[proofplan] The proof reduces to the unit ball $B(0,1)$ by scaling and then establishes a pointwise representation of $u(x) - u_B$ as a Riesz-type integral of $|\nabla u|$. For smooth $u$ on $\overline{B(0,1)}$, integrating along the segment from $y$ to $x$ gives $|u(x) - u(y)| \le |x-y| \int_0^1 |\nabla u(y + t(x-y))|\, d\mathcal{L}^1(t)$; averaging over $y \in B$, swapping integration order via Fubini, and bounding the resulting kernel produces the estimate $|u(x) - u_B| \le C \int_B |\nabla u(z)|\,|x-z|^{1-n}\, d\mathcal{L}^n(z)$. The right-hand side is the Riesz potential $I_1(|\nabla u|\mathbb{1}_B)$ evaluated at $x$, and a direct convolution-Hölder argument bounds its $L^p$ norm by $C(n,p)\|\nabla u\|_{L^p(B)}$. Density of $C^\infty(\overline{B(0,1)})$ in $W^{1,p}(B(0,1))$ ([Smooth Approximation Up to the Boundary](/theorems/3097)) then transfers the estimate to all $u \in W^{1,p}(B(0,1))$. Finally, the rescaling $v(z) := u(x_0 + rz)$ on $B(0,1)$ recovers the factor $r$ in front of the gradient norm. [/proofplan]

[step:Reduce to the unit ball $B(0,1)$ by rescaling] Set $B := B(x_0, r)$ and define the rescaled function \begin{align*} v: B(0,1) &\to \mathbb{R} \\ z &\mapsto u(x_0 + rz). \end{align*} By the chain rule and the bi-Lipschitz nature of the affine map $z \mapsto x_0 + rz$, $v \in W^{1,p}(B(0,1))$ with \begin{align*} \nabla v(z) = r\, \nabla u(x_0 + rz) \quad \mathcal{L}^n\text{-a.e. } z \in B(0,1). \end{align*} The change-of-variables formula gives \begin{align*} \|v\|_{L^p(B(0,1))}^p &= \int_{B(0,1)}|u(x_0 + rz)|^p\, d\mathcal{L}^n(z) = r^{-n}\int_B |u(x)|^p\, d\mathcal{L}^n(x) = r^{-n}\|u\|_{L^p(B)}^p, \\ \|\nabla v\|_{L^p(B(0,1))}^p &= r^p \int_{B(0,1)}|\nabla u(x_0 + rz)|^p\, d\mathcal{L}^n(z) = r^{p-n}\|\nabla u\|_{L^p(B)}^p. \end{align*} The ball average transforms by \begin{align*} v_{B(0,1)} = \frac{1}{\mathcal{L}^n(B(0,1))}\int_{B(0,1)}v\, d\mathcal{L}^n = \frac{r^{-n}}{\alpha_n}\int_B u\, d\mathcal{L}^n = \frac{1}{\alpha_n r^n}\int_B u\, d\mathcal{L}^n = u_B, \end{align*} where $\alpha_n := \mathcal{L}^n(B(0,1))$. Therefore $v - v_{B(0,1)}$ on $B(0,1)$ corresponds, after change of variables, to $u - u_B$ on $B$: \begin{align*} \|u - u_B\|_{L^p(B)}^p = r^n \|v - v_{B(0,1)}\|_{L^p(B(0,1))}^p. \end{align*} Suppose we have proved the unit-ball estimate \begin{align*} \|v - v_{B(0,1)}\|_{L^p(B(0,1))} \le C(n, p)\, \|\nabla v\|_{L^p(B(0,1))}. \end{align*} Then substituting the scaling identities, \begin{align*} \|u - u_B\|_{L^p(B)} = r^{n/p}\|v - v_{B(0,1)}\|_{L^p(B(0,1))} \le C(n,p)\, r^{n/p}\|\nabla v\|_{L^p(B(0,1))} = C(n,p)\, r\, \|\nabla u\|_{L^p(B)}, \end{align*} where the final equality uses $r^{n/p}\|\nabla v\|_{L^p(B(0,1))} = r^{n/p} \cdot r^{1 - n/p}\|\nabla u\|_{L^p(B)} = r\|\nabla u\|_{L^p(B)}$. So it suffices to prove the result for $r = 1$. [/step]

[step:Establish the pointwise representation $u(x) - u_B = \fint_B [u(x) - u(y)]\, d\mathcal{L}^n(y)$ on $B(0,1)$] For $u \in C^\infty(\overline{B(0,1)})$ and $x \in B(0,1)$, the average $u_B := \fint_{B(0,1)} u\, d\mathcal{L}^n = \alpha_n^{-1}\int_{B(0,1)} u\, d\mathcal{L}^n$ is a constant. Therefore \begin{align*} u(x) - u_B = u(x) - \fint_{B(0,1)} u(y)\, d\mathcal{L}^n(y) = \fint_{B(0,1)} \bigl[u(x) - u(y)\bigr]\, d\mathcal{L}^n(y). \end{align*} For each pair $(x, y)$ with $x, y \in B(0,1)$, the segment $\{y + t(x - y) : t \in [0, 1]\}$ joining $y$ to $x$ has Euclidean length $|x - y|$ and is contained in $B(0,1)$ by convexity of the ball. Apply the Fundamental Theorem of Calculus along this segment: \begin{align*} u(x) - u(y) = \int_0^1 \frac{d}{dt}\bigl[u(y + t(x-y))\bigr]\, d\mathcal{L}^1(t) = \int_0^1 \nabla u(y + t(x-y)) \cdot (x - y)\, d\mathcal{L}^1(t). \end{align*} Taking absolute values and applying $|a \cdot b| \le |a||b|$ with $|x - y| \le 2$ on $B(0,1)$, \begin{align*} |u(x) - u(y)| \le |x - y|\int_0^1 |\nabla u(y + t(x-y))|\, d\mathcal{L}^1(t). \end{align*} Integrating this pointwise bound over $y \in B(0,1)$ and dividing by $\alpha_n = \mathcal{L}^n(B(0,1))$, \begin{align*} |u(x) - u_B| \le \fint_{B(0,1)}|u(x) - u(y)|\, d\mathcal{L}^n(y) \le \fint_{B(0,1)} |x - y|\int_0^1 |\nabla u(y + t(x-y))|\, d\mathcal{L}^1(t)\, d\mathcal{L}^n(y). \end{align*}[/step]

[guided]**The averaging trick.** A constant subtracted from $u$ inside an average integrates to itself: \begin{align*} u_B = \fint_{B(0,1)} u\, d\mathcal{L}^n = \frac{1}{\alpha_n}\int_{B(0,1)} u(y)\, d\mathcal{L}^n(y), \end{align*} where $\alpha_n := \mathcal{L}^n(B(0,1))$. So \begin{align*} u(x) - u_B = u(x) - \fint_{B(0,1)} u(y)\, d\mathcal{L}^n(y) = \fint_{B(0,1)}\bigl[u(x) - u(y)\bigr]\, d\mathcal{L}^n(y), \end{align*} because $u(x)$ is constant in $y$, so $\fint_{B(0,1)} u(x)\, d\mathcal{L}^n(y) = u(x)$. This is the standard "average against $1$" trick: it converts the deviation $u(x) - u_B$ into an averaged increment $u(x) - u(y)$, which we can then estimate using the gradient. **Segment integration on a convex set.** $B(0,1)$ is convex: for any $x, y \in B(0,1)$ and $t \in [0,1]$, $y + t(x-y) \in B(0,1)$ (since $|y + t(x-y)| = |(1-t)y + tx| \le (1-t)|y| + t|x| < 1$). So the segment from $y$ to $x$ lies entirely in $B(0,1)$, and we may apply the Fundamental Theorem of Calculus (which is a standard one-variable result for $C^1$ functions, or smooth in our case) to the function $\phi(t) := u(y + t(x-y))$ on $[0, 1]$: \begin{align*} \phi(1) - \phi(0) = \int_0^1 \phi'(t)\, d\mathcal{L}^1(t). \end{align*} The chain rule gives $\phi'(t) = \nabla u(y + t(x-y)) \cdot (x - y)$, hence \begin{align*} u(x) - u(y) = \int_0^1 \nabla u(y + t(x-y)) \cdot (x - y)\, d\mathcal{L}^1(t). \end{align*} **Cauchy-Schwarz on the dot product.** Taking absolute values and using $|a \cdot b| \le |a||b|$ for vectors $a, b \in \mathbb{R}^n$ (the Cauchy-Schwarz inequality on $\mathbb{R}^n$ with the standard inner product), \begin{align*} |u(x) - u(y)| &\le \int_0^1 |\nabla u(y + t(x-y))|\,|x - y|\, d\mathcal{L}^1(t) \\ &= |x - y|\int_0^1 |\nabla u(y + t(x-y))|\, d\mathcal{L}^1(t), \end{align*} since $|x - y|$ is independent of $t$. **Averaging in $y$.** Integrating both sides over $y \in B(0,1)$ and dividing by $\alpha_n$, \begin{align*} |u(x) - u_B| &\stackrel{(\ast)}{\le} \fint_{B(0,1)}|u(x) - u(y)|\, d\mathcal{L}^n(y) \\ &\le \fint_{B(0,1)}|x - y|\int_0^1 |\nabla u(y + t(x-y))|\, d\mathcal{L}^1(t)\, d\mathcal{L}^n(y). \end{align*} The inequality $(\ast)$ is the triangle inequality moved inside the integral: $|u(x) - u_B| = \big|\fint(u(x) - u(y))\, d\mathcal{L}^n(y)\big| \le \fint|u(x) - u(y)|\, d\mathcal{L}^n(y)$.[/guided]

[step:Convert the segment integral to a Riesz-type pointwise bound on $|\nabla u|$]We rewrite the inner double integral by interchanging the order of integration (Fubini-Tonelli applies because the integrand is non-negative) and then changing variables via polar coordinates centred at $x$. **Polar substitution.** Introduce polar coordinates centred at $x$: write $y = x + s\omega$ with $s = |y - x| \in [0, 2]$ (since $y \in B(0,1)$ and $|x| < 1$, $|y - x| \le 2$) and $\omega \in S^{n-1}$. Then $d\mathcal{L}^n(y) = s^{n-1}\, d\mathcal{H}^{n-1}(\omega)\, d\mathcal{L}^1(s)$ and $|x - y| = s$. Now substitute $\tau := ts$ along the ray $y = x + s\omega$, with $z := y + t(x - y) = x + (1-t)s\omega$, hence the ray length from $x$ to $y + t(x-y)$ is $(1-t)s$. Set $\rho := (1-t)s \in [0, s]$, so $1 - t = \rho/s$, $dt = -d\rho/s$, $z = x + \rho\omega$, and the joint measure is \begin{align*} d\mathcal{L}^1(t)\, s^{n-1}\, d\mathcal{H}^{n-1}(\omega)\, d\mathcal{L}^1(s) = \frac{d\mathcal{L}^1(\rho)}{s}\, s^{n-1}\, d\mathcal{H}^{n-1}(\omega)\, d\mathcal{L}^1(s). \end{align*} Now, $|x-y|\,|\nabla u(y + t(x-y))| = s\,|\nabla u(x + \rho\omega)|$, with $s$ ranging from $\rho/(1) = \rho$ (when $t = 0$, $\rho = s$, but on integration over $t \in [0,1]$ for fixed $s$, $\rho \in [0, s]$) — alternatively, fix $\rho$ and note that $s$ ranges over $[\rho, R(x,\omega)]$ where $R(x, \omega)$ is the distance from $x$ to $\partial B(0,1)$ in direction $\omega$, which is at most $2$. So \begin{align*} \int_0^1 |x-y|\,|\nabla u(y + t(x-y))|\, d\mathcal{L}^1(t)\, d\mathcal{L}^n(y) \stackrel{\text{(over $y \in B$)}}{=} \int_{S^{n-1}}\int_0^{R(x,\omega)}\int_0^s s\,|\nabla u(x + \rho\omega)|\frac{d\mathcal{L}^1(\rho)\, s^{n-1}\, d\mathcal{L}^1(s)}{s}\, d\mathcal{H}^{n-1}(\omega). \end{align*} Simplify the inner factor: $s \cdot s^{n-1}/s = s^{n-1}$, so the integrand becomes $s^{n-1}|\nabla u(x + \rho\omega)|$. Integrating $s$ over $[\rho, R(x,\omega)] \subset [\rho, 2]$, \begin{align*} \int_\rho^{R(x,\omega)} s^{n-1}\, d\mathcal{L}^1(s) = \frac{R(x,\omega)^n - \rho^n}{n} \le \frac{2^n}{n}. \end{align*} Therefore \begin{align*} \fint_{B(0,1)}\int_0^1 |x-y|\,|\nabla u(y+t(x-y))|\, d\mathcal{L}^1(t)\, d\mathcal{L}^n(y) \le \frac{1}{\alpha_n}\cdot \frac{2^n}{n}\int_{S^{n-1}}\int_0^2 |\nabla u(x + \rho\omega)|\, d\mathcal{L}^1(\rho)\, d\mathcal{H}^{n-1}(\omega). \end{align*} Reverting to Cartesian coordinates $z = x + \rho\omega$, with $d\mathcal{L}^n(z) = \rho^{n-1}\, d\mathcal{L}^1(\rho)\, d\mathcal{H}^{n-1}(\omega)$, \begin{align*} \int_{S^{n-1}}\int_0^2 |\nabla u(x+\rho\omega)|\, d\mathcal{L}^1(\rho)\, d\mathcal{H}^{n-1}(\omega) = \int_{B(x, 2)}\frac{|\nabla u(z)|}{|z - x|^{n-1}}\, d\mathcal{L}^n(z). \end{align*} The integration domain $B(x, 2)$ contains $B(0,1)$ (since $|x| < 1$). We extend $\nabla u$ by zero outside $B(0,1)$ to get an integral over all of $\mathbb{R}^n$, or equivalently restrict the kernel to $B(0,1)$: \begin{align*} \int_{B(x,2)}\frac{|\nabla u(z)|}{|z-x|^{n-1}}\, d\mathcal{L}^n(z) = \int_{B(0,1)}\frac{|\nabla u(z)|}{|z - x|^{n-1}}\, d\mathcal{L}^n(z) + \int_{B(x,2)\setminus B(0,1)}0\, d\mathcal{L}^n(z) = \int_{B(0,1)}\frac{|\nabla u(z)|}{|z-x|^{n-1}}\, d\mathcal{L}^n(z). \end{align*} Putting it all together, for some constant $C_n := 2^n/(n\alpha_n)$ depending only on $n$, \begin{align*} |u(x) - u_B| \le C_n \int_{B(0,1)}\frac{|\nabla u(z)|}{|x-z|^{n-1}}\, d\mathcal{L}^n(z) \quad \text{for all } x \in B(0,1). \end{align*}[/step]

[guided]**Choosing polar coordinates.** The expression $|x-y|$ on $B(0,1)$ ranges over $[0, 2]$ — twice the radius of the ball — and we want to convert the segment-integral against $\nabla u$ into a kernel of the form $|x-z|^{-(n-1)}$, which is the standard Riesz potential of order $1$. Centre polar coordinates at $x$: every $y \in \mathbb{R}^n$ can be written $y = x + s\omega$ with $s = |y - x| \ge 0$ and $\omega := (y-x)/|y-x| \in S^{n-1}$ (for $y \ne x$). The Lebesgue measure decomposes as $d\mathcal{L}^n(y) = s^{n-1}\, d\mathcal{H}^{n-1}(\omega)\, d\mathcal{L}^1(s)$. **Substituting along the segment.** As $y$ varies and $t \in [0,1]$, the point $z := y + t(x-y) = x + (1-t)(y-x) = x + (1-t)s\omega$ slides along the ray from $x$ to $y$. Set $\rho := (1-t)s$; then $z = x + \rho\omega$, $\rho$ ranges from $0$ (at $t=1$) to $s$ (at $t=0$), and $1-t = \rho/s$, $dt = -d\rho/s$. So for fixed $(s, \omega)$, \begin{align*} \int_0^1 |\nabla u(y + t(x-y))|\, d\mathcal{L}^1(t) = \int_0^s |\nabla u(x + \rho\omega)|\frac{d\mathcal{L}^1(\rho)}{s}. \end{align*} **Combining the integrals.** The full integrand $|x - y|\cdot|\nabla u(y + t(x-y))|\cdot \mathbb{1}_{B(0,1)}(y)$ over $y \in B(0,1)$ and $t \in [0,1]$ yields, after Fubini and the substitution above, \begin{align*} \int_0^1\int_{B(0,1)} |x-y|\,|\nabla u(y+t(x-y))|\, d\mathcal{L}^n(y)\, d\mathcal{L}^1(t) = \int_{S^{n-1}}\int_0^{R(x,\omega)}\int_0^s s\,|\nabla u(x+\rho\omega)|\, s^{n-2}\, d\mathcal{L}^1(\rho)\, d\mathcal{L}^1(s)\, d\mathcal{H}^{n-1}(\omega), \end{align*} where $R(x, \omega) := \sup\{s : x + s\omega \in B(0,1)\} \le 2$ is the chord length from $x$ in direction $\omega$, and $s^{n-2}$ comes from $s^{n-1}/s$. The factor of $s^{n-1}$ is the Jacobian of polar coordinates, the factor $1/s$ comes from the substitution $\rho = (1-t)s$, and the explicit $s$ is from $|x-y|$. **Switching the order of $s$ and $\rho$.** Fubini-Tonelli applies (non-negative integrand) and we exchange the order of integration in $(s, \rho)$. The region is $0 \le \rho \le s \le R(x, \omega)$, which under swap becomes $0 \le \rho \le R(x, \omega)$, $\rho \le s \le R(x, \omega)$: \begin{align*} \int_0^{R(x,\omega)}\int_0^s s^{n-1}|\nabla u(x+\rho\omega)|\, d\mathcal{L}^1(\rho)\, d\mathcal{L}^1(s) = \int_0^{R(x,\omega)} |\nabla u(x+\rho\omega)|\int_\rho^{R(x,\omega)}s^{n-1}\, d\mathcal{L}^1(s)\, d\mathcal{L}^1(\rho). \end{align*} The inner $s$-integral evaluates to $(R(x,\omega)^n - \rho^n)/n \le R(x,\omega)^n/n \le 2^n/n$. Therefore \begin{align*} \le \frac{2^n}{n}\int_0^{R(x,\omega)}|\nabla u(x+\rho\omega)|\, d\mathcal{L}^1(\rho). \end{align*} **Reverting to Cartesian coordinates.** The angular-radial integral converts back: \begin{align*} \int_{S^{n-1}}\int_0^{R(x,\omega)}|\nabla u(x+\rho\omega)|\, d\mathcal{L}^1(\rho)\, d\mathcal{H}^{n-1}(\omega) &= \int_{B(0,1)} |\nabla u(z)|\, \frac{1}{|z-x|^{n-1}}\, d\mathcal{L}^n(z), \end{align*} using $d\mathcal{L}^n(z) = \rho^{n-1}\, d\mathcal{L}^1(\rho)\, d\mathcal{H}^{n-1}(\omega)$ where $\rho = |z - x|$ and the integration domain $\{x + \rho\omega : \rho \in [0, R(x,\omega)], \omega \in S^{n-1}\} = B(0,1)$ (since the rays from $x$ stay inside $B(0,1)$ exactly until they hit the boundary). **Final pointwise estimate.** Dividing by $\alpha_n$ for the average $\fint = \alpha_n^{-1}\int$, \begin{align*} |u(x) - u_B| \le \frac{2^n}{n\alpha_n}\int_{B(0,1)}\frac{|\nabla u(z)|}{|z-x|^{n-1}}\, d\mathcal{L}^n(z). \end{align*} Set $C_n := 2^n/(n\alpha_n)$, which depends only on the dimension $n$. The kernel $|z-x|^{-(n-1)}$ is locally integrable on $\mathbb{R}^n$ (since $n - 1 < n$), so the right-hand side is finite for every $x \in B(0,1)$ and indeed defines a bounded function of $x$ when $\nabla u \in L^1$ — but we will need an $L^p$ bound, addressed in the next step.[/guided]

[step:Bound the $L^p$ norm of the Riesz convolution by $\|\nabla u\|_{L^p}$]Define the truncated Riesz kernel \begin{align*} K: \mathbb{R}^n &\to [0, \infty] \\ w &\mapsto K(w) := \begin{cases} |w|^{-(n-1)}, & |w| < 2, \\ 0, & |w| \ge 2.\end{cases} \end{align*} The pointwise estimate of the previous step reads \begin{align*} |u(x) - u_B| \le C_n \int_{B(0,1)}\frac{|\nabla u(z)|}{|z-x|^{n-1}}\, d\mathcal{L}^n(z) \le C_n \bigl(K * |\nabla u|\mathbb{1}_{B(0,1)}\bigr)(x), \end{align*} since $|z - x| < 2$ for $z \in B(0,1)$ and $x \in B(0,1)$, so the kernel $K$ matches the integrand on the relevant set. We bound the $L^p$ norm of $K * f$ for $f \in L^p(\mathbb{R}^n)$ via Young's convolution inequality, which states that if $f \in L^p(\mathbb{R}^n)$ and $K \in L^1(\mathbb{R}^n)$, then $K * f \in L^p(\mathbb{R}^n)$ with \begin{align*} \|K * f\|_{L^p(\mathbb{R}^n)} \le \|K\|_{L^1(\mathbb{R}^n)}\|f\|_{L^p(\mathbb{R}^n)}. \end{align*} Verify $K \in L^1(\mathbb{R}^n)$: by polar coordinates, \begin{align*} \|K\|_{L^1(\mathbb{R}^n)} = \int_{\mathbb{R}^n} K(w)\, d\mathcal{L}^n(w) = \int_{B(0, 2)}|w|^{-(n-1)}\, d\mathcal{L}^n(w) = n\alpha_n \int_0^2 \rho^{-(n-1)}\rho^{n-1}\, d\mathcal{L}^1(\rho) = n\alpha_n \cdot 2 = 2n\alpha_n, \end{align*} which is finite (and depends only on $n$). Applied to $f := |\nabla u|\mathbb{1}_{B(0,1)} \in L^p(\mathbb{R}^n)$, \begin{align*} \|K * f\|_{L^p(\mathbb{R}^n)} \le 2n\alpha_n \|f\|_{L^p(\mathbb{R}^n)} = 2n\alpha_n \|\nabla u\|_{L^p(B(0,1))}. \end{align*} Restricting to $B(0,1) \subset \mathbb{R}^n$, \begin{align*} \|u - u_B\|_{L^p(B(0,1))} \le \|C_n (K * f)\|_{L^p(B(0,1))} \le C_n \|K * f\|_{L^p(\mathbb{R}^n)} \le C_n\cdot 2n\alpha_n\|\nabla u\|_{L^p(B(0,1))}. \end{align*} Setting $C(n, p) := C_n \cdot 2n\alpha_n = (2^n/(n\alpha_n))(2n\alpha_n) = 2^{n+1}$ — independent of $p$ in this case, but recorded as $C(n, p)$ for compatibility with the statement, \begin{align*} \|u - u_B\|_{L^p(B(0,1))} \le C(n, p)\|\nabla u\|_{L^p(B(0,1))}. \end{align*} For the case $p = \infty$, the bound becomes $\|u - u_B\|_{L^\infty(B(0,1))} \le C_n \|K\|_{L^1} \|\nabla u\|_{L^\infty(B(0,1))}$, by an analogous computation (with $\|K * f\|_{L^\infty} \le \|K\|_{L^1}\|f\|_{L^\infty}$).[/step]

[guided]**The Riesz kernel and Young's inequality.** The pointwise bound of the previous step says $|u(x) - u_B|$ is dominated by a convolution against the kernel $|w|^{-(n-1)}$ — but only over $B(0,1)$. To use the full machinery of convolutions on $\mathbb{R}^n$, define \begin{align*} K(w) := |w|^{-(n-1)}\mathbb{1}_{B(0,2)}(w), \end{align*} the truncation of the Riesz kernel of order $1$ to the ball of radius $2$. Truncating at radius $2$ is sufficient because $|z - x| < 2$ for $z, x \in B(0, 1)$, so the truncation does not change the relevant portion of the kernel. **Why $K \in L^1$.** The Riesz kernel $|w|^{-(n-1)}$ on $\mathbb{R}^n$ is **not** in $L^1$ globally — it has too much mass at infinity — but on a bounded region it is integrable, since $n - 1 < n$ and the singularity at $0$ is integrable in dimension $n$. Polar coordinates give \begin{align*} \int_{B(0,2)}|w|^{-(n-1)}\, d\mathcal{L}^n(w) = \int_{S^{n-1}}\int_0^2 \rho^{-(n-1)}\rho^{n-1}\, d\mathcal{L}^1(\rho)\, d\mathcal{H}^{n-1}(\omega) = \mathcal{H}^{n-1}(S^{n-1}) \cdot 2 = 2n\alpha_n, \end{align*} where we used $\mathcal{H}^{n-1}(S^{n-1}) = n\alpha_n$ (this is the standard relation between the $(n-1)$-dimensional surface area of the unit sphere and the $n$-dimensional volume of the unit ball). So $\|K\|_{L^1(\mathbb{R}^n)} = 2n\alpha_n$. **Young's inequality, version $1 + 1/p = 1/p$.** Young's convolution inequality states \begin{align*} \|K * f\|_{L^r(\mathbb{R}^n)} \le \|K\|_{L^a(\mathbb{R}^n)}\|f\|_{L^b(\mathbb{R}^n)} \quad \text{whenever } \frac{1}{r} = \frac{1}{a} + \frac{1}{b} - 1. \end{align*} With $a = 1$ and $b = p$, this gives $1/r = 0 + 1/p - 0 = 1/p$, i.e. $r = p$: \begin{align*} \|K * f\|_{L^p(\mathbb{R}^n)} \le \|K\|_{L^1(\mathbb{R}^n)}\|f\|_{L^p(\mathbb{R}^n)}. \end{align*} The hypotheses of Young require $1 \le a, b, r \le \infty$ and the constraint above; here $a = 1$, $b = p \in [1, \infty]$ both lie in this range. Apply with $f := |\nabla u|\mathbb{1}_{B(0,1)} \in L^p(\mathbb{R}^n)$: \begin{align*} \|K * f\|_{L^p(\mathbb{R}^n)} \le 2n\alpha_n\, \|\nabla u\|_{L^p(B(0,1))}. \end{align*} **Restricting to $B(0,1)$ and combining with the $C_n$ factor.** The pointwise bound was $|u(x) - u_B| \le C_n (K * f)(x)$ for $x \in B(0,1)$. Therefore \begin{align*} \|u - u_B\|_{L^p(B(0,1))} = \left(\int_{B(0,1)}|u - u_B|^p\, d\mathcal{L}^n\right)^{1/p} \le C_n\left(\int_{B(0,1)}(K*f)^p\, d\mathcal{L}^n\right)^{1/p} \le C_n\|K*f\|_{L^p(\mathbb{R}^n)}. \end{align*} The last inequality strictly enlarges the integration domain from $B(0,1)$ to $\mathbb{R}^n$, which is valid because $(K*f)^p \ge 0$. Combining, \begin{align*} \|u - u_B\|_{L^p(B(0,1))} \le C_n\cdot 2n\alpha_n\|\nabla u\|_{L^p(B(0,1))} = \frac{2^n}{n\alpha_n}\cdot 2n\alpha_n \|\nabla u\|_{L^p(B(0,1))} = 2^{n+1}\|\nabla u\|_{L^p(B(0,1))}. \end{align*} We have established the unit-ball estimate with $C(n,p) := 2^{n+1}$, which depends on $n$ only — but we write $C(n,p)$ for consistency with the theorem statement. **Validity for $p = \infty$.** For $p = \infty$, Young's inequality $\|K * f\|_{L^\infty} \le \|K\|_{L^1}\|f\|_{L^\infty}$ still holds (it is essentially the supremum bound on the convolution), and the same argument gives $\|u - u_B\|_{L^\infty(B(0,1))} \le 2^{n+1}\|\nabla u\|_{L^\infty(B(0,1))}$.[/guided]

[step:Extend from smooth functions to $W^{1,p}(B(0,1))$ by density] For $1 \le p < \infty$: by the [Smooth Approximation Up to the Boundary](/theorems/3097) theorem, $C^\infty(\overline{B(0,1)})$ is dense in $W^{1,p}(B(0,1))$ — the unit ball is a bounded Lipschitz domain. Given $u \in W^{1,p}(B(0,1))$, choose a sequence $u_m \in C^\infty(\overline{B(0,1)})$ with $u_m \to u$ in $W^{1,p}(B(0,1))$. The estimates of the previous step give \begin{align*} \|u_m - (u_m)_B\|_{L^p(B(0,1))} \le C(n, p)\|\nabla u_m\|_{L^p(B(0,1))}. \end{align*} The averages converge: $|(u_m)_B - u_B| = \alpha_n^{-1}|\int_B (u_m - u)\, d\mathcal{L}^n| \le \alpha_n^{-1}\mathcal{L}^n(B(0,1))^{1/p'}\|u_m - u\|_{L^p} = \alpha_n^{-1/p}\|u_m - u\|_{L^p}$ by Hölder's inequality with exponents $(p, p')$. Hence $(u_m)_B \to u_B$ as $m \to \infty$. Strong $L^p$ convergence and convergence of constants give $u_m - (u_m)_B \to u - u_B$ in $L^p(B(0,1))$, and similarly $\nabla u_m \to \nabla u$ in $L^p(B(0,1))$. Passing to the limit in the inequality (which is valid because both sides are continuous in the $L^p$ topology), \begin{align*} \|u - u_B\|_{L^p(B(0,1))} \le C(n,p)\|\nabla u\|_{L^p(B(0,1))}. \end{align*} For $p = \infty$: the same argument applies using density of $C^\infty(\overline{B(0,1)})$ in $W^{1,\infty}(B(0,1)) = \mathrm{Lip}(B(0,1))$ (more precisely, weak-$*$ density via the [$W^{1,\infty}$ Equals Lipschitz](/theorems/3129) characterisation), or directly: $u \in W^{1,\infty}$ has a Lipschitz representative, and the Lipschitz estimate $|u(x) - u(y)| \le \|\nabla u\|_{L^\infty}|x-y|$ yields $\|u - u_B\|_{L^\infty} \le \|u - u_B\|_{C^0} \le 2\|\nabla u\|_{L^\infty} = C(n,\infty)\|\nabla u\|_{L^\infty}$ — with $C(n,\infty)$ depending on the diameter of $B(0,1)$, equal to $2$. This completes the unit-ball case. [/step]

[step:Combine with the rescaling to obtain the general radius $r$] By Step 1, the estimate on $B(0,1)$ rescales to \begin{align*} \|u - u_B\|_{L^p(B)} = r^{n/p}\|v - v_{B(0,1)}\|_{L^p(B(0,1))} \le r^{n/p}C(n,p)\|\nabla v\|_{L^p(B(0,1))} = r^{n/p}\cdot C(n,p)\cdot r^{1 - n/p}\|\nabla u\|_{L^p(B)} = C(n,p)\, r\, \|\nabla u\|_{L^p(B)}, \end{align*} where we used $\|\nabla v\|_{L^p(B(0,1))} = r^{1 - n/p}\|\nabla u\|_{L^p(B)}$ from Step 1. Hence for the original $u \in W^{1,p}(B(x_0, r))$, \begin{align*} \|u - u_B\|_{L^p(B(x_0, r))} \le C(n, p)\, r\, \|\nabla u\|_{L^p(B(x_0, r))}, \end{align*} which is the desired Poincaré inequality on the ball $B(x_0, r)$. The constant $C(n, p)$ is independent of $x_0$ and $r$, as required. [/step]