Let $U$ be open in $\real^n$ , and $f\colon U \to \real$ , $g\colon U \to \real^m$ be twice continuously differentiable functions. Assume that $p \in U$ is a stationary point for $f$ on $M = g^{-1}(\{ 0 \})$ , and $\D g$ has full rank everywhere ¹ on $M$ . Then we know that $p$ is the solution to the Lagrange multiplier system \begin{equation}\label{lagrange} \D f(p) = \lambda \cdot \D g(p)\,, \end{equation}for a Lagrange multiplier vector .

Our aim is to develop an analogue of the second derivative test for the stationary point $p$ .

The most straightforward way to proceed is to consider a coordinate chart $\alpha\colon V \to M$ for the manifold $M$ , and consider the Hessian of the function $f \circ \alpha\colon V \to \real^n$ at $0 = \alpha^{-1}(p)$ . This Hessian is in fact just the Hessian form of $f \colon M \to \real$ expressed in the coordinates of the chart $\alpha$ . But the whole point of using Lagrange multipliers is to avoid calculating coordinate charts directly, so we find an equivalent expression for $\D^2 (f\circ \alpha)(0)$ in terms of $\D^2 f(p)$ without mentioning derivatives of $\alpha$ .

To do this, we differentiate $f \circ \alpha$ twice using the chain rule and product rule ². To reduce clutter, from now on we use the prime notation for derivatives rather than $\D$ . \begin{equation}\label{second-derivative} \begin{split} (f \circ \alpha)''(0) &= ((f' \circ \alpha) \cdot \alpha')'(0) \\ &= \bigl ( (f'' \circ \alpha) \cdot \alpha' \cdot \alpha' + (f' \circ \alpha) \cdot \alpha'' \bigr) (0) \\ &= f''(\alpha(0)) \cdot \alpha'(0) \cdot \alpha'(0) + f'(\alpha(0)) \cdot \alpha''(0) \\ &= f''(p) \cdot \alpha'(0) \cdot \alpha'(0) + f'(p) \cdot \alpha''(0)\,. \end{split} \end{equation} If we interpret $(f \circ \alpha)''(0)$ as a bilinear mapping of vectors $u,v \in \real^{n-m}$ , then formula () really means \begin{equation}\label{bilinear} (f \circ \alpha)''(0) \cdot (u, v) = f''(p) \cdot \bigl(\alpha'(0)\cdot u, \alpha'(0) \cdot v\bigr) + f'(p) \cdot \bigl(\alpha''(0) \cdot (u, v)\bigr)\,. \end{equation}To obtain the quadratic form, we set $v = u$ ; also we abbreviate the vector $\alpha'(0) \cdot u$ by $h$ , which belongs to the tangent space $\mathrm{T}_p M$ of $M$ at $p$ . So, \begin{equation}\label{quadratic} (f \circ \alpha)''(0) \cdot u^2 = f''(p) \cdot h^2 + f'(p) \cdot \bigl(\alpha''(0) \cdot u^2 \bigr)\,. \end{equation}Naïvely, we might think that $(f \circ \alpha)''(0)$ is simply $f''(p)$ restricted to the tangent space $\mathrm{T}_p M$ . This happens to be the first term in (), but there is also an additional contribution by the second term involving $\alpha''(0)$ ; intuitively, $\alpha''(0)$ is the curvature of the surface (manifold) $M$ , ``changing the geometry'' of the graph of $f$ .

But the second term of () still involves $\alpha$ . To eliminate it, we differentiate the equation $g \circ \alpha = 0$ twice. \begin{equation}\label{g-second-derivative} 0 = (g \circ \alpha)''(0) = g''(p) \cdot \alpha'(0) \cdot \alpha'(0) + g'(p) \cdot \alpha''(0)\,. \end{equation}(It is derived the same way as () but with $f$ replaced by $g$ .) Now we can substitute () and () in () to eliminate the term $f'(p) \cdot \alpha''(0)$ : \begin{equation}\label{ff-second-derivative} \begin{split} (f \circ \alpha)''(0) &= f''(p) \cdot \alpha'(0) \cdot \alpha'(0) + \lambda \cdot g'(p) \cdot \alpha''(0) \\ &= f''(p) \cdot \alpha'(0) \cdot \alpha'(0) - \lambda \cdot g''(p) \cdot \alpha'(0) \cdot \alpha'(0)\,, \end{split} \end{equation}or expressed as a quadratic form, \begin{equation}\label{ff-quadratic} (f \circ \alpha)''(0) \cdot u^2 = f''(p) \cdot h^2 - \lambda \cdot g''(p) \cdot h^2\,. \end{equation} Thus, to understand the nature of the stationary point $p$ , we can study the modified Hessian: \begin{equation}\label{lagrange-hessian} f''(p) - \lambda \cdot g''(p)\,, \quad \text{ restricted to $\mathrm{T}_p M$. } \end{equation}For example, if this bilinear form is positive definite, then $p$ is a local minimum, and if it is negative definite, then $p$ is a local maximum, and so on. All the tests that apply to the usual Hessian in $\real^n$ apply to the modified Hessian ().

In coordinates of $\real^n$ , the modified Hessian () takes the form \begin{equation}\label{lagrange-hessian-coord} \sum_{i=1}^n \sum_{j=1}^n \left( \left.\frac{\partial^2 f}{\partial x^i \partial x^j}\right|_p - \sum_{k=1}^m \lambda_k \, \left.\frac{\partial^2 g^k}{\partial x^i \partial x^j}\right|_p \right) h^i h^j\,. \end{equation}We emphasize that the vector $h$ can be restricted to lie in the tangent space $\mathrm{T}_p M$ , when studying the stationary point $p$ of $f$ restricted to $M$ .

In matrix form () can be written \begin{equation}\label{lagrange-hessian-matrix} B= \begin{bmatrix} \dfrac{\partial^2 f}{\partial x^i \partial x^j} - \sum\limits_{k=1}^m \lambda_k \, \dfrac{\partial^2 g^k}{\partial x^i \partial x^j} \end{bmatrix}_{ij}\,. \end{equation}But again, the test vector $h$ need only lie on $\mathrm{T}_p M$ , so if we want to apply positive/negative definiteness tests for matrices, they should instead be applied to the projected or reduced Hessian: \begin{equation} Z^\mathrm{t} B Z \end{equation}where the columns of the $n \times (n-m)$ matrix $Z$ form a basis for $\mathrm{T}_p M = \ker g'(p) \subset \real^n$ .

Footnotes

... everywhere¹

Actually, only $\D g(p)$ needs to have full rank, and the arguments presented here continue to hold in that case, although $M$ would not necessarily be a manifold then.

...http://planetmath.org/encyclopedia/ProductRule.html ²

Note that the ``product'' operation involved (second equality of ()) is the operation of composition of two linear mappings. Think hard about this if you are not sure; it took me several tries to get this formula right, since multi-variable iterated derivatives have a complicated structure.