Let \(f:\mathbb R^n\to\mathbb R\) be \(C^2\). Prove that every nondegenerate critical point of \(f\) is isolated.
Solution
Let \(x_0\in\mathbb R^n\) be a nondegenerate critical point of \(f\). Define \[ G=\nabla f:\mathbb R^n\to\mathbb R^n. \] Since \(f\) is \(C^2\), the map \(G\) is \(C^1\). Moreover \[ G(x_0)=0 \] because \(x_0\) is critical.
1. The derivative of \(G\) at \(x_0\) is invertible.
The Jacobian matrix of \(G=\nabla f\) is precisely the Hessian matrix of \(f\): \[ DG(x_0) =\left( \frac{\partial^2 f} {\partial x_i\partial x_j}(x_0) \right)_{i,j}. \] By nondegeneracy of the critical point, this matrix is nonsingular.
2. The gradient has no other zero sufficiently near \(x_0\).
By the inverse function theorem, there is an open neighborhood \(U\) of \(x_0\) such that \[ G|_U:U\to G(U) \] is one-to-one. If \(x\in U\) is another critical point, then \[ G(x)=0=G(x_0). \] Injectivity of \(G|_U\) forces \[ x=x_0. \] Therefore \(x_0\) is the only critical point in \(U\), so it is isolated.