Extreme values of multivariate functions

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A local maximum (or minimum) value of a function is the maximum (or minimum) value of the function in the neighbor of a point. More formally,

Definition (Local maximum, local minimum)

Let \(f(x,y)\) be a function on an open region \(U\subset \mathbb{R}^2\) and \(P=(a,b) \in U\).

  1. \(f(a,b)\) is said to be a local maximum value of the function \(f(x,y)\) if there exists \(\delta > 0\) such that, for all \((x,y)\in U\), if \((x,y)\in N_{\delta}(P)\cap U\) and \((x,y) \neq (a,b)\), then \(f(x,y) < f(a,b)\).
  2. \(f(a,b)\) is said to be a local minimum value of the function \(f(x,y)\) if there exists \(\delta > 0\) such that, for all \((x,y)\in U\), if \((x,y)\in N_{\delta}(P)\cap U\) and \((x,y) \neq (a,b)\), then \(f(x,y) > f(a,b)\).
Local maximum and minimum values are collectively called extreme values.

Theorem (Necessary condition for extreme values)

Let \(f(x,y)\) be a function on an open region \(U\), and \((a,b)\in U\). Suppose that \(f_x(a,b)\) and \(f_y(a,b)\) exist. If \(f(a,b)\) is an extreme value, then \(f_x(a,b) = f_y(a,b) = 0\).
Proof. If \(f(x,y)\) has an extreme value at \((a,b)\), then the univariate function \(f(x,b)\) of \(x\) has an extreme value at \(x = a\). By the corresponding theorem for univariate functions, \(f_x(a,b) = 0\). Similarly, \(f_y(a,b) = 0\). ■

The converse of this theorem is not necessarily true.

Example. Consider \(f(x,y) = x^2 - y^2\) (See Fig:NoExtreme below). We can see that \(f_x(0,0) = f_y(0,0) = 0\). However, the graph of \(f(x,0) = x^2\) is convex whereas that of \(f(0, y) = -y^2\) is concave. This means that, if we move from the origin along the \(x\)-axis, the value of \(f(x,y)\) increases, but if we move from the origin along the \(y\)-axis, the value decreases. Therefore, \(f(0,0)\)a  is neither a a local maximum nor a local minimum. □
Figrue Fig:NoExtreme. The graph of \(z=x^2 - y^2\). This function has neither a local maximum nor a local minimum at \((x,y) = (0,0)\). (See the above example.)

In the case of univariate functions, we can determine whether a function takes a local maximum or local minimum value at a given point by the sign of the second differential coefficient at that point. We have a corresponding theorem for the case of bivariate functions, but the situation is a bit more complicated. We remark the following theorem without proof.

Theorem (Criteria of extreme values)

Let \(f(x,y)\) be a function of class \(C^2\) on an open region \(U\). Suppose that \((a,b)\in U\) and \(f_x(a,b) = f_y(a,b) = 0\). Let us define the determinant \(D\) by
\[D = f_{xx}(a,b)f_{yy}(a,b) - [f_{xy}(a,b)]^2.\]
  1. If \(D > 0\),
    1. if \(f_{xx}(a,b) > 0\), then \(f(x,y)\) has a local minimum value at \((x,y) = (a,b)\);
    2. if \(f_{xx}(a,b) < 0\), then \(f(x,y)\) has a local maximum value at \((x,y) = (a,b)\).
  1. If \(D< 0\), \(f(x,y)\) does not have a local extremum at \((x,y) = (a,b)\).

Proof. Omitted (beyond the scope of this module). ■

Remark. Note that the "determinant" \(D\) in the above theorem is indeed the determinant of a matrix:\[D = \begin{vmatrix}f_{xx}(a,b) & f_{xy}(a,b)\\f_{yx}(a,b) & f_{yy}(a,b)\end{vmatrix}.\] This matrix is called the Hessian of the function. You may guess the Hessian of a general multivariate function \(f(x_1, x_2, \cdots, x_n)\). □

Remark. This theorem is only applicable to bivariate (two-variable) functions. For general multivariate functions, such as \(f(x_1, x_2, \cdots, x_n)\), the corresponding theorem is even more complicated (but it exists). That is, if all the eigenvalues of the Hessian are positive (or negative) at a given point, then the function takes a local minimum (or maximum) value at the point. (Note that the Hessian is always symmetric if the function is of class \(C^2\). Therefore, all the eigenvalues are guaranteed to be real.) □

Example. Let us find the extreme values of the function \(f(x,y) = x^3 + y^3 - 3(x+y)\) on \(\mathbb{R}^2\) (See the figure below).
The graph of \(z = x^3 + y^3 - 3(x+y)\).

Solving 
\[\begin{eqnarray*} f_x(x,y) &=& 3(x^2 - 1) = 0,\\ f_y(x,y) &=& 3(y^2 - 1) = 0, \end{eqnarray*}\]
we have
\[(x,y) = (1, 1), (-1, -1), (-1, 1), (1, -1).\]

From
\[\begin{eqnarray*} f_{xx}(x,y) &=& 6x,\\ f_{xy}(x,y) &=& 0,\\ f_{yy}(x,y) &=& 6y,\\ \end{eqnarray*}\]
we have 
\[D = f_{xx}(x,y)f_{yy}(x,y) - [f_{xy}(x,y)]^2 = 36xy.\]
  • When \((x,y) = (1, 1)\), we have \(D = 36 > 0\) and \(f_{xx}(1,1) = 6 > 0\) so that \(f(1,1) = -4\) is a local minimum.
  • When \((x,y) = (-1, -1)\), we have \(D = 36 > 0\) and \(f_{xx}(-1,-1) = -6 < 0\) so that \(f(-1,-1) = 4\) is a local maximum.
  • When \((x,y) = (-1, 1)\) or \((1, -1)\), we have \(D = -36 < 0\) so \(f(x,y)\) does not take extreme values.
□




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