The Implicit Function Theorem: A Proof

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We prove the Implicit Function Theorem for the bivariate case.

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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 \(F(x,y) = 0\)),
  • \(F_y(a,b) \neq 0\).
Then, there exist an open interval \(I\) on the \(x\)-axis such that \(a \in I\) and a univariate function \(y = \varphi(x)\) on \(I\) such that
  1. \(F(x,\varphi(x)) = 0\) for all \(x \in I\), that is, \(\varphi(x)\)  is an implicit function of \(F(x,y) = 0\), and
  2. \(b = \varphi(a)\).
Furthermore, the function \(\varphi(x)\) is differentiable on \(I\) and
\[\varphi'(x) = -\frac{F_x(x,\varphi(x))}{F_y(x,\varphi(x))}.\tag{Eq:IF}\]

Proof. Since \(F_y(a,b) \neq 0\), 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 \(\varphi(x)\).

Since \(F_y(a,b) > 0\) and \(F_y(x,y)\) is continuous, we can find a sufficiently small \(\varepsilon >0\) such that
\[|x - a| < \varepsilon, |y - b| < 2\varepsilon \implies F_y(x,y) > 0.\]
(That is, if \((x,y)\) is sufficiently close to \((a,b)\), then \(F_y(x,y) > 0\).) For all \(x_0\)satisfying \(|x_0 - a| < \varepsilon\), consider the univariate function \(g(y) = F(x_0, y)\). Then, on the interval \((b - 2\varepsilon, b + 2\varepsilon)\), \(g'(y) = F_y(x_0, y) > 0\) so \(g(y)\) is a strictly monotone increasing function. In particular, \(F(a,b) = 0\) implies \(F(a, b-\varepsilon) < 0\) and \(F(a, b + \varepsilon) > 0\).

Again, since \(F_y(x,y)\) is continuous, we can find a sufficiently small \(\varepsilon' > 0\) such that, for all \(x_0\) such that \(|x_0 - a| < \varepsilon'\), the following hold:
  • The univariate function \((x_0,y)\) of \(y\) is strictly monotone increasing on \((b - 2\varepsilon, b + 2\varepsilon)\),
  •  \(F(x_0, b - \varepsilon) < 0\) and \(F(x_0, b + \varepsilon) > 0\).
By the Intermediate Value Theorem, for all \(x_0\) such that \(|x_0 - a| < \varepsilon'\), there exists a \(y_0\) such that \(F(x_0, y_0) = 0\) in the range \(|y_0 - b| < \varepsilon\) (i.e., \(y_0 \in (b - \varepsilon, b + \varepsilon)\)). Moreover, there is exactly one such \(y_0\) as \(F(x_0,y)\) is strictly monotone increasing. Define the open interval \(I = (a - \varepsilon', a + \varepsilon')\). 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 \(y = \varphi(x)\) in this way, and this function satisfies the conditions (1) and (2) of the Theorem.

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

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

Suppose \(\lim_{h\to 0}\varphi(x_0 + h) \neq \varphi(x_0)\). Then we can find a sequence \({h_n}\) that converges to 0 (\(\lim_{n\to\infty}h_n = 0\)) and a positive real number \(\varepsilon'' > 0\) such that
  • For all \(n \in \mathbb{N}\), \(|\varphi(x_0 + h_n) - \varphi(x_0)| \geq \varepsilon''\).
By construction in Step 1, for all \(n\in\mathbb{N}\), \(\varphi(x_0 + h_n) \in [b - \varepsilon, b + \varepsilon]\). 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 \(\lim_{k\to\infty}\varphi(x_0 + h_{n_k}) = \alpha \in [b - \varepsilon, b + \varepsilon]\) where \(\alpha\) is a constant. Since \(|\varphi(x_0 + h_{n_k}) - \varphi(x_0)| \geq \varepsilon''\) for all \(k\), we have \(|\alpha - \varphi(x_0)|\geq \varepsilon''\) and, in particular, \(\alpha \neq \varphi(x_0)\).

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

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

It suffices to show that \(\lim_{h\to 0}\frac{\varphi(x + h) - \varphi(x)}{h}\) 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 \(y =\varphi(x)\) and \(k = \varphi(x + h) - \varphi(x)\), then, since \(F(x, \varphi(x)) = F(x + h, \varphi(x + h)) = 0\), we have
\[F_x(x + \theta h, \varphi(x) + \theta k)h + F_y(x + \theta h, \varphi(x) + \theta k)k = 0,\]
that is,
\[\frac{\varphi(x + h) - \varphi(x)}{h} = \frac{k}{h} = -\frac{F_x(x + \theta h, \varphi(x) + \theta k)}{F_y(x + \theta h, \varphi(x) + \theta k)}.\]
Since \(\varphi(x)\) 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
\[\lim_{h\to 0}\frac{\varphi(x + h) - \varphi(x)}{h} = -\frac{F_x(x,\varphi(x))}{F_y(x,\varphi(x))}.\tag{Eq:dphi}\]
Therefore, \(\varphi(x)\) is differentiable on \(I\) and its derivative is given by the right-hand side of (Eq:dphi).
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