Let \(f:\mathbb R^n\to\mathbb R\) be a function whose partial derivatives of order at most \(2\) are everywhere defined and continuous.
Let \(a\in\mathbb R^n\) be a critical point of \(f\), i.e. \[ \frac{\partial f}{\partial x_j}(a)=0,\qquad j=1,\ldots,n. \] Prove that \(a\) is a local minimum provided the Hessian matrix \[ \left(\frac{\partial^2f}{\partial x_i\partial x_j}\right) \] is positive definite at \(x=a\).
Assume the Hessian matrix is positive definite at every \(x\in\mathbb R^n\). Prove that \(f\) has at most one critical point.
Solution
Let \(H_f(x)\) denote the Hessian matrix of \(f\) at \(x\).
1. A critical point \(a\) with positive-definite Hessian \(H_f(a)\) is astrict local minimum.
Because \(H_f(a)\) is positive definite, there is a constant \(c>0\) such that \[ v^T H_f(a)v\ge 2c\norm{v}^2 \qquad\text{for all }v\in\RR^n. \] The Hessian depends continuously on \(x\), so there is a neighborhood \(U\) of \(a\) such that \[ v^T H_f(x)v\ge c\norm{v}^2 \qquad\text{for all }x\in U,\ v\in\RR^n. \]
Let \(h\) be small enough that the segment \(a+th\), \(0\le t\le1\), lies in \(U\), and define \[ g(t)=f(a+th). \] Then \[ g'(0)=\nabla f(a)\cdot h=0 \] because \(a\) is critical, while \[ g''(t)=h^T H_f(a+th)h\ge c\norm{h}^2. \] Taylor’s formula with integral remainder gives \[ \begin{aligned} f(a+h)-f(a) &=g(1)-g(0)\\ &=g'(0)+\int_0^1(1-t)g''(t)\,dt\\ &\ge c\norm{h}^2\int_0^1(1-t)\,dt\\ &=\frac c2\norm{h}^2, \end{aligned} \] which is strictly positive for \(h\ne0\).
2. If the Hessian is positive definite everywhere, there is at most onecritical point.
Suppose, toward a contradiction, that \(a\ne b\) are both critical points. Put \[ v=b-a\ne0 \] and define \[ g(t)=f(a+tv),\qquad 0\le t\le1. \] Then \[ g'(t)=\nabla f(a+tv)\cdot v \] and \[ g''(t)=v^T H_f(a+tv)v>0 \] for every \(t\in[0,1]\), because the Hessian is positive definite and \(v\ne0\). Therefore \(g'\) is strictly increasing on \([0,1]\). But criticality of \(a\) and \(b\) gives \[ g'(0)=\nabla f(a)\cdot v=0, \qquad g'(1)=\nabla f(b)\cdot v=0, \] contradicting strict increase.
3. Q.E.D.
Step 1 proves part 1, and step 2 proves part 2.