^{3/2} \tilde{w}_i^{7/3}, \end{aligned} $$

with $c_2$ a function.
**Proof of the Lemma.** See [3] with $c_2 = [\frac{1}{\sqrt{b_1}}\Delta_{\theta}(\sqrt{b_1}) + |\nabla_{\theta} \log(\sqrt{b_1})|^2]$.

**The moving-plane method:**

Let $\xi_i$ a number, we suppose $\xi_i \le t$, we set $t^{\xi_i} = 2\xi_i - t$ and $\tilde{w}_i^{\xi_i}(t, \theta) = \tilde{w}_i(t^{\xi_i}, \theta)$.

**Proposition 2.4** We have:

1) $\tilde{w}_i(\lambda_i, \theta) - \tilde{w}_i(\lambda_i + 4, \theta) \ge \tilde{k} > 0, \forall \theta \in \mathbb{S}_4$.

For all $\beta > 0$, there is $c_{\beta} > 0$ such that:

2) $\frac{1}{c_{\beta}}e^{3t/2} \le \tilde{w}_i(\lambda_i + t, \theta) \le c_{\beta}e^{3t/2}, \forall t \le \beta, \forall \theta \in \mathbb{S}_4$.

**Proof of the proposition.** See [3].

We set:

$$ \bar{Z}_i = -\partial_{tt}(\cdot) + \Delta_{\theta}(\cdot) + 2\nabla_{\theta}(\cdot) \cdot \nabla_{\theta} \log(\sqrt{b_1}) + (c + b_1^{-1/2}b_2 - c_2)(\cdot) $$

**Remark:** In the operator $\bar{Z}_i$ we have: $c + b_1^{-1/2}b_2 - c_2$ vérifie:

$$ c + b_1^{-1/2}b_2 - c_2 \ge k' > 0, \text{ pour } t \ll 0, $$

thus we can apply the Hopf maximum principle.

**Goal:**

Like in [2,3], we want to prove that:

$$ \bar{Z}_i(\tilde{w}_i^{\xi_i} - \tilde{w}_i) \le 0, \text{ si } \tilde{w}_i^{\xi_i} - \tilde{w}_i \le 0. $$

with, $\tilde{w}_i$, a particular function obtained from $w_i$.

We have:

**Lemma 2.5**

$$ b_1(y_i, t, \theta) = 1 - \frac{1}{6} \mathrm{Ricci}_{y_i}(\theta, \theta) e^{2t} + \dots, $$

$$ h(e^t \theta) = h(y_i) + \langle \nabla h(y_i) | \theta \rangle > e^t + \dots $$

and,

**Proposition 2.6**

$$ \bar{Z}_i(\tilde{w}_i^{\xi_i} - \tilde{w}_i) \le b_1^{-3/2}[(\tilde{w}_i^{\xi_i})^{7/3} - \tilde{w}_i^{7/3}]+ \\ +C|e^{2t}-e^{2t^{\xi_i}}| [|\nabla_{\theta}\tilde{w}_i^{\xi_i}| + |\nabla_{\theta}^2(\tilde{w}_i^{\xi_i})| + |\mathrm{Ricci}_{y_i}|[\tilde{w}_i^{\xi_i} + \tilde{w}_i^{\xi_i})^{7/3}] + |h|\tilde{w}_i^{\xi_i}] + C'|e^{3t^{\xi_i}}-e^{3t}|. $$
**Proof of the proposition.** See [3].

We have:

$$a(y_i, t, \theta) = \log J(y_i, e^t, \theta) = \log b_1, |\partial_t b_1(t)| + |\partial_{tt} b_1(t)| + |\partial_{tt} a(t)| \le Ce^{2t},$$
and,

$$|\partial_{\theta_j} b_1| + |\partial_{\theta_j, \theta_k} b_1| + \partial_{t, \theta_j} b_1| + |\partial_{t, \theta_j, \theta_k} b_1| \le Ce^{2t},$$

thus,

$$|\partial_t b_1(t^{\xi_i}) - \partial_t b_1(t)| \le C'|e^{2t} - e^{2t^{\xi_i}}|, \text{ sur } ]-\infty, \log \epsilon_1] \times \mathbb{S}_4, \forall x \in B(x_0, \epsilon_1)$$
locally,

$$\Delta_{\theta} = - \frac{1}{\sqrt{|\tilde{g}^k(e^t, \theta)|}} \partial_{\theta^l} [\tilde{g}^{\theta^l \theta^j}(e^t, \theta) \sqrt{|\tilde{g}^k(e^t, \theta)|} \partial_{\theta^j}].$$

Thus, in the chart]0, $\epsilon_1[\times U^k$, we have,

$$A_i = \left[ \left[ \frac{1}{\sqrt{|\tilde{g}^k|}} \partial_{\theta^l} (\tilde{g}^{\theta^l \theta^j} \sqrt{|\tilde{g}^k|} \partial_{\theta^j}) \right]^{\xi_i} - \frac{1}{\sqrt{|\tilde{g}^k|}} \partial_{\theta^l} (\tilde{g}^{\theta^l \theta^j} \sqrt{|\tilde{g}^k|} \partial_{\theta^j}) \right] (\tilde{w}_i^{\xi_i})$$

thus, $A_i = B_i + D_i$ with,

$$B_i = [\tilde{g}^{\theta^l \theta^j}(e^{t^{\xi_i}}, \theta) - \tilde{g}^{\theta^l \theta^j}(e^t, \theta)] \partial_{\theta^l \theta^j} \tilde{w}_i^{\xi_i}(t, \theta),$$

and,

$$D_i = \left[ \frac{1}{\sqrt{|\tilde{g}^k|(e^{t^{\xi_i}}, \theta)}} \partial_{\theta^l} [\tilde{g}^{\theta^l \theta^j}(e^{t^{\xi_i}}, \theta) \sqrt{|\tilde{g}^k|(e^{t^{\xi_i}}, \theta)}] - \frac{1}{\sqrt{|\tilde{g}^k|(e^t, \theta)}} \partial_{\theta^l} [\tilde{g}^{\theta^l \theta^j}(e^t, \theta) \sqrt{|\tilde{g}^k|(e^t, \theta)}] \right] \partial_{\theta^j} \tilde{w}_i^{\xi_i}(t, \theta),$$
thus,

$$A_i \le C_k |e^{2t} - e^{2t^{\xi_i}}| \left[ |\nabla_{\theta} \tilde{w}_i^{\xi_i}| + |\nabla_{\theta}^2 (\tilde{w}_i^{\xi_i})| \right],$$

We take $C = \max\{C_i, 1 \le i \le q\}$.

We have

$$c(y_i, t, \theta) = \left(\frac{3}{2}\right)^2 + \frac{3}{2} \partial_t a + h e^{2t}, \quad (\alpha_1)$$

$$b_2(t, \theta) = \partial_{tt}(\sqrt{b_1}) = \frac{1}{2\sqrt{b_1}} \partial_{tt} b_1 - \frac{1}{4(b_1)^{3/2}} (\partial_t b_1)^2, \quad (\alpha_2)$$

$$c_2 = \left[ -\frac{1}{\sqrt{b_1}} \Delta_{\theta} (\sqrt{b_1}) + |\nabla_{\theta} \log(\sqrt{b_1})|^2 \right], \quad (\alpha_3)$$
then,

$$ \partial_t c(y_i, t, \theta) = \frac{(n-2)}{2} \partial_{tt} a + 2e^{2t} h(e^t \theta) + e^{3t} \langle \nabla h(e^t \theta) | \theta \rangle, $$

and then,

$$ |\partial_t c_2| + |\partial_t b_1| + |\partial_t b_2| + |\partial_t c| \le K_1 e^{2t}, $$

We know $\theta_0 < v_i(y) \le 2^{3/2}$, by the elliptic estimates we obtain,

$$ \|\nabla v_i\|_{L^{\infty}(B(0,R))} + \|\nabla^2 v_i\|_{L^{\infty}(B(0,R))} \le C(R, n). \quad (\ast \ast \ast 2) $$

We are in dimension 5. We consider the function $\bar{w}_i(t, \theta) = \tilde{w}_i(t, \theta) - \frac{[u_i(y_i)]^{\alpha/3} \min_M u_i}{2} e^{2t}$.

With $t \le t_i = -\frac{2\alpha}{3} \log u_i(y_i)$, we have:

$$ \begin{aligned} \bar{w}_i(t, \theta) &= e^{2t} \left[ \sqrt{b_1}(t, \theta) e^{-t/2} u_i o \exp_{y_i}(e^t \theta) - \frac{[u_i(y_i)]^{\alpha/3} \min_M u_i}{2} \right] \ge \\ &\ge e^{2t} \frac{[u_i(y_i)]^{\alpha/3} \min_M u_i}{2} > 0, \end{aligned} $$

We set, $\mu_i = \frac{[u_i(y_i)]^{\alpha/3} \min_M u_i}{2}$.

Like in [2,3] we have:

**Lemma 2.7** There is $\nu < 0$ such that for $\lambda \le \nu$:

$$ \bar{w}_i^{\lambda}(t, \theta) - \bar{w}_i(t, \theta) \le 0, \quad \forall (t, \theta) \in [\lambda, t_i] \times \mathbb{S}_4, $$

**Proof of the lemma.** See [2,3].

We set: $\lambda_i = -\frac{2}{3} \log u_i(y_i)$, then,

**Lemma 2.8**

$$ \bar{w}_i(\lambda_i, \theta) - \bar{w}_i(\lambda_i + 4, \theta) > 0. $$

**Proof of the lemma.** See [2,3].

Let, $\xi_i = \sup\{\lambda \le \lambda_i + 2, \bar{w}_i^{\xi_i}(t, \theta) - \bar{w}_i(t, \theta) \le 0, \forall (t, \theta) \in [\xi_i, t_i] \times \mathbb{S}_4\}$.

$\xi_i$ exists (see [2]), We obtain:

$$ \tilde{w}_i^{\xi_i}(t, \theta) + |\nabla_{\theta} \tilde{w}_i^{\xi_i}(t, \theta)| + |\nabla_{\theta}^2 \tilde{w}_i^{\xi_i}(t, \theta)| \le C(R), \quad \forall (t, \theta) \in ]-\infty, \log R] \times \mathbb{S}_4, $$

We write:
$$ \bar{Z}_i(\bar{w}_i^{\xi_i} - \bar{w}_i) = \bar{Z}_i(\tilde{w}_i^{\xi_i} - \tilde{w}_i) - \mu_i \bar{Z}_i(e^{2t\xi_i} - e^{2t}), $$

$$ -\bar{Z}_i(e^{2t\xi_i} - e^{2t}) = \left[4 - \frac{9}{4} - \frac{3}{2}\partial_t a - he^{2t} + b_1^{-1/2}b_2 - c_2\right](e^{2t\xi_i} - e^{2t}) \le c_1(e^{2t\xi_i} - e^{2t}), $$

with $c_1 > 0$, car $|\partial_t a| + |\partial_t b_1| + |\partial_{tt} b_1| + |\partial_{t,\theta_j} b_1| + |\partial_{t,\theta_j,\theta_k} b_1| \le C'e^{2t} < 1$, for $t$ very small.

We obtain on $[\xi_i, t_i] \times \mathbb{S}_4$,

$$ \bar{Z}_i(\bar{w}_i^{\xi_i} - \bar{w}_i) \le c_2[(\tilde{w}_i^{\xi_i})^{7/3} - \tilde{w}_i^{7/3}] + [\mu_i c_1 - C'(R)](e^{2t\xi_i} - e^{2t}) \le 0, $$

Like in [2,3], we use the Hopf maximum principle to conclude that:

$$ \sup_{\theta \in \mathbb{S}_4} \bar{w}_i^{\xi_i}(t_i, \theta) \ge \inf_{\theta \in \mathbb{S}_4} \bar{w}_i(t_i, \theta), $$

Thus,

$$ e^{3t_i/2} \min_{B(x_0, \epsilon_1)} u_i \le \tilde{c} e^{3(\lambda_i - t_i)/2}, $$

Thus,

$$ [u_i(y_i)]^{1-2\alpha} \min_{B(x_0, \epsilon_1)} u_i = e^{3(2t_i - \lambda_i)/2} \min_{B(x_0, \epsilon_1)} u_i \le \tilde{c}, $$

which it is a contradiction for $\alpha = \frac{3}{7}$.

Thus,

$$ \exists R > 0, \exists c = c(M, g, R) > 0 \left[ \sup_{B(x_0, R)} u_i \right]^{1/7} \times \inf_M u_i \le c \quad \forall i $$

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