Inverse and implicit function theorems, extrema

In one sentence

Both theorems are local, both need a derivative that is continuous and invertible, and nearly every wrong option is produced by upgrading "local" to "global" or by forgetting which of the two variables the hypothesis is about.

Why the exam asks it

Because these are the first theorems where the word "local" carries real weight, and it is a word students read past. A map can have an invertible derivative at every single point of the plane and still fail to be injective — that fact is counterintuitive, has a two-line counterexample, and is therefore examined repeatedly.

The idea

The inverse function theorem

Let $f : \mathbb{R}^n \to \mathbb{R}^n$ be $C^1$ near $a$, with $Df(a)$ invertible. Then there are open sets $U \ni a$ and $V \ni f(a)$ such that $f : U \to V$ is a bijection with $C^1$ inverse, and $D(f^{-1})(f(a)) = [Df(a)]^{-1}$.

Three hypotheses, each examined.

$C^1$, not merely differentiable. In one dimension, $f(x) = x + 2x^2\sin(1/x)$ with $f(0) = 0$ has

$$f'(x) = 1 + 4x\sin(1/x) - 2\cos(1/x) \quad (x \ne 0), \qquad f'(0) = 1.$$

So the derivative at the origin exists, equals $1$, and is as invertible as a number can be. But whenever $\cos(1/x) = 1$ — which happens at $x = 1/(2\pi n)$, arbitrarily close to $0$ — the derivative is $1 + 0 - 2 = -1 < 0$. The function decreases on infinitely many intervals piling up at the origin, so it is injective on no neighbourhood of $0$.

The missing hypothesis is continuity of $f'$. This is the cleanest demonstration that "$C^1$" in the statement is load-bearing rather than decorative.

Invertible derivative, not merely nonzero somewhere. In $n$ dimensions "invertible" means $\det Df(a) \ne 0$, and it is needed at the point in question.

Local, and only local. Take

$$f(x,y) = (e^x\cos y,\; e^x \sin y).$$

Its Jacobian determinant is

$$\begin{vmatrix} e^x\cos y & -e^x\sin y \\ e^x \sin y & e^x\cos y\end{vmatrix} = e^{2x}(\cos^2 y + \sin^2 y) = e^{2x},$$

which is never zero. So every point has a neighbourhood on which $f$ is invertible. And yet $f$ is $2\pi$-periodic in $y$, so $f(x, y) = f(x, y + 2\pi)$ and $f$ is nowhere near injective on the plane. The theorem promised a neighbourhood; it never promised the domain.

The implicit function theorem

Let $F(x,y)$ be $C^1$ near $(a,b)$ with $F(a,b) = 0$ and $F_y(a,b) \ne 0$. Then near $(a,b)$ the equation $F = 0$ defines $y$ uniquely as a $C^1$ function of $x$, and $g'(a) = -F_x(a,b)/F_y(a,b)$.

The hypothesis is on the partial with respect to the variable you are solving for. Everything turns on that.

On the unit circle $F(x,y) = x^2 + y^2 - 1$: at $(0,1)$ we have $F_y = 2 \ne 0$, and indeed $y = \sqrt{1-x^2}$ nearby. At $(1,0)$ we have $F_y = 0$, and $y$ is genuinely not a function of $x$ there — just to the left of $x = 1$ there are two values of $y$, and to the right there are none. But $F_x = 2 \ne 0$ at that point, so $x = \sqrt{1 - y^2}$ is a perfectly good function of $y$. Nothing is singular about the circle at $(1,0)$; only the choice of which variable to solve for is bad.

The derivative formula is worth deriving rather than memorising: differentiate $F(x, g(x)) = 0$ by the chain rule to get $F_x + F_y g' = 0$.

Extrema

At an interior local extremum of a differentiable $f$, the gradient vanishes. The converse fails, as always, and the second-order test is:

• $\det H > 0$ and $f_{xx} > 0$: local minimum.
• $\det H > 0$ and $f_{xx} < 0$: local maximum.
• $\det H < 0$: saddle.
• $\det H = 0$: the test says nothing, and the paper likes this case.

Two facts from one variable that do not survive the move to two:

A unique critical point that is a local minimum need not be a global minimum. In one variable, on an interval, it is. In two,

$$f(x,y) = x^2 + y^2(1+x)^3$$

has $f_y = 2y(1+x)^3$, which vanishes only when $y = 0$ or $x = -1$; on $x = -1$, $f_x = -2 \ne 0$; so the only critical point is the origin, and there $f$ has a local minimum. Yet along the line $x = -2$,

$$f(-2, y) = 4 - y^2 \longrightarrow -\infty .$$

There is no global minimum at all. The single critical point is a local minimum and the function is unbounded below.

Existence of extrema needs compactness. A continuous function on a closed bounded set attains its bounds; on an unbounded or open set it need not. For constrained problems, Lagrange multipliers find candidates — they do not prove a maximum exists, and when the constraint set is not compact you must argue existence separately.

Lagrange multipliers

To extremise $f$ subject to $g = 0$: solve $\nabla f = \lambda \nabla g$ together with $g = 0$. The hypothesis usually skipped is that $\nabla g \ne 0$ at the point — where the constraint set has a singular point, the method can miss the extremum entirely.

Where intuition breaks

"$\det Df \ne 0$ everywhere, so $f$ is a bijection." $(e^x\cos y, e^x\sin y)$. Tempting because in one variable a nonvanishing derivative on an interval does give global injectivity — via the intermediate value theorem, which needs the domain to be connected in a way $\mathbb{R}^2$ subverts. The theorem is local; the word is in the name of the conclusion, not the hypothesis.

"$f'(a) \ne 0$, so $f$ is injective near $a$." $x + 2x^2\sin(1/x)$. A hypothesis dropped: the theorem asks for $C^1$, and this $f$ is differentiable everywhere with a derivative that is not continuous at $0$.

"$F(a,b) = 0$ and $\nabla F(a,b) \ne 0$, so $y$ is a function of $x$ near $(a,b)$." The circle at $(1,0)$: the gradient is $(2,0) \ne 0$, and $y$ is still not a function of $x$. The hypothesis is $F_y \ne 0$ specifically. A nonzero gradient guarantees some variable can be solved for — not the one you wanted.

"The only critical point is a local minimum, so it is the global minimum." $x^2 + y^2(1+x)^3$. True in one variable on an interval, false in two, and set precisely because the one-variable instinct is strong.

"$\det H = 0$, so the point is a saddle", or "so the test is inconclusive, therefore it is neither a max nor a min." Neither follows. $f = x^4 + y^4$ has $\det H = 0$ at the origin and a strict global minimum there; $f = x^4 - y^4$ has $\det H = 0$ and a saddle. Inconclusive means inconclusive — go back to the function.

"A continuous function on a closed set attains its maximum." Closed is not enough; $f(x) = x$ on $[0,\infty)$. Closed and bounded, in $\mathbb{R}^n$.

"Lagrange gives the maximum." It gives the stationary candidates. Compare their values, and check that a maximum exists at all — on a non-compact constraint set it may not.

The exam's angle

1. Read "local" and believe it. If an option says globally invertible, globally injective, or a global extremum, treat it as false until proved otherwise; that upgrade is the single most common way these questions are made wrong.
2. Check $C^1$, not just differentiability. If the question hands you a function with $\sin(1/x)$ in it, this is the point being tested.
3. For implicit functions, name the variable. Compute the partial with respect to the variable being solved for, and only that one.
4. For extrema: gradient zero, then Hessian, then — if $\det H = 0$ — go back to the function. Try restricting to lines through the point, or spot a perfect square.
5. Before answering that a maximum exists, check compactness. Closed and bounded, or an explicit argument.

The night before

• Inverse function theorem: $C^1$ and $\det Df(a) \ne 0$ $\Rightarrow$ locally a $C^1$-diffeomorphism. $D(f^{-1}) = [Df]^{-1}$.
• $(e^x\cos y, e^x\sin y)$: Jacobian $e^{2x} \ne 0$ everywhere, not injective. Local $\ne$ global.
• $x + 2x^2\sin(1/x)$: $f'(0) = 1$, injective on no neighbourhood of $0$. $C^1$ is needed.
• Implicit function theorem: hypothesis is $F_y \ne 0$ when solving for $y$. On the circle at $(1,0)$, $y$ fails and $x$ succeeds. $g' = -F_x/F_y$.
• Hessian: $\det H > 0$ with $f_{xx} > 0$ min, $f_{xx} < 0$ max; $\det H < 0$ saddle; $\det H = 0$ tells you nothing ($x^4 + y^4$ against $x^4 - y^4$).
• $x^2 + y^2(1+x)^3$: one critical point, a local min, unbounded below.
• Extrema exist on compact sets. Lagrange finds candidates, not existence.