Mixed Partials, Taylor Formulae, and Extrema: Examples and Counterexamples

For $y\mathrm{\ne }0$, $f(0,y)=0$, so $f_y(0,0)=0$; for $x\mathrm{\ne }0$, $f_y(x,0)=x$, hence $f_{xy}(0,0)=1$.

For $x\mathrm{\ne }0$, $f(x,0)=0$, so $f_x(0,0)=0$; for $y\mathrm{\ne }0$, $f_x(0,y)=-y$, hence $f_{yx}(0,0)=-1$.

Thus the two mixed partials exist and differ, while the continuity hypothesis in [L1] fails.

On a nonvertical line $y=mx$, $p(x,mx)=x^2(m-3x)(m-x)$, which is positive for sufficiently small nonzero $x$ when $m\mathrm{\ne }0$; on $y=0$ it is $3x^4>0$, and on the vertical line $x=0$ it is $y^2>0$.

Along the parabola $y=2x^2$, $p(x,2x^2)=-x^4<0$ for $x\mathrm{\ne }0$, whereas $p(0,0)=0$.

Hence $(0,0)$ is a strict linewise minimum but is not a local minimum in the sense of Local and strict local extrema for scalar fields on Euclidean open sets.

The flat function $r$ is smooth at $0$: every derivative is a polynomial in $1\mathrm{/}x$ times $e^{-1\mathrm{/}{x}^2}$ off $0$ and tends to $0$ by [L1].

Along $y=2r(x)$, $q(x,2r(x))=-r(x{)}^2<0$ for $x\mathrm{\ne }0$.

On each line $y=mx$, the factor $r(x)$ is smaller than every positive power of $\mathrm{\mid }x\mathrm{\mid }$, so $q(x,mx)>0$ for sufficiently small nonzero $x$; the same holds on $y=0$.

Thus $q$ is smooth and linewise strictly minimal at the origin but has no local minimum there.

The partial derivatives are $f_x=2x-3y^2(1-x{)}^2$ and $f_y=2y(1-x{)}^3$. Their simultaneous vanishing forces $(x,y)=(0,0)$.

At $x=2$, $f(2,y)=4-y^2$, which is negative for $\mathrm{\mid }y\mathrm{\mid }>2$, whereas $f(0,0)=0$.

At the origin the Hessian is $\mathrm{diag}(2,2)$, so [L1] makes it a strict local minimum.

Thus the unique critical point is a strict local minimum but not a global one.

The gradient is $(3x^2-3y^2,-6xy)$, and the Hessian entries are $6x,-6y,-6x$; both vanish at the origin.

Along the $x$-axis, $f(t,0)=t^3$, which has positive and negative values arbitrarily near $0$.

Hence the origin is a saddle even though its Hessian is zero, illustrating the inconclusive case [L1].

Each displayed function has zero gradient and zero Hessian at $(0,0)$.

$f_{+}$ is positive off the origin, so the origin is a strict local minimum; $f_{-}$ is negative off the origin, so it is a strict local maximum.

The values $f_s(t,0)=t^4$ and $f_s(0,t)=-t^4$ have opposite signs for $t\mathrm{\ne }0$, so the origin is a saddle.

These three different local types share the same zero Hessian, exactly as the inconclusive clause [L1] permits.

At $(0,0)$, $f=1$, $\mathrm{\nabla }f=(1,1)$, and $H_f=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)$.

Substitution into [L1] gives $1+x+y+\frac{1}{2}(x^2+2xy+2y^2)$, namely the claimed polynomial.

Thus the computed value, gradient, and Hessian yield the stated second-order approximation.

On the graph, $f(x,x+1)=2x^2+2x+1=2(x+1\mathrm{/}2{)}^2+1\mathrm{/}2$, so the unique constrained minimum is $a=(-1\mathrm{/}2,1\mathrm{/}2)$.

At $a$, $\mathrm{\nabla }f(a)=(-1,1)=\mathrm{\nabla }G(a)$. Hence the multiplier equation holds with $\lambda =1$, in accord with [L1].

This explicitly realizes the graph-constraint conclusion at the constrained minimum.

The constraint set $\{g=0\}$ is the singleton $(0,0)$, so $f$ has both a constrained local maximum and a constrained local minimum there.

At the origin, $\mathrm{\nabla }g=(0,0)$ while $\mathrm{\nabla }f=(1,0)$.

No scalar $\lambda$ can satisfy $\mathrm{\nabla }f(0,0)=\lambda \mathrm{\nabla }g(0,0)$. Thus a regularity hypothesis is necessary for the usual multiplier conclusion.