The Implicit Function Theorem: A Proof

We prove the Implicit Function Theorem for the bivariate case.

See also:

Theorem (Implicit function theorem)

Let $F(x,y)$ be a function of class $C^1$ on an open region $U\subset {\mathbb{R}}^2$. Suppose the point $P=(a,b)$ satisfies the following conditions:

• $F(a,b)=0$ (i.e., $P$ is a point on the curve $F(x,y)=0$),
• $F_y(a,b)\ne 0$.

Then, there exist an open interval $I$ on the $x$-axis such that $a\in I$ and a univariate function $y=\phi (x)$ on $I$ such that

1. $F(x,\phi (x))=0$ for all $x\in I$, that is, $\phi (x)$  is an implicit function of $F(x,y)=0$, and
2. $b=\phi (a)$.

Furthermore, the function $\phi (x)$ is differentiable on $I$ and

$$\begin{array}{}\text{(Eq:IF)}& {\phi }^{\prime }(x)=-\frac{F_x(x,\phi (x))}{F_y(x,\phi (x))}.\end{array}$$

Proof. Since $F_y(a,b)\ne 0$, either $F_y(a,b)>0$ or $F_y(a,b)<0$. In the following, we assume $F_y(a,b)>0$ (the other case is similar). We prove the theorem in three steps.

Step 1. Constructing a potential implicit function $\phi (x)$.

Since $F_y(a,b)>0$ and $F_y(x,y)$ is continuous, we can find a sufficiently small $\epsilon >0$ such that

$$|x-a|<\epsilon ,|y-b|<2\epsilon ⟹F_y(x,y)>0.$$

(That is, if $(x,y)$ is sufficiently close to $(a,b)$, then $F_y(x,y)>0$.) For all $x_0$satisfying $|x_0-a|<\epsilon$, consider the univariate function $g(y)=F(x_0,y)$. Then, on the interval $(b-2\epsilon ,b+2\epsilon )$, $g^{\prime }(y)=F_y(x_0,y)>0$ so $g(y)$ is a strictly monotone increasing function. In particular, $F(a,b)=0$ implies $F(a,b-\epsilon )<0$ and $F(a,b+\epsilon )>0$.

Again, since $F_y(x,y)$ is continuous, we can find a sufficiently small ${\epsilon }^{\prime }>0$ such that, for all $x_0$ such that $|x_0-a|<{\epsilon }^{\prime }$, the following hold:

• The univariate function $(x_0,y)$ of $y$ is strictly monotone increasing on $(b-2\epsilon ,b+2\epsilon )$,
•  $F(x_0,b-\epsilon )<0$ and $F(x_0,b+\epsilon )>0$.

By the Intermediate Value Theorem, for all $x_0$ such that $|x_0-a|<{\epsilon }^{\prime }$, there exists a $y_0$ such that $F(x_0,y_0)=0$ in the range $|y_0-b|<\epsilon$ (i.e., $y_0\in (b-\epsilon ,b+\epsilon )$). Moreover, there is exactly one such $y_0$ as $F(x_0,y)$ is strictly monotone increasing. Define the open interval $I=(a-{\epsilon }^{\prime },a+{\epsilon }^{\prime })$. For each $x_0\in I$, we can determine a unique $y_0$ as the (unique) solution of $F(x_0,y_0)=0$. In other words, we can define a function $y=\phi (x)$ in this way, and this function satisfies the conditions (1) and (2) of the Theorem.

Step 2. Showing that $\phi (x)$ is continuous on $I$.

It suffices to show that, for all $x_0\in I$, $\phi (x_0+h)\to \phi (x_0)$ as $h\to 0$. We prove this by contradiction. 

Suppose $\underset{h\to 0}{lim}\phi (x_0+h)\ne \phi (x_0)$. Then we can find a sequence $h_n$ that converges to 0 ($\underset{n\to \mathrm{\infty }}{lim}{h}_n=0$) and a positive real number ${\epsilon }^{″}>0$ such that

• For all $n\in \mathbb{N}$, $|\phi (x_0+h_n)-\phi (x_0)|\ge {\epsilon }^{″}$.

By construction in Step 1, for all $n\in \mathbb{N}$, $\phi (x_0+h_n)\in [b-\epsilon ,b+\epsilon ]$. Thus, by the Bolzano-Weierstrass Theorem:

"A sequence that is bounded in a closed interval contains a subsequence that converges to some value in the interval."

we can find a subsequence $\{h_{n_k}\}$ of $\{h_n\}$ such that $\underset{k\to \mathrm{\infty }}{lim}\phi (x_0+h_{n_k})=\alpha \in [b-\epsilon ,b+\epsilon ]$ where $\alpha$ is a constant. Since $|\phi (x_0+h_{n_k})-\phi (x_0)|\ge {\epsilon }^{″}$ for all $k$, we have $|\alpha -\phi (x_0)|\ge {\epsilon }^{″}$ and, in particular, $\alpha \ne \phi (x_0)$.

Now, consider a point sequence on ${\mathbb{R}}^2$: $\{(x_0+h_{n_k},\phi (x_0+h_{n_k}))\}$. This point sequence converges to $(x_0,\alpha )$. Since $F(x,y)$ is continuous, $F(x_0+h_{n_k},\phi (x_0+h_{n_k}))$ converges to $F(x_0,\alpha )$ as $k\to \mathrm{\infty }$. However, for all $k$, we have $F(x_0+h_{n_k},\phi (x_0+h_{n_k}))=0$ and, in particular, $F(x_0,\alpha )=0$. We also have $F(x_0,\phi (x_0))=0$ by construction. On the other hand, if any $y_0\in [b-\epsilon ,b+\epsilon ]$ satisfies $F(x_0,y_0)=0$, it must be unique (for a given $x_0$). Therefore, it must be that $\alpha =\phi (x_0)$, which is a contradiction. Thus, $\phi (x)$ is continuous on $I$.

Step 3. Showing that $\phi (x)$ is differentiable on $I$.

It suffices to show that $\underset{h\to 0}{lim}\frac{\phi (x+h)-\phi (x)}{h}$ exists. By Taylor's Theorem, we can write

$$F(x+h,y+k)=F(x,y)+F_x(x+\theta h,y+\theta k)h+F_y(x+\theta h,y+\theta k)k$$

for some $\theta \in (0,1)$.  (See also: Taylor's theorem and asymptotic expansion)

If we set $y=\phi (x)$ and $k=\phi (x+h)-\phi (x)$, then, since $F(x,\phi (x))=F(x+h,\phi (x+h))=0$, we have

$$F_x(x+\theta h,\phi (x)+\theta k)h+F_y(x+\theta h,\phi (x)+\theta k)k=0,$$

that is,

$$\frac{\phi (x+h)-\phi (x)}{h}=\frac{k}{h}=-\frac{F_x(x+\theta h,\phi (x)+\theta k)}{F_y(x+\theta h,\phi (x)+\theta k)}.$$

Since $\phi (x)$ is continuous, we have $k\to 0$ as $h\to 0$. By the continuity of $F_x(x,y)$ and $F_y(x,y)$, we have

$$\begin{array}{}\text{(Eq:dphi)}& \underset{h\to 0}{lim}\frac{\phi (x+h)-\phi (x)}{h}=-\frac{F_x(x,\phi (x))}{F_y(x,\phi (x))}.\end{array}$$

Therefore, $\phi (x)$ is differentiable on $I$ and its derivative is given by the right-hand side of (Eq:dphi).

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