Point of local extremum implies critical point for a function of multiple variables

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This article describes a test that can be used to determine whether a point in the domain of a function gives a point of local, endpoint, or absolute (global) maximum or minimum of the function, and/or to narrow down the possibilities for points where such maxima or minima occur.
View a complete list of such tests
This article describes an analogue for functions of multiple variables of the following term/fact/notion for functions of one variable: point of local extremum implies critical point

Statement

Suppose f is a function of a vector variable \overline{x}. Suppose \overline{c} is a point in the interior of the domain of f, i.e., f is defined in an open ball containing \overline{c}.

Suppose further that \overline{c} is a point of local extremum for f, i.e., f attains a local extreme value (maximum or minimum) at \overline{c}.

Then the following are true:

Facts used

  1. Point of local extremum implies critical point (the single variable case)
  2. Relation between gradient vector and directional derivatives

Proof

Directional derivative version

Given: A function f of a vector variable \overline{x}. A point \overline{c} in the interior of the domain where it attains a local extreme value. A unit vector \overline{u}.

To prove: The directional derivative \nabla_{\overline{u}}(f)(\overline{c}) equals zero or it doesn't exist.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Consider the following function of one variable: g(t) := f(\overline{c} + t\overline{u}). In other words, g is the restriction of f to a line through \overline{c} along the direction of \overline{u}.
2 g has a local extreme value at t = 0, which corresponds to the value f(\overline{c}). Step (1) Since f has a local extreme value at \overline{c}, g (a function of one variable) has a local extreme value of the same type at t = 0.
3 g has a critical point at 0, i.e., g'(0) = 0 or g'(0) doesn't exist. Fact (1) Step (2) Fact-step combination direct
4 The directional derivative \nabla_{\overline{u}}(f)(\overline{c}) equals zero or it doesn't exist. Definition of directional derivative Step (3) One of the definitions of directional derivative is as the ordinary derivative g'(0). Thus, this follows directly from Step (3).

Gradient vector version

Given: A function f of a vector variable \overline{x}. A point \overline{c} in the interior of the domain where it attains a local extreme value.

To prove: The gradient vector (\nabla f)(\overline{c}) is either zero or it doesn't exist.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 For every unit vector \overline{u}, the directional derivative \nabla_{\overline{u}}(f)(\overline{c}) equals zero or it doesn't exist. Directional derivative version of proof
2 If the gradient vector (\nabla f)(\overline{c}) exists, then for every unit vector \overline{u}, the directional derivative \nabla_{\overline{u}}(f)(\overline{c}) exists and equals \overline{u} \cdot (\nabla f)(\overline{c}). Fact (2)
3 If the gradient vector (\nabla f)(\overline{c}) exists, then for every unit vector \overline{u}, we have \overline{u} \cdot (\nabla f)(\overline{c}) = 0 Steps (1), (2) Step-combination direct
4 If the gradient vector (\nabla f)(\overline{c}) exists, it must equal the zero vector. Any vector whose dot product with every vector is zero must be the zero vector. Step (3) Step-fact combination direct