[proofplan] We prove Morrey's Inequality by establishing two independent estimates on smooth functions $u \in C_c^1(\mathbb{R}^n)$ and then combining them. The Holder seminorm bound is obtained by averaging the fundamental-theorem representation of $u(x) - u(y)$ over the intersection of two balls, using polar coordinates to convert the gradient integral into a Riesz-potential estimate, then applying Holder's inequality to extract the $L^p$ gradient norm. The $L^\infty$ bound follows by adding and subtracting a ball average and estimating each piece. Combining the two estimates yields control of the full $C^{0,\gamma}$ norm by the $W^{1,p}$ norm. [/proofplan] [step:Bound the ball-averaged oscillation by a Riesz potential of the gradient] Fix $x \in \mathbb{R}^n$ and $r > 0$. Let $u \in C_c^1(\mathbb{R}^n)$. [claim:Ball-average estimate] For any $x \in \mathbb{R}^n$ and $r > 0$: \begin{align*} \frac{1}{\alpha_n r^n} \int_{B(x,r)} |u(x) - u(y)| \, d\mathcal{L}^n(y) \le \frac{1}{n \alpha_n} \int_{B(x,r)} \frac{|\nabla u(z)|}{|x - z|^{n-1}} \, d\mathcal{L}^n(z) \end{align*} where $\alpha_n = \mathcal{L}^n(B(0,1))$ is the volume of the unit ball in $\mathbb{R}^n$. [/claim] [proof] Fix a direction $w \in \partial B(0,1)$ and a radius $s \in (0, r)$. By the [Fundamental Theorem of Calculus](/theorems/632) along the line segment from $x$ to $x + sw$: \begin{align*} u(x + sw) - u(x) = \int_0^s \nabla u(x + tw) \cdot w \, d\mathcal{L}^1(t). \end{align*} Taking the absolute value and applying the Cauchy-Schwarz inequality with $|w| = 1$: \begin{align*} |u(x + sw) - u(x)| \le \int_0^s |\nabla u(x + tw)| \, d\mathcal{L}^1(t). \end{align*} We integrate over the unit sphere $\partial B(0,1)$ with respect to the surface measure $\sigma$, multiply by $s^{n-1}$, and integrate over $s \in [0, r]$ with respect to $\mathcal{L}^1$: \begin{align*} \int_0^r \int_{\partial B(0,1)} |u(x + sw) - u(x)| \, s^{n-1} \, d\sigma(w) \, d\mathcal{L}^1(s) \le \int_0^r s^{n-1} \int_{\partial B(0,1)} \int_0^s |\nabla u(x + tw)| \, d\mathcal{L}^1(t) \, d\sigma(w) \, d\mathcal{L}^1(s). \end{align*} On the left-hand side, the polar coordinates decomposition $d\mathcal{L}^n(y) = s^{n-1} \, d\sigma(w) \, d\mathcal{L}^1(s)$ under the substitution $y = x + sw$ gives \begin{align*} \text{LHS} = \int_{B(x,r)} |u(x) - u(y)| \, d\mathcal{L}^n(y). \end{align*} On the right-hand side, since $0 \le t \le s \le r$ and the integrand is non-negative, we enlarge the upper limit of the inner integral from $s$ to $r$. After this enlargement, the integrand no longer depends on $s$, so the $s$-integral factors: \begin{align*} \text{RHS} \le \left( \int_0^r s^{n-1} \, d\mathcal{L}^1(s) \right) \int_{\partial B(0,1)} \int_0^r |\nabla u(x + tw)| \, d\mathcal{L}^1(t) \, d\sigma(w) = \frac{r^n}{n} \int_{\partial B(0,1)} \int_0^r |\nabla u(x + tw)| \, d\mathcal{L}^1(t) \, d\sigma(w). \end{align*} Converting back to Cartesian coordinates via $z = x + tw$ with $d\mathcal{L}^n(z) = t^{n-1} \, d\sigma(w) \, d\mathcal{L}^1(t)$, we multiply and divide by $t^{n-1}$: \begin{align*} \int_{\partial B(0,1)} \int_0^r |\nabla u(x + tw)| \, d\mathcal{L}^1(t) \, d\sigma(w) = \int_{\partial B(0,1)} \int_0^r \frac{|\nabla u(x + tw)|}{t^{n-1}} \, t^{n-1} \, d\mathcal{L}^1(t) \, d\sigma(w) = \int_{B(x,r)} \frac{|\nabla u(z)|}{|x - z|^{n-1}} \, d\mathcal{L}^n(z). \end{align*} Dividing both sides by $\alpha_n r^n = \mathcal{L}^n(B(x,r))$ yields the claim. [/proof] [guided] The aim is to express the average oscillation of $u$ over a ball in terms of a weighted integral of $|\nabla u|$. The weight $|x - z|^{-(n-1)}$ is the Riesz kernel, which arises naturally from converting between polar and Cartesian coordinates. Fix $w \in \partial B(0,1)$ and $s \in (0, r)$. By the [Fundamental Theorem of Calculus](/theorems/632) along the ray from $x$ in direction $w$: \begin{align*} u(x + sw) - u(x) = \int_0^s \nabla u(x + tw) \cdot w \, d\mathcal{L}^1(t). \end{align*} Taking absolute values and using $|w| = 1$: \begin{align*} |u(x + sw) - u(x)| \le \int_0^s |\nabla u(x + tw)| \, d\mathcal{L}^1(t). \end{align*} To convert this radial estimate into a volume integral, we integrate over all directions $w \in \partial B(0,1)$ and all radii $s \in [0, r]$, weighting by the polar Jacobian $s^{n-1}$. On the left, the substitution $y = x + sw$ with $d\mathcal{L}^n(y) = s^{n-1} \, d\sigma(w) \, d\mathcal{L}^1(s)$ converts the iterated integral into $\int_{B(x,r)} |u(x) - u(y)| \, d\mathcal{L}^n(y)$. On the right, the constraint $t \le s$ couples the $s$ and $t$ integrals. The key step is to enlarge the domain: since the integrand is non-negative and $s \le r$, replacing the upper limit $s$ by $r$ only increases the integral. After this enlargement, the inner integral no longer depends on $s$, so the double integral factors as $\frac{r^n}{n} \cdot \int_{\partial B(0,1)} \int_0^r |\nabla u(x + tw)| \, d\mathcal{L}^1(t) \, d\sigma(w)$. The remaining iterated integral converts back to Cartesian via $z = x + tw$. The Jacobian $t^{n-1}$ in $d\mathcal{L}^n(z) = t^{n-1} \, d\sigma(w) \, d\mathcal{L}^1(t)$ introduces the factor $1/t^{n-1} = 1/|x - z|^{n-1}$: \begin{align*} \int_{\partial B(0,1)} \int_0^r |\nabla u(x + tw)| \, d\mathcal{L}^1(t) \, d\sigma(w) = \int_{B(x,r)} \frac{|\nabla u(z)|}{|x - z|^{n-1}} \, d\mathcal{L}^n(z). \end{align*} Combining and dividing by $\alpha_n r^n$: \begin{align*} \frac{1}{\alpha_n r^n} \int_{B(x,r)} |u(x) - u(y)| \, d\mathcal{L}^n(y) \le \frac{1}{n \alpha_n} \int_{B(x,r)} \frac{|\nabla u(z)|}{|x - z|^{n-1}} \, d\mathcal{L}^n(z). \end{align*} [/guided] [/step] [step:Apply Holder's inequality to extract the $L^p$ gradient norm with a power of $r$] [claim:Holder reduction of the Riesz potential] For any $x \in \mathbb{R}^n$ and $r > 0$, with $C_1 = \frac{1}{n\alpha_n}\left(n\alpha_n \frac{p-1}{p-n}\right)^{(p-1)/p}$: \begin{align*} \frac{1}{\alpha_n r^n} \int_{B(x,r)} |u(x) - u(y)| \, d\mathcal{L}^n(y) \le C_1 \, r^{1 - n/p} \left( \int_{B(x,r)} |\nabla u(z)|^p \, d\mathcal{L}^n(z) \right)^{1/p}. \end{align*} [/claim] [proof] We apply [Holder's inequality](/pages/1222) to the Riesz-potential integral from the previous step with the conjugate pair $(p, q)$ where $q = p/(p-1)$, pairing $|\nabla u(z)|$ in $L^p$ with $|x - z|^{-(n-1)}$ in $L^q$: \begin{align*} \int_{B(x,r)} \frac{|\nabla u(z)|}{|x - z|^{n-1}} \, d\mathcal{L}^n(z) \le \left( \int_{B(x,r)} |\nabla u(z)|^p \, d\mathcal{L}^n(z) \right)^{1/p} \left( \int_{B(x,r)} |x - z|^{-(n-1)q} \, d\mathcal{L}^n(z) \right)^{1/q}. \end{align*} We evaluate the second factor using polar coordinates centered at $x$: substituting $\rho = |x - z|$ with $d\mathcal{L}^n(z) = \rho^{n-1} \, d\sigma(w) \, d\mathcal{L}^1(\rho)$ and $\int_{\partial B(0,1)} d\sigma = n\alpha_n$: \begin{align*} \int_{B(x,r)} |x - z|^{-(n-1)q} \, d\mathcal{L}^n(z) = n\alpha_n \int_0^r \rho^{n - 1 - (n-1)q} \, d\mathcal{L}^1(\rho). \end{align*} Substituting $q = p/(p-1)$, the exponent becomes $n - 1 - (n-1)p/(p-1) = (1 - n)/(p - 1)$. Since $p > n$, this exponent satisfies $(1-n)/(p-1) > -1$, so the integral converges: \begin{align*} n\alpha_n \int_0^r \rho^{(1-n)/(p-1)} \, d\mathcal{L}^1(\rho) = n\alpha_n \cdot \frac{p - 1}{p - n} \, r^{(p-n)/(p-1)}. \end{align*} Raising to the power $1/q = (p-1)/p$: \begin{align*} \left( \int_{B(x,r)} |x - z|^{-(n-1)q} \, d\mathcal{L}^n(z) \right)^{1/q} = \left( n\alpha_n \frac{p-1}{p-n} \right)^{(p-1)/p} r^{1 - n/p}. \end{align*} Combining with the ball-average estimate from the previous step and dividing by $\alpha_n r^n$: \begin{align*} \frac{1}{\alpha_n r^n} \int_{B(x,r)} |u(x) - u(y)| \, d\mathcal{L}^n(y) \le \frac{1}{n\alpha_n} \left( n\alpha_n \frac{p-1}{p-n} \right)^{(p-1)/p} r^{1 - n/p} \left( \int_{B(x,r)} |\nabla u(z)|^p \, d\mathcal{L}^n(z) \right)^{1/p}. \end{align*} [/proof] [/step] [step:Establish the Holder seminorm bound via ball-intersection averaging] Fix $x, y \in \mathbb{R}^n$ with $x \neq y$ and set $r = |x - y|$. Define the intersection $W = B(x, r) \cap B(y, r)$ and the average \begin{align*} (u)_W = \frac{1}{\mathcal{L}^n(W)} \int_W u(z) \, d\mathcal{L}^n(z). \end{align*} By the triangle inequality: \begin{align*} |u(x) - u(y)| \le |u(x) - (u)_W| + |u(y) - (u)_W|. \end{align*} For the first term: \begin{align*} |u(x) - (u)_W| = \left| \frac{1}{\mathcal{L}^n(W)} \int_W (u(x) - u(z)) \, d\mathcal{L}^n(z) \right| \le \frac{1}{\mathcal{L}^n(W)} \int_W |u(x) - u(z)| \, d\mathcal{L}^n(z). \end{align*} Since $W \subseteq B(x, r)$, we enlarge the integration domain: \begin{align*} \frac{1}{\mathcal{L}^n(W)} \int_W |u(x) - u(z)| \, d\mathcal{L}^n(z) \le \frac{1}{\mathcal{L}^n(W)} \int_{B(x,r)} |u(x) - u(z)| \, d\mathcal{L}^n(z). \end{align*} The Lebesgue measure of $W$ satisfies $\mathcal{L}^n(W) = \gamma_n r^n$ for a dimensional constant $\gamma_n > 0$ (the measure of the intersection of two unit balls with centres at distance $1$, scaled by $r^n$). Writing $\mathcal{L}^n(W) = (\gamma_n / \alpha_n) \cdot \alpha_n r^n$ and applying the estimate from the previous step: \begin{align*} |u(x) - (u)_W| \le \frac{\alpha_n}{\gamma_n} C_1 \, r^{1 - n/p} \left( \int_{B(x,r)} |\nabla u(z)|^p \, d\mathcal{L}^n(z) \right)^{1/p}. \end{align*} Since $B(x, r) \subseteq \mathbb{R}^n$, we enlarge the gradient integral to all of $\mathbb{R}^n$: \begin{align*} |u(x) - (u)_W| \le \frac{\alpha_n C_1}{\gamma_n} \, r^{1 - n/p} \|\nabla u\|_{L^p(\mathbb{R}^n)}. \end{align*} By the identical argument centered at $y$ (since $W \subseteq B(y, r)$ as well): \begin{align*} |u(y) - (u)_W| \le \frac{\alpha_n C_1}{\gamma_n} \, r^{1 - n/p} \|\nabla u\|_{L^p(\mathbb{R}^n)}. \end{align*} Summing and setting $C_2 = 2\alpha_n C_1 / \gamma_n$: \begin{align*} |u(x) - u(y)| \le C_2 \, |x - y|^{1 - n/p} \|\nabla u\|_{L^p(\mathbb{R}^n)}. \end{align*} [/step] [step:Bound the $L^\infty$ norm by the $W^{1,p}$ norm] Fix $x \in \mathbb{R}^n$. Adding and subtracting the ball average over $B(x, 1)$ and applying the triangle inequality: \begin{align*} |u(x)| \le |u(x) - (u)_{B(x,1)}| + |(u)_{B(x,1)}|. \end{align*} For the first term, setting $r = 1$ in the estimate from the second step: \begin{align*} |u(x) - (u)_{B(x,1)}| \le C_1 \left( \int_{B(x,1)} |\nabla u(z)|^p \, d\mathcal{L}^n(z) \right)^{1/p} \le C_1 \|\nabla u\|_{L^p(\mathbb{R}^n)}. \end{align*} For the second term, applying Holder's inequality with exponents $(p, p/(p-1))$ over $B(x, 1)$: \begin{align*} |(u)_{B(x,1)}| \le \frac{1}{\alpha_n} \int_{B(x,1)} |u(z)| \, d\mathcal{L}^n(z) \le \frac{1}{\alpha_n} \left( \int_{B(x,1)} |u(z)|^p \, d\mathcal{L}^n(z) \right)^{1/p} \left( \alpha_n \right)^{(p-1)/p} = \alpha_n^{-1/p} \|u\|_{L^p(\mathbb{R}^n)}. \end{align*} Combining and taking the supremum over $x \in \mathbb{R}^n$, with $C_3 = \max\{C_1, \alpha_n^{-1/p}\}$: \begin{align*} \|u\|_{L^\infty(\mathbb{R}^n)} \le C_3 \left( \|\nabla u\|_{L^p(\mathbb{R}^n)} + \|u\|_{L^p(\mathbb{R}^n)} \right) = C_3 \|u\|_{W^{1,p}(\mathbb{R}^n)}. \end{align*} [/step] [step:Combine the Holder seminorm and $L^\infty$ bounds into the full $C^{0,\gamma}$ estimate] The [Holder norm](/pages/1016) is \begin{align*} \|u\|_{C^{0,\gamma}(\mathbb{R}^n)} = \|u\|_{L^\infty(\mathbb{R}^n)} + \sup_{x \neq y} \frac{|u(x) - u(y)|}{|x - y|^\gamma} \end{align*} where $\gamma = 1 - n/p$. Substituting the estimates from the previous two steps: \begin{align*} \|u\|_{C^{0,\gamma}(\mathbb{R}^n)} \le C_3 \|u\|_{W^{1,p}(\mathbb{R}^n)} + C_2 \|\nabla u\|_{L^p(\mathbb{R}^n)}. \end{align*} Since $\|\nabla u\|_{L^p(\mathbb{R}^n)} \le \|u\|_{W^{1,p}(\mathbb{R}^n)}$, defining $C = C_2 + C_3$: \begin{align*} \|u\|_{C^{0,\gamma}(\mathbb{R}^n)} \le C \|u\|_{W^{1,p}(\mathbb{R}^n)}. \end{align*} This completes the proof for $u \in C_c^1(\mathbb{R}^n)$. The result extends to all $u \in W^{1,p}(\mathbb{R}^n)$ by [density](/pages/1225) of $C_c^\infty(\mathbb{R}^n)$ in $W^{1,p}(\mathbb{R}^n)$: given $u_m \to u$ in $W^{1,p}$, the inequality gives $\|u_m - u_l\|_{C^{0,\gamma}} \le C \|u_m - u_l\|_{W^{1,p}} \to 0$, so $(u_m)$ is Cauchy in $C^{0,\gamma}(\mathbb{R}^n)$, and the limit coincides with $u$ after modification on a set of $\mathcal{L}^n$-measure zero. [/step]