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The Hessian determinant of each $C^{2}$ holomorphic component is nonpositive; a nondegenerate critical point is a saddle

Statement

Let $f = u + i v$ be holomorphic with $C^{2}$ components. For either component $w \in {u, v}$ ,

$\det H_{w} = - w_{x x}^{2} - w_{x y}^{2} \leq 0.$

If $a$ is a critical point of $w$ and $\det H_{w} (a) \neq 0$ , then the determinant is negative and $a$ is neither a local maximum nor a local minimum, hence is a saddle in the Hessian-test sense. If the determinant is zero, the Hessian test is inconclusive.

Facts & Assumptions

Given: A holomorphic $f = u + i v$ whose components are $C^{2}$ , and one component $w$ .

[L1]

Each such component is harmonic: $w_{x x} + w_{y y} = 0$ (The $C^{2}$ real and imaginary parts of a holomorphic function satisfy Laplace's equation and form a harmonic-conjugate pair).

[F1]

The Hessian is the matrix of second partial derivatives, and a critical point is one at which the gradient vanishes (The Hessian matrix and critical points of a scalar field).

[L2]

For a $C^{2}$ function, $w_{x y} = w_{y x}$ (Clairaut--Schwarz theorem for continuous second partial derivatives).

[L3]

At a critical point of a $C^{2}$ function of two variables, a negative Hessian determinant gives neither a local minimum nor a local maximum, while determinant zero gives no conclusion (The two-variable Hessian determinant test).

Proof

technique · direct

1.1

By [F1] and [L2], $\det H_{w} = w_{x x} w_{y y} - w_{x y}^{2}$ . By [L1], $w_{y y} = - w_{x x}$ , so $\det H_{w} = - w_{x x}^{2} - w_{x y}^{2} \leq 0$ .

F1L1L2algebra

2.1

If the determinant at a critical point is nonzero, step 1.1 makes it negative, and [L3] gives neither a local maximum nor a local minimum. If it is zero, [L3] gives no conclusion.

step 1.1givenL3∎

Depends on

Used by

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Sources

J. Lebl, Guide to Cultivating Complex Analysis, Exercise 2.1.7 (standard reference, not scraped)

The Hessian determinant of each C2C2 holomorphic component is nonpositive; a nondegenerate critical point is a saddle