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MATH
[1251]

Theorems from Local extrema

All theorems from Local extrema.

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The multivariable Fermat condition

If all first partial derivatives of ff exist at an interior local maximum or local minimum x0\mathbf{x}_0, then

∇f(x0)=0.{\nabla f(\mathbf{x}_0)=\mathbf{0}}.

Differentiability is sufficient for the partial derivatives to exist, but is stronger than what this coordinatewise condition needs. The condition is necessary, not a classification: a stationary point may be a local maximum, a local minimum, a saddle, or a point requiring higher-order analysis.

The second-order partial derivative test

Let x0=(x0,y0)\mathbf{x}_0=(x_0,y_0) be a stationary point. Assume that ff has continuous second-order partial derivatives on a neighbourhood of x0\mathbf{x}_0 (so ff is C2C^2 there). Define

D=det⁡Hf(x0)=fxx(x0)fyy(x0)−fxy(x0)2.D =\det H_f(\mathbf{x}_0) =f_{xx}(\mathbf{x}_0)f_{yy}(\mathbf{x}_0) -f_{xy}(\mathbf{x}_0)^2.

Then:

  • if D>0D>0 and fxx(x0)>0f_{xx}(\mathbf{x}_0)>0, x0\mathbf{x}_0 is a local minimum;
  • if D>0D>0 and fxx(x0)<0f_{xx}(\mathbf{x}_0)<0, x0\mathbf{x}_0 is a local maximum;
  • if D<0D<0, x0\mathbf{x}_0 is a saddle point;
  • if D=0D=0, the test is inconclusive.

At a stationary point, the quadratic Taylor term is

12hTHf(x0)h.\frac12\mathbf{h}^{\mathsf T}H_f(\mathbf{x}_0)\mathbf{h}.

For a symmetric Hessian, this quadratic form is positive definite if it is positive for every nonzero h\mathbf{h}, negative definite if it is negative for every nonzero h\mathbf{h} and indefinite if it takes both positive and negative values. The test uses DD and fxxf_{xx} to tell these cases apart.

When D>0D>0, the quadratic form has one sign in all non-zero directions and fxxf_{xx} determines whether that sign is positive or negative. When D<0D<0, the form changes sign, which is the algebraic signature of a saddle. If D=0D=0, higher-order terms or direct comparisons along curves are required.

To see the determinant rule when D>0D>0, write a=fxx(x0)a=f_{xx}(\mathbf{x}_0), b=fxy(x0)b=f_{xy}(\mathbf{x}_0) and c=fyy(x0)c=f_{yy}(\mathbf{x}_0). Then a≠0a\ne0 and the quadratic part satisfies

ah2+2bhk+ck2=a(h+bak)2+Dak2.ah^2+2bhk+ck^2 =a\left(h+\frac ba k\right)^2+\frac Da k^2.

Both coefficients on the right have the sign of aa, so small nonzero displacements change ff in the same direction.

For intuition, a symmetric Hessian has principal directions described by its eigenvectors. Its eigenvalues give the curvatures in those directions: both positive suggests a local minimum, both negative a local maximum and opposite signs a saddle. In two variables, DD is the product of those two eigenvalues, while fxxf_{xx} picks the sign when they agree.

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