Critical points and extrema

The questions are different

A local minimum compares $f(a)$ to nearby values in the domain; an absolute minimum compares it to every value in the domain. Replace $\le$ by $<$ away from $a$ for a strict minimum. Maxima are defined by reversing the inequalities. Neither local extremum need be absolute, and an absolute extremum may fail to exist. On a nonempty closed bounded set, continuity guarantees both absolute extrema, but locating them still requires checking all relevant candidates.

Theorem

First-order necessary condition

If $f$ is differentiable on an open neighbourhood of an interior point $a$ and has a local extremum at $a$, then $\nabla f(a)=0$.

Proof

Restrict to coordinate lines

For every coordinate direction $e_i$, the one-variable function $t\mapsto f(a+te_i)$ has an interior local extremum at $t=0$. Its derivative there is zero. Hence all entries of $\nabla f(a)$ vanish.

This is necessary, not sufficient: $f(x,y)=x^2-y^2$ has zero gradient at the origin but takes both positive and negative values arbitrarily close. If differentiability fails, a local extremum can occur where the gradient does not exist, as for $f(x,y)=\sqrt{x^2+y^2}$ at zero. Boundary points need not satisfy the interior condition.

The second derivative test

Theorem

Hessian classification

Let $f$ be $C^2$ near a stationary point $a$. If $h^TH_f(a)h>0$ for every nonzero $h$, then $a$ is a strict local minimum. If the quadratic form is negative for every nonzero $h$, it is a strict local maximum. If it takes both signs, $a$ is a saddle point, hence neither local minimum nor maximum. Semidefinite or zero forms make this test inconclusive.

Proof

Why the signs persist

Taylor's theorem gives $f(a+h)-f(a)=\tfrac12h^TH_f(a)h+o(\|h\|^2)$. If the Hessian is positive definite, its continuous quadratic form has a positive minimum $m$ on the compact unit sphere. Thus the quadratic term is at least $m\|h\|^2/2$, which dominates the little-$o$ remainder for small nonzero $h$. The negative case is analogous. For an indefinite form, choose one fixed direction of each sign; the quadratic term dominates along each sufficiently short ray.

For a symmetric $2\times2$ Hessian at a stationary point, set $D=f_{xx}f_{yy}-f_{xy}^2$. If $D>0$ and $f_{xx}>0$ there is a strict minimum; if $D>0$ and $f_{xx}<0$ a strict maximum; if $D<0$ a saddle. If $D=0$ the test is inconclusive. These are facts about a stationary point, not a license to classify an arbitrary point.

Worked example

Classifying two stationary points

Let $f(x,y)=x^3-3x+y^2$. The equations $3x^2-3=0$ and $2y=0$ give $(1,0)$ and $(-1,0)$. The Hessian is $\mathrm{diag}(6x,2)$. At $(1,0)$ it is positive definite, so this point is a strict local minimum with value $-2$. At $(-1,0)$ it is indefinite, so this is a saddle with value $2$. Neither statement alone decides global extrema; in fact the function is unbounded above and below as $x\to\pm\infty$.

Worked example

An inconclusive Hessian still permits a proof

At the origin, $f=x^4+y^4$ and $g=x^4-y^4$ both have zero Hessian. Nevertheless $f>0$ away from zero, so it has a strict minimum. The function $g$ has positive values along the $x$-axis and negative values along the $y$-axis, so the origin is a saddle. Higher-order terms resolve what the second-order test cannot.

Absolute extrema on a region

1. Establish existence where possible: continuity on a nonempty compact region guarantees extrema.
2. Find interior stationary points and nondifferentiable points in the region.
3. Optimize on every boundary piece, including endpoints, corners and singularities. Parametrize the boundary or use the multiplier method when its hypotheses hold.
4. Evaluate $f$ at every candidate and compare the actual values. A boundary test can be necessary even when an interior point is a strict local extremum.

Worked example

Interior and boundary compete

On the closed disk $x^2+y^2\le1$, take $f(x,y)=x^2+y^2+2x$. Continuity and compactness guarantee extrema. Interior stationarity requires $(2x+2,2y)=0$, giving $(-1,0)$, which is on the boundary, not in the interior. On the boundary $x^2+y^2=1$, $f=1+2x$, so its minimum is $-1$ at $(-1,0)$ and maximum is $3$ at $(1,0)$. Both are absolute by the preceding existence and candidate analysis.

Method comparison

Use first derivatives to generate candidates, not to certify their type. Use a definite Hessian to obtain a strict local conclusion efficiently; use signs along curves or higher-order terms when the Hessian is inconclusive. For absolute extrema, local classification is optional but a complete boundary and value comparison is essential. On an open or unbounded domain, examine limiting behavior and possible nonattainment; compactness cannot be assumed.

Practice

1. Classify every stationary point of $f=x^2+2xy+3y^2-2x$.
2. Find the absolute extrema of $f=x+y$ on $[0,2]\times[0,1]$.
3. Determine the nature of the origin for $f=x^4+y^6$ and $g=x^4-y^6$; explain why the Hessian test alone does not answer.
4. Give a continuous function on a noncompact domain with no absolute maximum and explain why the compactness theorem does not apply.

Full solutions

1. The equations $2x+2y-2=0$, $2x+6y=0$ yield $(x,y)=(3/2,-1/2)$. The Hessian $\begin{pmatrix}2&2\\2&6\end{pmatrix}$ has $D=8>0$ and $f_{xx}=2>0$, so this is a strict local minimum. Completing the square gives $f=(x+y-1)^2+2(y+1/2)^2-3/2$, proving it is the unique absolute minimum, value $-3/2$.
2. Since $0\le x\le2$ and $0\le y\le1$, $0\le x+y\le3$. The minimum is $0$ at $(0,0)$ and the maximum is $3$ at $(2,1)$.
3. Both Hessians vanish at zero, so the test is inconclusive. For the first function the value is positive off the origin, giving a strict minimum. For the second, $(t,0)$ gives $t^4>0$ and $(0,t)$ gives $-t^6<0$ for nonzero $t$, giving a saddle.
4. $f(x)=x$ on $\mathbb R$ is continuous but unbounded above; $\mathbb R$ is not compact. Alternatively, $f(x)=x$ on $(0,1)$ is bounded but does not attain its supremum.

Synthesis

Stationarity is the bridge from local optimality to equations. The Hessian examines second-order behavior but can be inconclusive. A global problem additionally needs existence and a complete accounting of boundary and exceptional points.