Green-kernel proof of the stable Bernstein theorem in R^4

The Green-kernel proof of the stable Bernstein theorem in \(\mathbb{R}^4\) is due to Cabré–Catino–Mari–Mastrolia–Roncoroni [CCM+26]. The key point is that this proof does not use \(\mu\)-bubbles. Instead, it combines:

the positive supersolution furnished by stability;
the minimal positive Green kernel \(\mathscr G\) of \(-\Delta^M\);
a sharp pointwise estimate for \(|\nabla\mathscr G|\) obtained from Bochner–Kato and a simple convexity lemma;
a weighted Schoen–Simon–Yau inequality;
a logarithmic cut-off in the variable \(\mathscr G\).

Theorem 5.3.1

Let \(M^3\hookrightarrow\mathbb{R}^4\) be a connected, complete, two-sided, stable minimal immersion. Then \(M\) is an affine hyperplane.

Reduction to bounded curvature. It is enough to prove the theorem under the auxiliary assumption

\[ |A|\in L^\infty(M). \]

Indeed, if the theorem were false, then \(|A|\) is nonzero somewhere. If \(|A|\) is globally bounded, we are already in the bounded-curvature case. If not, one performs the standard point-picking argument on intrinsic balls \(B_j^M(p_0)\): choose \(q_j\) and a scale \(\lambda_j=|A(q_j)|\to\infty\) so that, after replacing the immersion by

\[ F_j(x)=\lambda_j\bigl(F(x)-F(q_j)\bigr), \]

the rescaled hypersurfaces have \(|A_j|(q_j)=1\) and uniformly bounded curvature on larger and larger intrinsic balls around \(q_j\). Stability and minimality are scale invariant, so a smooth local compactness theorem gives a complete, two-sided, stable minimal limit in \(\mathbb{R}^4\) with bounded second fundamental form and \(|A|(0)=1\). The bounded-curvature case below would force this limit to be flat, a contradiction.

Green kernel and stability supersolution. Let now \(M^n\hookrightarrow\mathbb{R}^{n+1}\) be complete, two-sided, stable, and minimal, with \(n\geq3\), and assume \(|A|\in L^\infty(M)\). We keep \(n\) arbitrary until the last step. Stability means

\begin{equation} \tag{5.3.1} \label{eq:green-stability} \int_M |A|^2\phi^2\,d\mu \leq \int_M |\nabla\phi|^2\,d\mu, \qquad \phi\in \operatorname{Lip}_c(M). \end{equation}

Equivalently, the operator \(-\Delta-|A|^2\) is nonnegative. By the positive solution characterization of nonnegative Schrödinger operators, there exists a positive \(C^2\) function \(u\) such that

\begin{equation} \tag{5.3.2} \label{eq:green-u-supersolution} \Delta u+|A|^2u\leq0 \qquad\text{on }M. \end{equation}

We next explain why the Green kernel exists. The needed potential-theoretic input is the following.

Definition 5.3.2

Let \((N^m,g)\) be a connected, complete, noncompact Riemannian manifold. We say that \(N\) is non-parabolic if the Laplacian \(-\Delta^N\) admits a positive minimal Green kernel: for some, equivalently for every, point \(o\in N\) there is a function \(G(o,\cdot)\gt{}0\) on \(N\setminus\{o\}\) such that

\[ \Delta^N G(o,\cdot)=-\delta_o, \]

in the distributional sense, and \(G\) is minimal among all positive fundamental solutions. If no such positive Green kernel exists, \(N\) is called parabolic.

Equivalently, \(N\) is parabolic if every positive superharmonic function on \(N\) is constant; equivalently, the Brownian motion on \(N\) is recurrent. Another useful equivalent criterion is the capacity criterion below.

Definition 5.3.3

For a compact set \(K\Subset N\), define its \(2\)-capacity by

\[ \operatorname{Cap}_2(K) := \inf\left\{ \int_N |\nabla\phi|^2\,d\mu: \phi\in C_c^\infty(N),\quad \phi\geq1 \text{ on a neighborhood of }K \right\}. \]

Then \(N\) is non-parabolic if and only if \(\operatorname{Cap}_2(K)\gt{}0\) for some nonempty compact set \(K\); it is parabolic if and only if \(\operatorname{Cap}_2(K)=0\) for every compact \(K\).

Proposition 5.3.4

Let \((N^m,g)\) be complete and noncompact, with \(m\gt{}2\). Suppose that there is a constant \(S\gt{}0\) such that

\begin{equation} \tag{5.3.3} \label{eq:sobolev-nonparabolic} \left(\int_N |f|^{\frac{2m}{m-2}}\,d\mu\right)^{\frac{m-2}{m}} \leq S\int_N |\nabla f|^2\,d\mu \qquad \forall f\in C_c^\infty(N). \end{equation}

Then \(N\) is non-parabolic.

Proof. Choose a compact set \(K\Subset N\) with \(|K|\gt{}0\). If \(\phi\in C_c^\infty(N)\) and \(\phi\geq1\) near \(K\), then

\[ |K| \leq \int_K |\phi|^{\frac{2m}{m-2}}\,d\mu \leq \int_N |\phi|^{\frac{2m}{m-2}}\,d\mu. \]

Raising both sides to the power \((m-2)/m\) and using (5.3.3), we get

\[ |K|^{\frac{m-2}{m}} \leq \left(\int_N |\phi|^{\frac{2m}{m-2}}\,d\mu\right)^{\frac{m-2}{m}} \leq S\int_N |\nabla\phi|^2\,d\mu. \]

Taking the infimum over all admissible \(\phi\) gives

\[ \operatorname{Cap}_2(K) \geq S^{-1}|K|^{\frac{m-2}{m}}\gt{}0. \]

By the capacity criterion, \(N\) is non-parabolic. ◻

Proposition 5.3.5

Let \(M^n\hookrightarrow\mathbb{R}^{n+1}\) be a complete minimal hypersurface with \(n\gt{}2\). Then \(M\) is non-parabolic.

Proof. We use the sharp Michael–Simon Sobolev inequality of Brendle [Bre21]; in the present minimal hypersurface case it gives the following \(W^{1,2}\) Sobolev consequence:

\begin{equation} \tag{5.3.4} \label{eq:minimal-W12-sobolev} \left(\int_M |f|^{\frac{2n}{n-2}}\,d\mu\right)^{\frac{n-2}{n}} \leq C_n \int_M |\nabla f|^2\,d\mu . \end{equation}

Since \(n\gt{}2\), Proposition 5.3.4 applies and \(M\) is non-parabolic. ◻

For the theorem, Proposition 5.3.5 applies to our \(M^n\), since \(n\geq3\). Hence there exists a minimal positive Green kernel \(\mathscr G\) of \(-\Delta\) with pole \(o\in M\):

\[ \Delta\mathscr G=-\delta_o,\qquad \mathscr G\gt{}0,\qquad \mathscr G(x)\to0\quad\text{as }x\to\infty. \]

For a.e. \(s\gt{}0\), the level set

\[ \Sigma_s:=\{\mathscr G=s\}, \]

is smooth, and the Green flux identity gives

\begin{equation} \tag{5.3.5} \label{eq:green-flux} \int_{\Sigma_s}|\nabla\mathscr G|\,d\sigma=1. \end{equation}

Indeed, applying the divergence theorem to \(\{\mathscr G\gt{}s\}\setminus B_\varepsilon(o)\) and letting \(\varepsilon\to0\) gives exactly the mass of the pole.

Ricci lower bound from the Gauss equation. Let \(\{e_i\}_{i=1}^n\) diagonalize \(A\) at a point, with principal curvatures \(\kappa_i\) and \(\sum_i\kappa_i=0\). The traced Gauss equation gives

\[ \mathrm{Ric}(e_i,e_i) = \sum_{j\neq i}A_{ii}A_{jj} = -\kappa_i^2. \]

Thus, for every unit vector \(v\),

\[ \mathrm{Ric}(v,v)=-|A(v,\cdot)|^2. \]

Since \(A\) is trace-free,

\[ \kappa_i^2 = \left(\sum_{j\neq i}\kappa_j\right)^2 \leq (n-1)\sum_{j\neq i}\kappa_j^2 =(n-1)(|A|^2-\kappa_i^2), \]

and hence \(\kappa_i^2\leq\frac{n-1}{n}|A|^2\). Therefore

\begin{equation} \tag{5.3.6} \label{eq:green-ricci-lower} \mathrm{Ric}\geq -\frac{n-1}{n}|A|^2g. \end{equation}

The key convexity lemma. We shall use the following elementary computation several times.

Lemma 5.3.6

Let \(w,v\gt{}0\) be \(C^2\) functions on a domain \(\Omega\subset M\) satisfying

\[ \Delta w+Ww\geq0,\qquad \Delta v+Vv\leq0. \]

For \(\delta\geq0\), set

\[ \xi_\delta=w^{1+\delta}v^{-\delta}. \]

Then

\begin{equation} \tag{5.3.7} \label{eq:green-magic} \Delta\xi_\delta \geq \bigl(\delta V-(1+\delta)W\bigr)\xi_\delta +\delta(1+\delta) \left|\nabla\log\frac{w}{v}\right|^2\xi_\delta . \end{equation}

Proof. Since \(\xi_\delta=e^{\log\xi_\delta}\),

\[ \Delta\xi_\delta = \xi_\delta\left(\Delta\log\xi_\delta+ |\nabla\log\xi_\delta|^2\right). \]

Now

\[ \log\xi_\delta=(1+\delta)\log w-\delta\log v. \]

Using

\[ \Delta\log w=\frac{\Delta w}{w}-|\nabla\log w|^2, \qquad \Delta\log v=\frac{\Delta v}{v}-|\nabla\log v|^2, \]

we get

\[ \begin{aligned} \Delta\log\xi_\delta &=(1+\delta)\Delta\log w-\delta\Delta\log v \\ &\geq \delta V-(1+\delta)W -(1+\delta)|\nabla\log w|^2 +\delta|\nabla\log v|^2 . \end{aligned} \]

Also,

\[ \begin{aligned} |\nabla\log\xi_\delta|^2 &=(1+\delta)^2|\nabla\log w|^2 +\delta^2|\nabla\log v|^2 \\ &\quad -2\delta(1+\delta) \left\langle\nabla\log w,\nabla\log v\right\rangle . \end{aligned} \]

Adding the last two displays gives

\[ \Delta\log\xi_\delta+|\nabla\log\xi_\delta|^2 \geq \delta V-(1+\delta)W +\delta(1+\delta) \left|\nabla\log w-\nabla\log v\right|^2, \]

which is (5.3.7). ◻

Pointwise gradient estimate for the Green kernel. We first record the local differential inequality for the Green kernel.

Proposition 5.3.7

Let

\[ q=|\nabla\mathscr G|, \qquad \alpha=\frac{n-2}{n-1}, \qquad w=q^\alpha . \]

On the open set \(\{q\gt{}0\}\setminus\{o\}\), the function \(w\) satisfies

\begin{equation} \tag{5.3.8} \label{eq:green-gradient-subsolution} \Delta w+\frac{n-2}{n}|A|^2w\geq0. \end{equation}

Proof. Recall first the standard Bochner formula: for every smooth function \(f\) on a Riemannian manifold,

\[ \frac12\Delta |\nabla f|^2 = |\nabla^2 f|^2 +\left\langle\nabla f,\nabla\Delta f\right\rangle +\mathrm{Ric}(\nabla f,\nabla f). \]

We apply this to \(f=\mathscr G\) on \(M\setminus\{o\}\). Since \(\Delta\mathscr G=0\) away from the pole, the middle term vanishes. Hence

\[ \frac12\Delta q^2 = |\nabla^2\mathscr G|^2+\mathrm{Ric}(\nabla\mathscr G,\nabla\mathscr G). \]

The refined Kato inequality for harmonic functions gives

\[ |\nabla^2\mathscr G|^2\geq \frac{n}{n-1}|\nabla q|^2. \]

Together with the Ricci lower bound (5.3.6), this gives

\[ \frac12\Delta q^2 \geq \frac{n}{n-1}|\nabla q|^2 -\frac{n-1}{n}|A|^2q^2. \]

Since

\[ \frac12\Delta q^2=q\Delta q+|\nabla q|^2, \]

we subtract \(|\nabla q|^2\) and obtain

\begin{equation} \tag{5.3.9} \label{eq:green-q-ineq} q\Delta q \geq \frac{1}{n-1}|\nabla q|^2 -\frac{n-1}{n}|A|^2q^2. \end{equation}

Now compute the Laplacian of \(w=q^\alpha\):

\[ \begin{aligned} \Delta w &=\alpha q^{\alpha-1}\Delta q +\alpha(\alpha-1)q^{\alpha-2}|\nabla q|^2 \\ &\geq \alpha q^{\alpha-2} \left[ \frac{1}{n-1}|\nabla q|^2 -\frac{n-1}{n}|A|^2q^2 \right] +\alpha(\alpha-1)q^{\alpha-2}|\nabla q|^2, \end{aligned} \]

where we used (5.3.9) in the second line. The coefficient of \(|\nabla q|^2\) is

\[ \alpha\left(\frac1{n-1}+\alpha-1\right) = \alpha\left(\frac1{n-1}+\frac{n-2}{n-1}-1\right)=0. \]

Hence

\[ \Delta w \geq -\alpha\frac{n-1}{n}|A|^2q^\alpha = -\frac{n-2}{n}|A|^2w, \]

which is exactly (5.3.8). ◻

Apply Lemma 5.3.6 to

\[ W=\frac{n-2}{n}|A|^2,\qquad V=|A|^2,\qquad v=u, \qquad \delta\geq0. \]

For

\[ \xi_\delta=w^{1+\delta}u^{-\delta}, \]

the coefficient of the zeroth-order \(|A|^2\xi_\delta\) term in Lemma 5.3.6 is

\[ \delta V-(1+\delta)W = \left[ \delta-(1+\delta)\frac{n-2}{n} \right]|A|^2 = \frac{2\delta-(n-2)}{n}|A|^2 . \]

Thus this coefficient is nonnegative when \(\delta\geq\frac{n-2}{2}\), and it vanishes exactly at the critical choice

\[ \delta=\frac{n-2}{2}. \]

With this choice, \(1+\delta=\frac n2\), and

\[ \xi:=w^{n/2}u^{-(n-2)/2} = |\nabla\mathscr G|^{\frac{n(n-2)}{2(n-1)}}u^{-\frac{n-2}{2}} \]

is subharmonic on \(\{q\gt{}0\}\setminus U\), where

\[ U:=\{\mathscr G\gt{}1\}. \]

Because \(|A|\) is bounded, (5.3.6) gives a lower Ricci bound, and the Cheng–Yau gradient estimate [CY75] applied outside \(U\) gives

\[ |\nabla\mathscr G|\leq C\mathscr G. \]

We package the remaining two comparison-principle arguments into one lemma. First \(\mathscr G\) is compared with the stability supersolution \(u\); this auxiliary estimate is used only to remove the negative power of \(u\) and prove \(\xi\to0\) at infinity. Then \(\xi\) is compared directly with the harmonic barrier \(\mathscr G\).

Lemma 5.3.8

There is a constant \(C_0\gt{}0\) such that

\[ \xi\leq C_0\mathscr G \qquad\text{on }M\setminus U. \]

Proof. We first prove the auxiliary comparison

\[ \mathscr G\leq Cu \qquad\text{on }M\setminus U. \]

Since \(\mathscr G\to0\) at infinity, the set \(\bar U=\{\mathscr G\geq1\}\) is compact. Choose \(C\) so large that

\[ \mathscr G\leq Cu \qquad\text{on }\partial U. \]

This is possible because \(\mathscr G=1\) on \(\partial U\) and \(u\gt{}0\) there. Let \(\Omega_R\) be a smooth exhaustion of \(M\) with \(\bar U\Subset\Omega_R\), and let \(\mathscr G_R\) be the Dirichlet Green function on \(\Omega_R\) with pole \(o\). On the annular region \(\Omega_R\setminus \bar U\), the function \(\mathscr G_R\) is harmonic and vanishes on \(\partial\Omega_R\). Also,

\[ \Delta(Cu)=C\Delta u\leq -C|A|^2u\leq0, \]

so \(Cu\) is superharmonic. On the boundary of \(\Omega_R\setminus \bar U\) we have

\[ \mathscr G_R\leq \mathscr G=1\leq Cu \quad\text{on }\partial U, \qquad \mathscr G_R=0\leq Cu \quad\text{on }\partial\Omega_R. \]

The comparison principle on the bounded annulus therefore gives

\[ \mathscr G_R\leq Cu \qquad\text{on }\Omega_R\setminus U. \]

Letting \(R\to\infty\) and using \(\mathscr G_R\uparrow\mathscr G\), we obtain

\[ \mathscr G\leq Cu \qquad\text{on }M\setminus U. \]

This is where the first comparison is used. Combining it with \(|\nabla\mathscr G|\leq C\mathscr G\), we get

\[ \xi \leq C\mathscr G^{\frac{n(n-2)}{2(n-1)}}u^{-\frac{n-2}{2}} \leq C\mathscr G^{\frac{n-2}{2(n-1)}}\to0 \qquad\text{as }x\to\infty. \]

Now we perform the second comparison, this time between the subharmonic quantity \(\xi\) and a multiple of the harmonic Green kernel \(\mathscr G\). Since \(\bar U\) is compact and \(\xi\) is continuous up to \(\partial U\), while \(\mathscr G=1\) on \(\partial U\), we can choose \(C_0\) so large that

\[ \xi\leq C_0\mathscr G \qquad\text{on }\partial U. \]

Set

\[ h:=\xi-C_0\mathscr G . \]

On \(M\setminus U\), the pole \(o\) is not present, so \(\mathscr G\) is harmonic. Also \(\xi\) is subharmonic there, in the weak sense obtained from the preceding pointwise computation by the standard regularization \(q_\tau=(q^2+\tau)^{1/2}\) and then letting \(\tau\downarrow0\). Hence

\[ \Delta h=\Delta\xi-C_0\Delta\mathscr G\geq0 \qquad\text{on }M\setminus U. \]

Thus \(h\) is subharmonic. Moreover,

\[ h\leq0\qquad\text{on }\partial U. \]

We now explain how the boundary condition at infinity is used. Since \(\xi(x)\to0\) and \(\mathscr G(x)\to0\) as \(x\to\infty\), for every \(\varepsilon\gt{}0\) there is a compact set \(K_\varepsilon\supset \bar U\) such that

\[ h(x)=\xi(x)-C_0\mathscr G(x)\leq \xi(x)\leq\varepsilon \qquad\text{on }M\setminus K_\varepsilon . \]

Choose a smooth bounded domain \(\Omega_\varepsilon\) with \(K_\varepsilon\Subset\Omega_\varepsilon\). On the bounded annular domain \(\Omega_\varepsilon\setminus\bar U\), the maximum principle gives

\[ h\leq\max\{0,\varepsilon\}=\varepsilon. \]

Indeed, the inner boundary gives \(h\leq0\), and the outer boundary lies in \(M\setminus K_\varepsilon\), where \(h\leq\varepsilon\). Letting \(\varepsilon\downarrow0\), we obtain \(h\leq0\) on \(M\setminus U\), namely \(\xi\leq C_0\mathscr G\). ◻

Taking the power \(2/n\) yields the Green-kernel gradient estimate

\begin{equation} \tag{5.3.10} \label{eq:green-gradient-estimate} |\nabla\mathscr G|^{\frac{n-2}{n-1}} \leq C u^{\frac{n-2}{n}}\mathscr G^{\frac2n} \qquad\text{on }M\setminus U. \end{equation}

Weighted Schoen–Simon–Yau inequality. Set

\[ \beta=\frac{n-2}{n}. \]

Simons’ identity and the refined Kato inequality for \(A\) give, on \(\{|A|\gt{}0\}\),

\begin{equation} \tag{5.3.11} \label{eq:green-simons-kato} |A|\Delta |A| \geq \frac2n|\nabla|A||^2-|A|^4. \end{equation}

Let \(\widetilde w=|A|^\beta\). Then

\[ \begin{aligned} \Delta\widetilde w &=\beta |A|^{\beta-1}\Delta|A| +\beta(\beta-1)|A|^{\beta-2}|\nabla|A||^2 \\ &\geq \beta |A|^{\beta-2} \left[ \frac2n|\nabla|A||^2-|A|^4 \right] +\beta(\beta-1)|A|^{\beta-2}|\nabla|A||^2 . \end{aligned} \]

Again the gradient coefficient vanishes because

\[ \frac2n+\beta-1 = \frac2n+\frac{n-2}{n}-1=0. \]

Hence

\begin{equation} \tag{5.3.12} \label{eq:green-A-subsolution} \Delta\widetilde w+\frac{n-2}{n}|A|^2\widetilde w\geq0. \end{equation}

For \(0\lt{}t\leq1\), (5.3.2) gives

\[ \Delta u^t = t u^{t-1}\Delta u+t(t-1)u^{t-2}|\nabla u|^2 \leq -t|A|^2u^t. \]

Apply Lemma 5.3.6 to

\[ w=\widetilde w,\qquad v=u^t,\qquad W=\frac{n-2}{n}|A|^2,\qquad V=t|A|^2. \]

For \(\delta\geq0\), define

\[ z=|A|^{\beta(1+\delta)}u^{-t\delta}. \]

Then

\[ \Delta z \geq \left(t\delta-(1+\delta)\frac{n-2}{n}\right)|A|^2z. \]

Adding \(|A|^2z\) to both sides,

\begin{equation} \tag{5.3.13} \label{eq:green-z-subsolution} \Delta z+|A|^2z \geq \gamma |A|^2z, \qquad \gamma:= \frac2n+\left(t-\frac{n-2}{n}\right)\delta . \end{equation}

Assume \(\gamma\gt{}0\). Using (5.3.1) with test function \(z\psi\) and integrating by parts gives

\[ \begin{aligned} 0 &\leq \int_M |\nabla(z\psi)|^2-|A|^2z^2\psi^2 \\ &= \int_M z^2|\nabla\psi|^2 -\int_M z\psi^2(\Delta z+|A|^2z). \end{aligned} \]

Together with (5.3.13),

\[ \gamma\int_M |A|^2z^2\psi^2 \leq \int_M z^2|\nabla\psi|^2. \]

Put

\[ m:=1+\beta(1+\delta). \]

Taking \(\psi=\varphi^m\) and using Young’s inequality,

\begin{equation} \tag{5.3.14} \label{eq:green-weighted-ssy} \int_M |A|^{2m}u^{-2t\delta}\varphi^{2m} \leq C\int_M u^{-2t\delta}|\nabla\varphi|^{2m}, \qquad \varphi\in\operatorname{Lip}_c(M). \end{equation}

Indeed, the right-hand side before Young is

\[ C\int_M |A|^{2(m-1)}u^{-2t\delta}\varphi^{2m-2}|\nabla\varphi|^2, \]

and this is bounded by

\[ \frac12\int_M |A|^{2m}u^{-2t\delta}\varphi^{2m} + C\int_M u^{-2t\delta}|\nabla\varphi|^{2m}. \]

The parameter choice and the logarithmic cut-off. We now choose \(\varphi=\eta(\mathscr G)\). By the coarea formula,

\begin{equation} \tag{5.3.15} \label{eq:green-coarea} \begin{aligned} \int_M u^{-2t\delta}|\nabla\varphi|^{2m} &= \int_M u^{-2t\delta}|\eta'(\mathscr G)|^{2m} |\nabla\mathscr G|^{2m} \\ &= \int_0^\infty |\eta'(s)|^{2m} \left[ \int_{\Sigma_s} u^{-2t\delta}|\nabla\mathscr G|^{2m-1}\,d\sigma \right]ds . \end{aligned} \end{equation}

The estimate (5.3.10) gives

\[ u^{-t\delta}|\nabla\mathscr G|^{\beta(1+\delta)} \leq C u^{-t\delta+ \frac{(n-1)(n-2)}{n^2}(1+\delta)} \mathscr G^{\frac{2(n-1)}{n^2}(1+\delta)}. \]

We therefore impose

\begin{equation} \tag{5.3.16} \label{eq:green-choice-1} t\delta= \frac{(n-1)(n-2)}{n^2}(1+\delta). \end{equation}

Then

\[ u^{-2t\delta}|\nabla\mathscr G|^{2m-1} = |\nabla\mathscr G| \left( u^{-t\delta}|\nabla\mathscr G|^{\beta(1+\delta)} \right)^2 \leq C|\nabla\mathscr G| \mathscr G^{\frac{4(n-1)}{n^2}(1+\delta)}. \]

On \(\Sigma_s\) this becomes

\[ u^{-2t\delta}|\nabla\mathscr G|^{2m-1} \leq C|\nabla\mathscr G|s^{\frac{4(n-1)}{n^2}(1+\delta)}. \]

Using the flux identity (5.3.5), (5.3.15) becomes

\[ \int_M u^{-2t\delta}|\nabla\eta(\mathscr G)|^{2m} \leq C\int_0^\infty |\eta'(s)|^{2m} s^{\frac{4(n-1)}{n^2}(1+\delta)}\,ds. \]

For a logarithmic cut-off to close, we need

\begin{equation} \tag{5.3.17} \label{eq:green-choice-2} \frac{4(n-1)}{n^2}(1+\delta)\geq 2m-1 = 1+\frac{2(n-2)}{n}(1+\delta). \end{equation}

This inequality has a nonnegative solution \(\delta\) only for \(n=3\). In that case,

\[ \beta=\frac13,\qquad \delta=\frac72,\qquad t=\frac27, \qquad m=\frac52, \qquad \gamma=\frac12. \]

Then (5.3.14) reads

\begin{equation} \tag{5.3.18} \label{eq:green-final-ssy-n3} \int_M |A|^5u^{-2}\varphi^5 \leq C\int_M u^{-2}|\nabla\varphi|^5. \end{equation}

Also, the Green gradient estimate becomes

\[ |\nabla\mathscr G|^{1/2} \leq Cu^{1/3}\mathscr G^{2/3}. \]

Therefore

\[ u^{-1}|\nabla\mathscr G|^{3/2} \leq C\mathscr G^2, \]

and, on \(\Sigma_s\),

\[ u^{-2}|\nabla\mathscr G|^4 = |\nabla\mathscr G| \left(u^{-1}|\nabla\mathscr G|^{3/2}\right)^2 \leq C|\nabla\mathscr G|s^4. \]

Hence

\[ \int_M u^{-2}|\nabla\eta(\mathscr G)|^5 \leq C\int_0^\infty |\eta'(s)|^5s^4\,ds. \]

For \(R\gt{}1\), choose the logarithmic cut-off

\[ \eta_R(s)= \begin{cases} 0,&0\leq s\leq R^{-2},\\[2mm] 2+\dfrac{\log s}{\log R},&R^{-2}\lt{}s\lt{}R^{-1},\\[3mm] 1,&s\geq R^{-1}. \end{cases} \]

Then \(|\eta_R'(s)|=(s\log R)^{-1}\) on \((R^{-2},R^{-1})\) and \(0\) elsewhere. Putting \(\varphi=\eta_R(\mathscr G)\) in (5.3.18),

\[ \begin{aligned} \int_{\{\mathscr G\geq R^{-1}\}} |A|^5u^{-2} &\leq C\int_M u^{-2}|\nabla\eta_R(\mathscr G)|^5 \\ &\leq C\int_{R^{-2}}^{R^{-1}} \frac{s^4}{s^5(\log R)^5}\,ds \\ &= \frac{C}{(\log R)^5}\int_{R^{-2}}^{R^{-1}}\frac{ds}{s} = \frac{C}{(\log R)^4}. \end{aligned} \]

Letting \(R\to\infty\), the sets \(\{\mathscr G\geq R^{-1}\}\) exhaust \(M\) up to the end, and the right-hand side tends to zero. Hence

\[ |A|^5u^{-2}\equiv0. \]

Since \(u\gt{}0\), we conclude \(A\equiv0\). Thus the immersion is totally geodesic, and the connected complete image is an affine hyperplane in \(\mathbb{R}^4\). This proves Theorem 5.3.1.

Green-kernel proof of the stable Bernstein theorem in R^4

Gaoming Wang

Assistant Professor