Proof of the Implicit Function Theorem

The setup - monotonicity is everything

Let's start with a picture, not a proof. Imagine $F$ is temperature in a room. At our special point $(x_0, y_0)$, the temperature is exactly zero - $F(x_0, y_0) = 0$. And as you walk north (increase $y$), the temperature rises - $F_y > 0$.

So just south of our point, the temperature is negative. Just north, it's positive. Now pick any vertical line through a nearby $x$-value. As you walk north along it, you cross from cold to hot, passing through zero exactly once. That crossing IS the implicit function $y = f(x)$.

That's the entire proof in one breath. Let's formalize it.

Step 1: Find a rectangle where $F_y > 0$

Assume $F_y(x_0, y_0) > 0$. (If it's negative, just replace $F$ by $-F$ - everything works the same.) Since $F_y$ is continuous, there's a rectangle

throughout which $F_y > 0$. Inside this rectangle, for each fixed $x$, the function $y \mapsto F(x, y)$ is strictly increasing.

Step 2: Signs at the boundary

At our base point, $F(x_0, y_0) = 0$ and $F$ is strictly increasing in $y$. So:

• At the bottom of the rectangle: $F(x_0, y_0 - \eta) \lt 0$
• At the top: $F(x_0, y_0 + \eta) > 0$

Since $F$ is continuous in $x$ as well, for $x$ sufficiently close to $x_0$ (shrink $\delta$ if needed), we still have:

• $F(x, y_0 - \eta) \lt 0$ - still cold at the bottom
• $F(x, y_0 + \eta) > 0$ - still hot at the top

Step 3: Existence and uniqueness

For each such $x$, look at $y \mapsto F(x, y)$ on $[y_0 - \eta,\; y_0 + \eta]$:

1. Intermediate Value Theorem: $F$ is continuous, negative at the bottom, positive at the top - so it crosses zero at least once.
2. Strict monotonicity: $F_y > 0$ means $F$ is strictly increasing in $y$ - so it crosses zero at most once.

At least one + at most one = exactly one. Call this unique zero $y = f(x)$.

Drag the slider to move $x$. Each vertical slice crosses $F=0$ exactly once.

The squeeze - bounding the solution curve

We know $y = f(x)$ exists and is unique near $(x_0, y_0)$. But is $f$ continuous? Differentiable? Here's where the squeeze argument earns its name.

The implicit derivative is bounded

Define the "slope function":

Since $F_y > 0$ throughout our rectangle $E$, and both $F_x$ and $F_y$ are continuous, $g$ is continuous on a closed, bounded set. So $g$ achieves a minimum $m$ and maximum $M$ on $E$.

What does this mean? Wherever the solution curve lives inside $E$, its slope $f'(x) = g(x, f(x))$ is squeezed between $m$ and $M$. The curve is trapped between two straight lines:

Shrink $\delta$ to stay inside the rectangle

We need the bounding lines to stay within the rectangle. Choose $\delta$ small enough that $|M| \cdot \delta \lt \eta$ and $|m| \cdot \delta \lt \eta$. Then both lines (and therefore the solution curve between them) remain within the vertical bounds $|y - y_0| \lt \eta$.

Continuity and differentiability, for free

Continuity: As $x \to x_0$, the two bounding lines both pass through $(x_0, y_0)$, and $f(x)$ is squeezed between them. So $f(x) \to y_0 = f(x_0)$. That's continuity.

Differentiability: As $\delta \to 0$, the bounding slopes $m$ and $M$ can be made as close together as we like (since $g$ is continuous and they both approach $g(x_0, y_0)$). In the limit, both bounds converge to the single slope $-F_x(x_0,y_0)/F_y(x_0,y_0)$. The solution curve, squeezed between them, must have this same slope. That's differentiability, with

Shrink $\delta$ - the bounding lines narrow, squeezing the curve toward a single tangent slope.

What the theorem actually guarantees

Let's take stock. From a single condition - $F_y(x_0, y_0) \neq 0$ - we proved four things:

The proof generalizes to systems of $m$ equations $F_1 = 0, \ldots, F_m = 0$ in $n + m$ variables by induction on $m$. The single condition $F_y \neq 0$ is replaced by: the Jacobian determinant $\partial(F_1, \ldots, F_m) / \partial(y_1, \ldots, y_m) \neq 0$. Same logic, bigger notation.