Lipschitz Continuity of Bounded Rational Functions

This text is fully human written and the result of my weekends thought process about a totally different problem. Neither should it contain anything super surprising for any heartfelt mathematician. I still found the result surprising, since it illustrates a huge difference in the expressiveness of uni- and multivariate polynomials.

Preliminaries

Definition 1. A function \(f\colon D \to \mathbb{R}\) is bounded if there exists a bound \(b \in \mathbb{R}\) such that \(|f(x)| \leq b\) for all \(x \in D\).

Definition 2. A function \(f\colon D \to \mathbb{R}\) is Lipschitz continuous if there exists a \(L \in \mathbb{R}\) such that \(|f(x) - f(y)| \leq L \, \lVert x - y \rVert\) holds for all \(x, y \in D\). The smallest \(L\) that satisfies this condition is called the Lipschitz modulus of \(f\).

Proposition 1. A differentiable function is Lipschitz continuous if and only if its first derivative is bounded.

Since a derivative must be bounded on a compact domain, we also have:

Proposition 2. A continuously differentiable function is Lipschitz continuous if the domain is compact.

The Univariate Case

Definition 3. A rational function is a function \(f\colon D \to \mathbb{R}\) with domain \(D \subseteq \mathbb{R}\) such that there exist polynomials \(P, Q \in \mathbb{R}[X]\) with \(f(x) = \frac{P(x)}{Q(x)}\) for all \(x \in D\).

For a normal form, it can be assumed that the polynomials \(P, Q\) are co-prime, that is, \(\gcd(P,Q) = 1\). In particular, \(Q\) cannot have a root on \(D\): suppose \(Q(x) = 0\) for some \(x \in D\), then \(P(x) \neq 0\). The function \(f\) would diverge at this point. Thus, it can even be assumed that \(Q(x) > 0\) for all \(x \in D\).

Note that a rational functions \(f = \frac{P(x)}{Q(x)}\) always is differentiable; its derivative \(f'\) simply is given by the well-known quotient rule, $$ f'(x) ~=~ \frac{P'(x) Q(x) + P(x) Q'(x)}{Q(x)^2} $$ for all \(x \in D\). This again is a rational function as \(P', Q'\) are polynomials and \(Q(x)^2 = Q^2(x)\).

Proposition 3. A rational function \(f = \frac{P(x)}{Q(x)}\) is bounded if and only if \(\deg P \leq \deg Q\).

It's a simple consequence that if \(f\) is bounded, then so must be \(f'\): The denominator \(Q(x)^2\) has the same roots as \(Q(x)\), and \(\deg P' Q + P Q' = \max \, \{ \deg P' Q, \deg P Q' \} \leq \deg P Q \leq \deg Q^2\) as differentiating decrements the degree by one. From Proposition 1 we thus obtain:

Proposition 4. A bounded rational function is Lipschitz continuous.

While this to me was rather intuitive, it should rather be surprising as the next section will show. The converse statement (Lipschitz implies bounded) obviously is false, as can already be seen from \(f(x) = x\).

The Multivariate Case

Definition 4. A multivariate rational function is a function \(f\colon D \to \mathbb{R}\) with domain \(D \subseteq \mathbb{R}^d\) such that there exist multivariate polynomials \(P, Q \in \mathbb{R}[X_1, \ldots, X_d]\) with \(f(x) = \frac{P(x)}{Q(x)}\) for all \(x \in D\). Again, \(P\) and \(Q\) can be assumed co-prime.

In contrast to the nice result from Proposition 4, the multivariate case is much wilder.

Example 1. Consider $$ f(x,y) = \frac{xy}{x^2 y^2 + 1} ~. $$ The function is multivariate rational. Its also bounded, as by the arithmetic-geometric mean inequality we have $$ x^2 y^2 + 1 \geq 2 \sqrt{x^2 y^2 \cdot 1} = 2 |xy| ~. $$ However, considering the partial derivative $$ f_y(x,y) = \frac{x \, (x^2 y^2 + 1) + xy \, (2 x^2 y)}{(x^2 y^2 + 1)^2} = \frac{3 x^3 y^2 + x}{x^4 y^4 + 2 x^2 y^2 + 1} $$ and projecting onto the subvariety \(y=0\) (this means we are just looking at \(f\) on the \(x\)-axis), we have \(f_y(x, 0) = x\) which diverges as \(x \to \pm\infty\). Analogous to Proposition 1, this means that \(f\) is not Lipschitz continuous.

What happened here? Introducing the fresh variable \(z = xy\) and letting $$ g(z) = \frac{z}{z^2 + 1} ~, $$ we have that \(g(xy) = f(x,y)\) for all \(x, y \in \mathbb{R}\). In this light, we can view \(f\) as squeezing the function \(g\) along the \(y\)-axis, causing an arbitrarily steep wall. Just look at \(g(x), g(2x), g(3x), \ldots\) The \(y\)-component of the function \(f\) compressed its \(x\)-component and vice versa; the compression actually happens in all directions. In certain directions, the denominator does not grow fast enough to swallow the numerator. This observation motivates the next definition.

Definition 5. A multivariate rational function \(f\colon D \to \mathbb{R}, z \mapsto \frac{P(z)}{Q(z)}\) is radially bounded if there exists a \(c > 0\) such that \(Q(z) \geq c \, \lVert z \rVert^{\deg P}\) for all \(z \in D\).

We immediately see that this is equivalent to boundedness in the univariate case, as was illustrated in Proposition 3. However, this new condition is stronger in the multivariate case:

Proposition 6. A radially bounded (multivariate) rational function is Lipschitz continuous.

In the univariate case there simply wasn't enough space for both properties to differ. Revisiting the function from Example 1, we have the norm \(\lVert z \rVert = \sqrt{x^2 + y^2}\) which simply evaluates to \(|x|\) on the subvariety \(y = 0\). Since \(Q(x,0) = 1 < c \, x^2\) for any \(c\), the function there indeed is not radially bounded.

Example 2. The function $$ g(x,y) = \frac{xy}{x^2 + y^2 + 1} $$is radially bounded.

I am still astonished that a function can stay bounded without being radially bounded.