In the analysis of [functions](/page/Function), there is a substantial gap between [continuity](/page/Continuity) and differentiability, as a continuous function can be nowhere [differentiable](/page/Derivative). However, physical models often involve functions that satisfy a **Lipschitz condition**, meaning their rate of change is globally bounded. Intuitively, the graph of a Lipschitz function cannot be infinitely steep. **Rademacher's Theorem** bridges this gap by asserting that this boundedness of the slope is sufficient to guarantee differentiability "almost everywhere." This result is the cornerstone of Geometric Measure Theory and the analysis of [Sobolev spaces](/page/Sobolev%20Space), as it allows us to define derivatives (like the Jacobian in the Area Formula) for functions that are not strictly smooth, provided we ignore a set of measure zero. ## Formal Definition We first define the class of Lipschitz functions and the concept of differentiability in Euclidean space. [definition:Lipschitz Continuity] Let $U \subseteq \mathbb{R}^n$ be an [open set](/page/Open%20Set). A function $f: U \to \mathbb{R}^m$ is called **Lipschitz continuous** (or simply Lipschitz) if there exists a constant $L \ge 0$ such that $|f(x) - f(y)| \le L |x - y|$ for all $x, y \in U$. The smallest such $L$ is denoted by $\text{Lip}(f)$. [/definition] [definition:Total Differentiability] A function $f: U \to \mathbb{R}^m$ is **differentiable** at a point $x \in U$ if there exists a [linear map](/page/Linear%20Map) $A: \mathbb{R}^n \to \mathbb{R}^m$ (denoted by $Df(x)$ or $J f(x)$) such that $\lim_{y \to x} \frac{|f(y) - f(x) - A(y-x)|}{|y - x|} = 0$. [/definition] We now state the main theorem. [theorem:Rademacher's Theorem] Let $U \subseteq \mathbb{R}^n$ be an open set and let $f: U \to \mathbb{R}^m$ be a Lipschitz function. Then $f$ is differentiable at $\mathcal{L}^n$-almost every point $x \in U$. That is, the set of points where $f$ fails to be differentiable has Lebesgue measure zero: $\mathcal{L}^n(\{ x \in U : f \text{ is not differentiable at } x \}) = 0$. [/theorem] ## Examples We examine specific Lipschitz functions to illustrate the theorem and the nature of the singular [sets](/page/Set) where differentiability fails. [example:Distance Function] Let $K \subset \mathbb{R}^n$ be a nonempty [closed set](/page/Closed%20Set). Define the signed distance function $d_K: \mathbb{R}^n \to \mathbb{R}$ by $d_K(x) := \inf_{y \in K} |x - y|$. We claim $d_K$ is 1-Lipschitz. To prove this, fix $x, z \in \mathbb{R}^n$. For any $\epsilon > 0$, there exists $y_z \in K$ such that $|z - y_z| < d_K(z) + \epsilon$. Using the triangle inequality, we see that $d_K(x) \le |x - y_z| \le |x - z| + |z - y_z| < |x - z| + d_K(z) + \epsilon$. Rearranging yields $d_K(x) - d_K(z) < |x - z| + \epsilon$. Since this holds for all $\epsilon > 0$ and the argument is symmetric in $x$ and $z$, we conclude $|d_K(x) - d_K(z)| \le |x - z|$, so $\text{Lip}(d_K) \le 1$. By Rademacher's Theorem, $d_K$ is differentiable almost everywhere. At any point $x \notin K$ where $d_K$ is differentiable, there exists a unique point $\Pi_K(x) \in K$ (the metric projection) such that $|x - \Pi_K(x)| = d_K(x)$, and the gradient is explicitly given by $\nabla d_K(x) = \frac{x - \Pi_K(x)}{|x - \Pi_K(x)|}$. Consequently, $|\nabla d_K(x)| = 1$ almost everywhere outside $K$. The set where differentiability fails is precisely the set of points $x$ for which the closest point in $K$ is **not** unique, often called the *skeleton* or *cut locus*. For example, if $K = \{-1, 1\} \subset \mathbb{R}$, then $d_K(x) = \min(|x-1|, |x+1|)$, which is differentiable everywhere except at $x=0$. [/example] [example:Folded Plane] Consider the function $f: \mathbb{R}^2 \to \mathbb{R}$ defined by $f(x_1, x_2) := |x_1| + x_2$. This function is Lipschitz because it is the sum of two Lipschitz functions ($x \mapsto |x_1|$ and $x \mapsto x_2$). We check differentiability by computing partial derivatives. The partial derivative with respect to $x_2$ is $\frac{\partial f}{\partial x_2} = 1$ everywhere. For $x_1$, we have $\frac{\partial f}{\partial x_1} = \text{sgn}(x_1)$, which is $1$ for $x_1 > 0$ and $-1$ for $x_1 < 0$. At any point $(0, x_2)$ on the vertical axis, the [limit](/page/Limit) $\lim_{h \to 0} \frac{f(h, x_2) - f(0, x_2)}{h} = \lim_{h \to 0} \frac{|h|}{h}$ does not exist. Since a partial derivative fails to exist, the total derivative cannot exist at any point on the line $\{ (0, x_2) : x_2 \in \mathbb{R} \}$. The singular set is a line in $\mathbb{R}^2$. Since $\mathcal{L}^2(\text{line}) = 0$, the function is differentiable almost everywhere, consistent with Rademacher's Theorem. [/example] ## Key Results Rademacher's Theorem allows us to identify Lipschitz functions with the Sobolev space $W^{1,\infty}$. This identification is crucial for applying [calculus of variations](/page/Calculus%20of%20Variations) techniques to Lipschitz candidates. [theorem:Lipschitz Sobolev Identity] Let $U \subseteq \mathbb{R}^n$ be an open, bounded set with $C^1$ [boundary](/page/Boundary). Let $f: U \to \mathbb{R}$. Then $f$ is Lipschitz continuous on $U$ if and only if $f \in W^{1, \infty}(U)$. This means $f \in L^\infty(U)$ and its weak derivatives $\partial_i f$ exist and belong to $L^\infty(U)$. Furthermore, the Lipschitz constant corresponds to the norm of the gradient: $\text{Lip}(f) = \|\nabla f\|_{L^\infty(U)} = \text{ess sup}_{x \in U} |\nabla f(x)|$. [/theorem] ## Problems [problem] Let $f: [0, 1] \to \mathbb{R}$ be the Cantor function. 1. Is $f$ Lipschitz continuous? 2. Does $f$ satisfy the conclusion of Rademacher's Theorem (differentiable a.e.)? [/problem] [solution] **Part 1: Lipschitz Continuity** The Cantor function is continuous and monotone increasing, but it is **not** Lipschitz continuous. Suppose for contradiction that $f$ is Lipschitz with constant $L$. Then $f$ would be absolutely continuous. The [Fundamental Theorem of Calculus](/theorems/632) would apply, implying $f(1) - f(0) = \int_0^1 f'(x) \, d\mathcal{L}^1(x)$. The Cantor function is locally constant on the complement of the Cantor set $C$. Since $C$ has measure zero ($\mathcal{L}^1(C) = 0$), the classical derivative $f'(x) = 0$ almost everywhere. Substituting this into the [integral](/page/Integral) yields $f(1) - f(0) = \int_0^1 0 \, d\mathcal{L}^1(x) = 0$. However, by construction, $f(0) = 0$ and $f(1) = 1$, implying $1 = 0$, a contradiction. Thus, $f$ cannot be Lipschitz. **Part 2: Differentiability** Despite not being Lipschitz, $f$ **is** differentiable almost everywhere. The derivative exists and equals 0 on the open set $U = [0, 1] \setminus C$. Since $\mathcal{L}^1(U) = 1$, the set of points where $f$ is differentiable has full measure. This demonstrates that Rademacher's Theorem provides a *sufficient* condition, but not a *necessary* one. Lipschitz functions are a strict subset of functions differentiable a.e. [/solution] ## References 1. L.C. Evans, *Partial Differential Equations* (1998). 2. L.C. Evans and R.F. Gariepy, *Measure Theory and Fine Properties of Functions* (2015). 3. H. Federer, *Geometric Measure Theory* (1969).