Proof of Bernstein Theorem

Now, we are ready to prove the Bernstein Theorem up to the dimension \(n\leq 7\).

More analysis on the stationary cones

Recall that the Jacobi operator is defined as

for any smooth function \(\varphi\) on the regular part of \(M\). If \(M^n\subset\mathbb{R}^{n+1}\) is a stationary cone with isolated singular point \(0\), then we can rewrite the Jacobi operator as

where \(\Sigma=M \cap \mathbb{S}^{n}\). This is called the link of the stationary cone, which is a minimal hypersurface in \(\mathbb{S}^{n}\).

So it is important to study the operator \(\Delta^\Sigma + |A_\Sigma|^{2}\) on the link \(\Sigma\). Suppose \(\Sigma\) is not totally geodesic, we define \(\lambda_1(\Sigma)\) to be the first eigenvalue of the operator \(-\Delta^\Sigma - |A_\Sigma|^{2}\) by

Proof. We use the following Simons inequality for the link \(\Sigma\):

Now, we directly integrate the inequality over \(\Sigma\) to get

Using \(|A_\Sigma|\) as a test function in the Rayleigh quotient gives

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Proof. We compute

where in the last step we used that \(\varphi\) is an eigenfunction of the operator \(-\Delta^\Sigma - |A_\Sigma|^2\) with eigenvalue \(\lambda_1(\Sigma) \leq -(n-1)\), and that \(u\gt{}0\). The proof is finished by analyzing the resulting ODE inequality. ◻

We consider \(W(t)=V(t^{-1/\beta})t^{\gamma/\beta}\).

Then, we can choose \(\beta\) and \(\gamma\) such that \(W''(t)\leq 0\). To see this, we compute

So we have

The choice of \(\gamma = -\kappa_n\) and \(\beta = 2\sqrt{\frac{(n-2)^{2}}{4}-(n-1)}\) satisfies the above equations, and hence \(W''(t)\leq 0\). Since \(t=r^{-\beta}\) and \(W(t)=V(r)r^{\kappa_n}\), this concavity gives the desired monotonicity of \(V(r)r^{\kappa_n}\) in the \(r\) variable.

We are ready to prove the Bernstein Theorem. We need to study the blow-down limit of the graph of \(u\). This is due to De Giorgi [DG65].

Proof. Suppose \(M\) is the graph of \(u\). Then it is a minimal hypersurface in \(\mathbb{R}^{n+1}\), and it is minimizing area in \(\mathbb{R}^{n+1}\).

Let \(P\) be the region \(\{ (x,y) \in \mathbb{R}^n \times \mathbb{R}: y\lt{}u(x) \}\). By the previous result, we know \(P\) is perimeter minimizer in \(\mathbb{R}^{n+1}\). We consider the blow-down limit of \(P\) defined as follows. Define \(P_r:=\frac{1}{r}P\). Up to a subsequence, \(P_r\) converges to a set \(P_\infty\) in \(L^1_{\mathrm{loc}}(\mathbb{R}^{n+1})\). By compactness, \(P_\infty\) is a perimeter minimizer in \(\mathbb{R}^{n+1}\). The boundary of \(P_r\), denoted by \(M_r\), converges to the boundary of \(P_\infty\) in the varifold sense; denote the limit by \(V_\infty\).

By the previous regularity result, we know \(\mathrm{sing}\,\|V_\infty\|\) is empty or discrete. If it is empty, then \(V_\infty\) is a density-one hyperplane in \(\mathbb{R}^{n+1}\). If it is discrete, then \(V_\infty\) can only have an isolated singular point at the vertex \(0\), since it is a stationary cone.

Now, we consider \(P'_r:=P_r+e_{n+1}\). This is again a perimeter minimizer in \(\mathbb{R}^{n+1}\), and we have \(P_r \subset P'_r\). Up to a subsequence, \(P'_r\) converges to a set \(P'_\infty\), and \(P'_\infty = P_\infty + e_{n+1}\). In particular, \(P_\infty \subset P'_\infty\). By the strong maximum principle, we have either \(P_\infty = P'_\infty\), or their boundaries are disjoint. In the first case, we are done, since \(V_\infty\) is translation invariant along \(e_{n+1}\) direction, it can only be a density-one hyperplane in \(\mathbb{R}^{n+1}\). In the second case, we know that \(V_\infty+\lambda e_{n+1}\) is disjoint from \(V_\infty\) for any \(\lambda\gt{}0\). Hence, we can construct a positive Jacobi field \(u\) on \(M_\infty\) by \(u:=\left< \nu_{M_\infty}, e_{n+1} \right>\) where \(M_\infty:=\mathrm{reg}\|V_\infty\|\). Now, for the function \(V(r):=\int_{ \Sigma} \varphi(x) u(r x) d\Sigma(x)\), we have

where \(C\) is independent of \(r\). On the other hand, we have \(\kappa_7=2\), and hence

which implies \(V(r)\to +\infty\) as \(r\to 0\). This contradiction implies \(\Sigma\) is totally geodesic, and hence \(M_\infty\) is a density-one hyperplane in \(\mathbb{R}^{n+1}\). ◻

Proof. By Allard’s regularity theorem, \(M_r\) converges smoothly to a minimal hypersurface \(M_\infty\) in \(\mathbb{R}^{n+1}\).

In particular, if we denote \(A_r\) to be the second fundamental form of \(M_r\), then for any fixed \(x\in M=M_1\), we have

On the other hand, we have

which implies \(A(x)=0\). Hence, \(M\) is flat, so \(u\) is an affine function. ◻