A large part of PDE theory is about **regularity**: if a function $u$ satisfies an equation such as \begin{align*} -\Delta u = f \quad \text{in } U, \end{align*} what can be said about the smoothness of $u$ in terms of the smoothness of $f$? Two common regularity languages are: 1. **Sobolev regularity**, which is based on [integrability](/page/Integral) of derivatives (useful for weak solutions). For example, a weak formulation of $-\Delta u=f$ asks for $u\in W^{1,2}(U)$ such that \begin{align*} \int_U \nabla u(x)\cdot \nabla \varphi(x)\,\mathrm{d}x = \int_U f(x)\varphi(x)\,\mathrm{d}x \quad \text{for all }\varphi\in C_c^\infty(U). \end{align*} 2. **Hölder regularity**, which measures smoothness in a **pointwise** way, through uniform control of oscillations. This is the natural setting of **Schauder theory**, where one proves estimates of the form “if $f$ is Hölder continuous, then $u$ has Hölder continuous derivatives.” Hölder spaces are therefore a standard way to express that a PDE solution is not only [differentiable](/page/Derivative), but that its derivatives vary in a quantitatively controlled manner. --- ## Formal Definition [definition:HolderSpace] Let $n\in \mathbb{N}_0$ and let $U\subseteq \mathbb{R}^n$ be an open, bounded [set](/page/Set). Fix an integer $k\in \mathbb{N}_0$ and an exponent $\alpha\in (0,1]$. A function in this space is a map \begin{align*} u:\overline U &\to \mathbb{R}\\ x &\mapsto u(x). \end{align*} Define the Hölder seminorm map \begin{align*} [\cdot]_{C^{0,\alpha}(\overline U)}: C^0(\overline U) &\to [0,\infty]\\ f &\mapsto \sup_{\substack{x,y\in \overline U\\ x\neq y}} \frac{|f(x)-f(y)|}{|x-y|^\alpha}. \end{align*} For a multi-index $\beta=(\beta_1,\dots,\beta_n)\in \mathbb{N}_0^n$ with order $|\beta|=\beta_1+\cdots+\beta_n$, define the partial derivative operator \begin{align*} D^\beta u := \partial_1^{\beta_1}\cdots \partial_n^{\beta_n}u \quad \text{on }U. \end{align*} The Hölder space $C^{k,\alpha}(\overline U)$ consists of all [functions](/page/Function) $u\in C^k(\overline U)$ such that the norm \begin{align*} \|u\|_{C^{k,\alpha}(\overline U)} := \sum_{|\beta|\le k}\|D^\beta u\|_{L^\infty(U)} + \sup_{|\beta|=k}\,[D^\beta u]_{C^{0,\alpha}(\overline U)} \end{align*} is finite, where \begin{align*} \|D^\beta u\|_{L^\infty(U)} := \sup_{x\in U}|D^\beta u(x)|. \end{align*} [/definition] [quotetheorem:73] --- ## Examples [example:BasicHolders] **Example 1 (Smooth functions).** If $m\in \mathbb{N}_0$ and $u\in C^{m}(\overline U)$ with $m\ge k+1$, then $u\in C^{k,\alpha}(\overline U)$ for every $\alpha\in (0,1]$. The reason is that each $k$-th derivative is continuously differentiable on $\overline U$, hence locally Lipschitz on the compact set $\overline U$, which implies $\alpha$-Hölder [continuity](/page/Continuity). **Example 2 (A function that is Hölder but not Lipschitz at the origin).** Let $U=B_1(0)\subseteq \mathbb{R}^n$ and define \begin{align*} u:\overline U &\to \mathbb{R}\\ x &\mapsto |x|^\alpha. \end{align*} Then $u\in C^{0,\alpha}(\overline U)$. Moreover, if $\beta\in (\alpha,1]$, then $u\notin C^{0,\beta}(\overline U)$ because the quotient $|u(x)-u(0)|/|x-0|^\beta = |x|^{\alpha-\beta}$ becomes unbounded as $x\to 0$. **Example 3 (Lipschitz implies Hölder of any smaller exponent).** Let $L>0$, and let $u:\overline U\to \mathbb{R}$ satisfy \begin{align*} |u(x)-u(y)|\le L|x-y| \quad \text{for all }x,y\in \overline U. \end{align*} Fix $\alpha\in (0,1]$. Since $\overline U$ is bounded, let $R:=\sup_{x,y\in \overline U}|x-y|<\infty$. Then for $x\neq y$, \begin{align*} \frac{|u(x)-u(y)|}{|x-y|^\alpha} \le L|x-y|^{1-\alpha} \le LR^{1-\alpha}, \end{align*} so $[u]_{C^{0,\alpha}(\overline U)}\le LR^{1-\alpha}$ and therefore $u\in C^{0,\alpha}(\overline U)$. [/example] --- ## Key Results [theorem:HolderImpliesUniformContinuity] Let $U\subseteq \mathbb{R}^n$ be open and bounded, let $\alpha\in (0,1]$, and let $u\in C^{0,\alpha}(\overline U)$. Then $u$ is uniformly continuous on $\overline U$. [/theorem] [proof] **Step 1 (Introduce the Hölder constant).** Since $u\in C^{0,\alpha}(\overline U)$, the seminorm is finite, and we define the constant \begin{align*} H := [u]_{C^{0,\alpha}(\overline U)} \in [0,\infty). \end{align*} By definition of the supremum, for every $x,y\in \overline U$ with $x\neq y$, \begin{align*} \frac{|u(x)-u(y)|}{|x-y|^\alpha} \le H, \end{align*} which implies \begin{align*} |u(x)-u(y)| \le H|x-y|^\alpha. \end{align*} **Step 2 (Verify uniform continuity).** Let $\varepsilon>0$ be arbitrary. Choose \begin{align*} \delta := \left(\frac{\varepsilon}{H+1}\right)^{1/\alpha} > 0. \end{align*} If $x,y\in \overline U$ satisfy $|x-y|<\delta$, then \begin{align*} |u(x)-u(y)| \le H|x-y|^\alpha < H\delta^\alpha = H\frac{\varepsilon}{H+1} < \varepsilon. \end{align*} This proves uniform continuity of $u$ on $\overline U$. [/proof] [theorem:SchauderForPoisson] Let $U\subseteq \mathbb{R}^n$ be bounded with [boundary](/page/Boundary) $\partial U$ of class $C^{2,\alpha}$ for some $\alpha\in (0,1)$. Let \begin{align*} f:\overline U &\to \mathbb{R}\\ x &\mapsto f(x) \end{align*} satisfy $f\in C^{0,\alpha}(\overline U)$, and let \begin{align*} g:\partial U &\to \mathbb{R}\\ x &\mapsto g(x) \end{align*} satisfy $g\in C^{2,\alpha}(\partial U)$. If $u\in C^{2}(\overline U)$ solves the Dirichlet problem \begin{align*} -\Delta u &= f \quad \text{in }U,\\ u &= g \quad \text{on }\partial U, \end{align*} then $u\in C^{2,\alpha}(\overline U)$ and there exists a constant $C>0$ (depending on $U$, $n$, and $\alpha$) such that \begin{align*} \|u\|_{C^{2,\alpha}(\overline U)} \le C\Big(\|u\|_{L^\infty(U)}+\|f\|_{C^{0,\alpha}(\overline U)}+\|g\|_{C^{2,\alpha}(\partial U)}\Big). \end{align*} [/theorem] This estimate captures the main reason Hölder spaces appear in elliptic PDE: the equation turns Hölder control of the data $f$ (and boundary data $g$) into Hölder control of the second derivatives of $u$. --- ## Problems [problem] Let $U=B_1(0)\subseteq \mathbb{R}^n$ and define \begin{align*} u:\overline U &\to \mathbb{R}\\ x &\mapsto x_1, \end{align*} where $x=(x_1,\dots,x_n)$. Fix $\alpha\in (0,1]$. 1. Compute $[u]_{C^{0,\alpha}(\overline U)}$. 2. Show that $u\in C^{0,\alpha}(\overline U)$ and give an explicit upper bound on $\|u\|_{C^{0,\alpha}(\overline U)}$. [/problem] [solution] **Step 1 (Estimate the Hölder seminorm).** Take $x,y\in \overline U$ with $x\neq y$. Then \begin{align*} |u(x)-u(y)| = |x_1-y_1| \le |x-y|. \end{align*} Therefore \begin{align*} \frac{|u(x)-u(y)|}{|x-y|^\alpha} \le \frac{|x-y|}{|x-y|^\alpha} = |x-y|^{1-\alpha}. \end{align*} Since $x,y\in \overline U=B_1(0)$, we have $|x-y|\le 2$, so \begin{align*} |x-y|^{1-\alpha}\le 2^{1-\alpha}. \end{align*} Taking the supremum over all distinct $x,y\in \overline U$ yields \begin{align*} [u]_{C^{0,\alpha}(\overline U)} \le 2^{1-\alpha}. \end{align*} **Step 2 (Show the bound is sharp).** Choose $x=(1,0,\dots,0)$ and $y=(-1,0,\dots,0)$. Then $|x-y|=2$ and $|u(x)-u(y)|=2$, so \begin{align*} \frac{|u(x)-u(y)|}{|x-y|^\alpha} = \frac{2}{2^\alpha} = 2^{1-\alpha}. \end{align*} Hence \begin{align*} [u]_{C^{0,\alpha}(\overline U)} = 2^{1-\alpha}. \end{align*} **Step 3 (Bound the full $C^{0,\alpha}$ norm).** The $L^\infty$ part is \begin{align*} \|u\|_{L^\infty(U)} = \sup_{x\in B_1(0)} |x_1| = 1. \end{align*} Thus \begin{align*} \|u\|_{C^{0,\alpha}(\overline U)} = \|u\|_{L^\infty(U)} + [u]_{C^{0,\alpha}(\overline U)} = 1+2^{1-\alpha}. \end{align*} In particular, $u\in C^{0,\alpha}(\overline U)$. [/solution] --- ## References - Gilbarg, D. and Trudinger, N. S., *Elliptic Partial Differential Equations of Second Order* (2001). - Evans, L. C., *Partial Differential Equations* (2010). - Krylov, N. V., *Lectures on Elliptic and Parabolic Equations in Hölder Spaces* (1996). - Lieberman, G. M., *Second Order Parabolic Differential Equations* (1996).