Discussion
Let \(f(.)\) be a twice differentiable function from \(\mathbb{R}^{2} \rightarrow \mathbb{R}\). If \(\mathbf{p}, \mathbf{x}_{0} \in \mathbb{R}^{2}\) where \(\|\mathbf{p}\|\) is sufficiently small (here \(\|.\|\) is the Euclidean norm or distance function), then
\[ \begin{aligned} f(\mathbf{x}_{0} + \mathbf{p}) &= f(\mathbf{x}_{0}) + \nabla f(\mathbf{x}_{0})^{T}\mathbf{p} \\ &+ \tfrac{1}{2}\mathbf{p}^{T}\nabla^{2}f(\boldsymbol{\psi})\mathbf{p} \end{aligned} \]
where \(\boldsymbol{\psi} \in \mathbb{R}^{2}\) is a point on the line segment joining \(\mathbf{x}_{0}\) and \(\mathbf{x}_{0} + \mathbf{p}\). If \(\mathbf{x}_{0}\) is a strict local minimum of \(f(\mathbf{x})\), then which one of the following statements is TRUE?
Answer: B
The true statement is \(\nabla f(\mathbf{x}_{0})^{T}\mathbf{p} = 0\) and \(\mathbf{p}^{T}\nabla^{2}f(\boldsymbol{\psi})\mathbf{p} \gt 0\).
At a local minimum of a differentiable function the gradient is zero, so the first-order term \(\nabla f(\mathbf{x}_{0})^{T}\mathbf{p}\) vanishes for every direction. A strict minimum means \(f(\mathbf{x}_{0} + \mathbf{p}) \gt f(\mathbf{x}_{0})\) for every small non-zero p. With the first-order term gone, the expansion leaves
\[ \begin{aligned} &f(\mathbf{x}_{0} + \mathbf{p}) - f(\mathbf{x}_{0}) \\ &= \tfrac{1}{2}\mathbf{p}^{T}\nabla^{2}f(\boldsymbol{\psi})\mathbf{p} \end{aligned} \]
so the quadratic term must be positive.
A positive gradient term (option A) contradicts the zero gradient at a minimum. A zero quadratic term would leave f unchanged, so the minimum would not be strict, and a negative one would make x0 a point where f decreases, a maximum or saddle.
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