## Problems

[problem] **Smooth Cutoff Functions (Partitions of Unity Construction)** Let $U$ and $V$ be open subsets of $\mathbb{R}^n$ such that $V \Subset U$ (meaning $\overline{V}$ is compact and $\overline{V} \subset U$). Construct a smooth cutoff function $\zeta \in C_c^\infty(U)$ satisfying the following properties: 1. $0 \le \zeta(x) \le 1$ for all $x \in U$. 2. $\zeta(x) = 1$ for all $x \in V$. 3. $\operatorname{supp}(\zeta) \subset U$. [/problem]

[solution] **Step 1: Geometric Setup and Intermediate Sets** Since $\overline{V}$ is compact and $U^c = \mathbb{R}^n \setminus U$ is closed, and they are disjoint, the distance between them is strictly positive. We define: \begin{align*} d := \operatorname{dist}(\overline{V}, \mathbb{R}^n \setminus U) > 0 \end{align*} We define an intermediate open set $W$ as the $d/2$-neighborhood of $\overline{V}$: \begin{align*} W := \{ x \in \mathbb{R}^n : \operatorname{dist}(x, \overline{V}) < d/2 \} \end{align*} By construction, we have the inclusions $V \subset \overline{V} \subset W$. Furthermore, for any $y \in \overline{W}$, we have $\operatorname{dist}(y, \overline{V}) \le d/2$. Thus, the triangle inequality implies: \begin{align*} \operatorname{dist}(y, \mathbb{R}^n \setminus U) \ge \operatorname{dist}(\overline{V}, \mathbb{R}^n \setminus U) - \operatorname{dist}(y, \overline{V}) \ge d - d/2 = d/2 > 0 \end{align*} This confirms that $\overline{W} \subset U$, and consequently $V \Subset W \Subset U$. We now define two auxiliary distances to safeguard our mollification parameter: \begin{align*} a &:= \operatorname{dist}(\overline{V}, \mathbb{R}^n \setminus W) > 0 \\ b &:= \operatorname{dist}(\overline{W}, \mathbb{R}^n \setminus U) > 0 \end{align*} **Step 2: The Rough Cutoff** We define the characteristic function of the intermediate set $W$ on the entire space $\mathbb{R}^n$: \begin{align*} \chi_W: \mathbb{R}^n &\to \{0, 1\} \\ x &\mapsto \begin{cases} 1 & \text{if } x \in W \\ 0 & \text{if } x \notin W \end{cases} \end{align*} Note that $\chi_W \in L^\infty(\mathbb{R}^n)$ with compact support $\overline{W}$. **Step 3: Mollification** We choose a standard mollifier $\eta \in C_c^\infty(\mathbb{R}^n)$ supported in $B(0,1)$. For a parameter $\varepsilon > 0$ to be fixed, we define $\eta_\varepsilon(x) = \varepsilon^{-n}\eta(x/\varepsilon)$. We define the candidate function $\zeta$ by the convolution on $\mathbb{R}^n$: \begin{align*} \zeta: \mathbb{R}^n &\to \mathbb{R} \\ x &\mapsto (\eta_\varepsilon * \chi_W)(x) = \int_{\mathbb{R}^n} \eta_\varepsilon(x - y)\chi_W(y) \, d\mathcal{L}^n(y) \end{align*} Since $\eta_\varepsilon \in C_c^\infty(\mathbb{R}^n)$, properties of convolution imply that $\zeta \in C^\infty(\mathbb{R}^n)$. Furthermore, since $0 \le \chi_W \le 1$ and $\int \eta_\varepsilon = 1$, we satisfy the bounds $0 \le \zeta \le 1$. **Step 4: Ensuring $\zeta \equiv 1$ on $V$** We choose $\varepsilon$ such that $0 < \varepsilon < a$. Let $x \in V$. The support of the integrand is restricted to $y \in B(x, \varepsilon)$. Since $x \in V$ and $\varepsilon < \operatorname{dist}(\overline{V}, \mathbb{R}^n \setminus W)$, the ball $B(x, \varepsilon)$ is entirely contained within $W$. Therefore, for all $y \in B(x, \varepsilon)$, we have $\chi_W(y) = 1$. The integral becomes: \begin{align*} \zeta(x) = \int_{B(x,\varepsilon)} \eta_\varepsilon(x - y) \cdot 1 \, d\mathcal{L}^n(y) = 1 \end{align*} **Step 5: Ensuring Support is Compact in $U$** We further restrict $\varepsilon$ such that $0 < \varepsilon < b$. If $\operatorname{dist}(x, W) > \varepsilon$, then for all $y \in W$, $|x - y| > \varepsilon$. This implies that the supports of $\eta_\varepsilon(x - \cdot)$ and $\chi_W$ are disjoint, so $\zeta(x) = 0$. Thus, the support of $\zeta$ satisfies: \begin{align*} \operatorname{supp}(\zeta) \subseteq \{ x \in \mathbb{R}^n : \operatorname{dist}(x, W) \le \varepsilon \} \end{align*} Let $x \in \operatorname{supp}(\zeta)$. By the definition of $b$, every point in $W$ is at distance at least $b$ from $U^c$. Since $\varepsilon < b$, the triangle inequality ensures $x$ is still at a positive distance from $U^c$: \begin{align*} \operatorname{dist}(x, U^c) \ge \operatorname{dist}(\overline{W}, U^c) - \varepsilon = b - \varepsilon > 0 \end{align*} Thus $\operatorname{supp}(\zeta) \subset U$. Since $\operatorname{supp}(\zeta)$ is a closed, bounded subset of $\mathbb{R}^n$, it is compact. We conclude $\zeta \in C_c^\infty(U)$. [/solution]

[problem] **Interpolation of Hölder Norms** Assume parameters satisfying $0 < \beta < \gamma \le 1$. Let $U \subset \mathbb{R}^n$ be an open set. Prove the interpolation inequality: \begin{align*} \|u\|_{C^{0,\gamma}(U)} \le \|u\|_{C^{0,\beta}(U)}^{\frac{1-\gamma}{1-\beta}} \|u\|_{C^{0,1}(U)}^{\frac{\gamma-\beta}{1-\beta}} \end{align*} for any $u \in C^{0,1}(U)$, where the Hölder norm is defined as $\|u\|_{C^{0,\alpha}(U)} = \|u\|_{L^\infty(U)} + [u]_{C^{0,\alpha}(U)}$. [/problem]

[solution] **Step 1: Exponent Analysis** We define the interpolation weight $\theta \in (0,1)$ by: \begin{align*} \theta = \frac{1-\gamma}{1-\beta} \end{align*} Computing the complementary exponent $1-\theta$: \begin{align*} 1 - \theta = 1 - \frac{1-\gamma}{1-\beta} = \frac{1 - \beta - 1 + \gamma}{1-\beta} = \frac{\gamma - \beta}{1-\beta} \end{align*} We observe that $\gamma$ is the convex combination of $\beta$ and $1$: \begin{align*} \beta\theta + 1(1-\theta) = \gamma \end{align*} **Step 2: Interpolation of the Seminorm** We estimate the Hölder seminorm $[u]_{C^{0,\gamma}(U)}$. Let $x, y \in U$ with $x \neq y$. We decompose the denominator $|x-y|^\gamma$ using the identity $\gamma = \beta\theta + (1-\theta)$: \begin{align*} \frac{|u(x) - u(y)|}{|x - y|^\gamma} &= \frac{|u(x) - u(y)|^\theta \, |u(x) - u(y)|^{1-\theta}}{|x - y|^{\beta\theta} \, |x - y|^{1-\theta}} \\ &= \left( \frac{|u(x) - u(y)|}{|x - y|^\beta} \right)^\theta \left( \frac{|u(x) - u(y)|}{|x - y|} \right)^{1-\theta} \end{align*} By the definition of the supremum, for any non-negative functions $f, g$, $\sup(fg) \le (\sup f)(\sup g)$. Applying this to the expression above: \begin{align*} [u]_{C^{0,\gamma}(U)} &= \sup_{x \neq y} \left[ \left( \frac{|u(x) - u(y)|}{|x - y|^\beta} \right)^\theta \left( \frac{|u(x) - u(y)|}{|x - y|} \right)^{1-\theta} \right] \\ &\le \left( \sup_{x \neq y} \frac{|u(x) - u(y)|}{|x - y|^\beta} \right)^\theta \left( \sup_{x \neq y} \frac{|u(x) - u(y)|}{|x - y|} \right)^{1-\theta} \\ &= [u]_{C^{0,\beta}(U)}^\theta \, [u]_{C^{0,1}(U)}^{1-\theta} \end{align*} **Step 3: Discrete Hölder Inequality for the Full Norm** We express the full norm as the sum of the $L^\infty$ norm and the seminorm. We apply the **Discrete Hölder Inequality** to vectors in $\mathbb{R}^2$. Let $p = \frac{1}{\theta}$ and $q = \frac{1}{1-\theta}$. Note that $\frac{1}{p} + \frac{1}{q} = 1$. Define the vectors $a, b \in \mathbb{R}^2$ as: \begin{align*} a &= \left( \|u\|_{L^\infty(U)}^\theta, \ [u]_{C^{0,\beta}(U)}^\theta \right) \\ b &= \left( \|u\|_{L^\infty(U)}^{1-\theta}, \ [u]_{C^{0,1}(U)}^{1-\theta} \right) \end{align*} The standard Hölder inequality $\sum a_i b_i \le (\sum a_i^p)^{1/p} (\sum b_i^q)^{1/q}$ yields: \begin{align*} \|u\|_{L^\infty(U)} + [u]_{C^{0,\gamma}(U)} &\le \|u\|_{L^\infty(U)}^{\theta} \|u\|_{L^\infty(U)}^{1-\theta} + [u]_{C^{0,\beta}(U)}^\theta [u]_{C^{0,1}(U)}^{1-\theta} \\ &\le \left( (\|u\|_{L^\infty(U)}^\theta)^{1/\theta} + ([u]_{C^{0,\beta}(U)}^\theta)^{1/\theta} \right)^\theta \\ &\quad \times \left( (\|u\|_{L^\infty(U)}^{1-\theta})^{1/(1-\theta)} + ([u]_{C^{0,1}(U)}^{1-\theta})^{1/(1-\theta)} \right)^{1-\theta} \end{align*} Simplifying the exponents: \begin{align*} \|u\|_{C^{0,\gamma}(U)} &\le \left( \|u\|_{L^\infty(U)} + [u]_{C^{0,\beta}(U)} \right)^\theta \left( \|u\|_{L^\infty(U)} + [u]_{C^{0,1}(U)} \right)^{1-\theta} \\ &= \|u\|_{C^{0,\beta}(U)}^\theta \|u\|_{C^{0,1}(U)}^{1-\theta} \end{align*} **Step 4: Conclusion** Substituting the value of $\theta = \frac{1-\gamma}{1-\beta}$ and $1-\theta = \frac{\gamma-\beta}{1-\beta}$ into the result from Step 3, we obtain: \begin{align*} \|u\|_{C^{0,\gamma}(U)} \le \|u\|_{C^{0,\beta}(U)}^{\frac{1-\gamma}{1-\beta}} \|u\|_{C^{0,1}(U)}^{\frac{\gamma-\beta}{1-\beta}} \end{align*} [/solution]