[proofplan] The BV Poincaré inequality on a ball asserts $\int_B |u - u_B| \, d\mathcal{L}^n \le C(n) r |Du|(B)$ for $u \in BV(B(x_0, r))$. The strategy is to deduce the BV statement from its smooth ($W^{1,1}$) counterpart by approximation: for smooth $u$, the standard Poincaré inequality on balls (the $p = 1$ case of the [Poincaré Inequality on Balls](/theorems/3103)) gives the bound with $|\nabla u|$ in place of $|Du|$. For general $u \in BV(B)$, the [BV Smooth Approximation](/theorems/3131) theorem produces smooth approximations $u_k \in C^\infty(B) \cap BV(B)$ converging strictly to $u$, meaning $u_k \to u$ in $L^1(B)$ and $|Du_k|(B) \to |Du|(B)$. Applying the smooth Poincaré bound to each $u_k$ and passing to the limit on both sides yields the BV bound. The two technical pieces are (i) the $L^1$ convergence of the means $u_{k,B} \to u_B$ and (ii) the fact that for smooth functions $|Du_k|(B) = \int_B |\nabla u_k| \, d\mathcal{L}^n$. [/proofplan]

[step:Recall the Poincaré inequality on balls for smooth functions]The [Poincaré Inequality on Balls](/theorems/3103) with $p = 1$ states: there exists $C(n) > 0$ such that for every ball $B = B(x_0, r) \subset \mathbb{R}^n$ and every $w \in W^{1,1}(B)$, \begin{align*} \int_B |w - w_B| \, d\mathcal{L}^n \le C(n) \, r \int_B |\nabla w| \, d\mathcal{L}^n, \end{align*} where $w_B = \fint_B w \, d\mathcal{L}^n$. In particular, the inequality applies to every $w \in C^\infty(B) \cap W^{1,1}(B)$, and in that case $\int_B |\nabla w| \, d\mathcal{L}^n = |Dw|(B)$ since the distributional gradient of a smooth $W^{1,1}$ function coincides with its classical gradient.[/step]

[guided]The [Poincaré Inequality on Balls](/theorems/3103) is established for $W^{1, p}$ functions with $1 \le p \le \infty$. The case relevant here is $p = 1$: for every ball $B = B(x_0, r)$ and every $w \in W^{1, 1}(B)$, \begin{align*} \int_B |w - w_B| \, d\mathcal{L}^n = \|w - w_B\|_{L^1(B)} \le C(n) \, r \, \|\nabla w\|_{L^1(B)} = C(n) \, r \int_B |\nabla w| \, d\mathcal{L}^n. \end{align*} The mean is $w_B = \fint_B w \, d\mathcal{L}^n = \mathcal{L}^n(B)^{-1} \int_B w \, d\mathcal{L}^n$. The constant $C(n)$ is independent of the radius $r$; the factor $r$ on the right is the explicit scaling factor. For $w \in C^\infty(B) \cap W^{1, 1}(B)$, the distributional gradient $Dw$ coincides with the classical gradient, so $Dw$ is the absolutely continuous measure $\nabla w \cdot \mathcal{L}^n$ and \begin{align*} |Dw|(B) = \int_B |\nabla w| \, d\mathcal{L}^n. \end{align*} Therefore the smooth-function form of the Poincaré inequality reads \begin{align*} \int_B |w - w_B| \, d\mathcal{L}^n \le C(n) \, r \, |Dw|(B) \quad \text{for } w \in C^\infty(B) \cap W^{1, 1}(B). \end{align*} This is the form that admits a direct extension to BV via approximation.[/guided]

[step:Approximate $u \in BV(B)$ strictly by smooth functions]Let $u \in BV(B)$. By the [BV Smooth Approximation](/theorems/3131) theorem, there exists a sequence $(u_k)$ with $u_k \in C^\infty(B) \cap BV(B)$ such that \begin{align*} u_k \to u \quad \text{in } L^1(B), \qquad |Du_k|(B) \to |Du|(B) \text{ as } k \to \infty. \end{align*} Since $u_k \in C^\infty(B) \cap BV(B)$ and $|Du_k|(B) = \int_B |\nabla u_k| \, d\mathcal{L}^n < \infty$, each $u_k$ also belongs to $W^{1, 1}(B)$. Therefore the smooth Poincaré inequality from Step 1 applies: \begin{align*} \int_B |u_k - (u_k)_B| \, d\mathcal{L}^n \le C(n) \, r \, |Du_k|(B), \qquad k \in \mathbb{N}, \end{align*} where $(u_k)_B = \fint_B u_k \, d\mathcal{L}^n$.[/step]

[guided]The [BV Smooth Approximation](/theorems/3131) theorem furnishes a sequence in $C^\infty(B) \cap BV(B)$ that converges to $u$ both in $L^1$ and in total variation; this is the notion of strict convergence in $BV$. Specifically, the conclusion of the theorem applied to the open set $B = B(x_0, r) \subseteq \mathbb{R}^n$ with $u \in BV(B)$ produces $u_k \in C^\infty(B) \cap BV(B)$ with \begin{align*} \|u_k - u\|_{L^1(B)} \to 0, \qquad |Du_k|(B) \to |Du|(B). \end{align*} The hypothesis of [BV Smooth Approximation](/theorems/3131) is $u \in BV(\Omega)$ for $\Omega = B$ open, met directly. *Why $u_k \in W^{1, 1}(B)$.* By construction $u_k \in C^\infty(B)$, so its distributional gradient is the absolutely continuous measure $Du_k = \nabla u_k \cdot \mathcal{L}^n$, hence $|Du_k|(B) = \int_B |\nabla u_k| \, d\mathcal{L}^n$. Since $|Du_k|(B) \to |Du|(B) < \infty$, the sequence $(|Du_k|(B))_k$ is bounded, hence $\|\nabla u_k\|_{L^1(B)} = |Du_k|(B) < \infty$ for each $k$. Combined with $\|u_k\|_{L^1(B)} < \infty$ (from $u_k \to u$ in $L^1(B)$, hence the $L^1$ norms are bounded), we conclude $u_k \in W^{1, 1}(B)$. *Smooth Poincaré applied.* The smooth Poincaré inequality from Step 1 applies to each $u_k$: \begin{align*} \int_B |u_k - (u_k)_B| \, d\mathcal{L}^n \le C(n) \, r \, |Du_k|(B), \qquad k \in \mathbb{N}, \end{align*} where $(u_k)_B = \fint_B u_k \, d\mathcal{L}^n$ is the average of $u_k$ over $B$. The constant $C(n)$ is the dimensional constant from [Poincaré Inequality on Balls](/theorems/3103), independent of $k$ and $r$.[/guided]

[step:Pass to the limit on both sides]Take $k \to \infty$ in the inequality of Step 2. *Right-hand side.* By construction, $|Du_k|(B) \to |Du|(B)$, so \begin{align*} C(n) \, r \, |Du_k|(B) \to C(n) \, r \, |Du|(B). \end{align*} *Left-hand side.* Since $u_k \to u$ in $L^1(B)$, the means converge: \begin{align*} (u_k)_B = \frac{1}{\mathcal{L}^n(B)} \int_B u_k \, d\mathcal{L}^n \to \frac{1}{\mathcal{L}^n(B)} \int_B u \, d\mathcal{L}^n = u_B, \end{align*} where the convergence follows from $|\int_B u_k - u_B| \le \frac{1}{\mathcal{L}^n(B)} \|u_k - u\|_{L^1(B)} \to 0$. By the triangle inequality, \begin{align*} \int_B \big| u_k - (u_k)_B \big| \, d\mathcal{L}^n &\ge \int_B \big| u - u_B \big| \, d\mathcal{L}^n - \int_B \big| (u_k - u) - ((u_k)_B - u_B) \big| \, d\mathcal{L}^n \\ &\ge \int_B |u - u_B| \, d\mathcal{L}^n - \|u_k - u\|_{L^1(B)} - \mathcal{L}^n(B) \, |(u_k)_B - u_B|. \end{align*} Both error terms tend to $0$, so \begin{align*} \liminf_{k \to \infty} \int_B \big| u_k - (u_k)_B \big| \, d\mathcal{L}^n \ge \int_B |u - u_B| \, d\mathcal{L}^n. \end{align*} Combining with the inequality from Step 2, \begin{align*} \int_B |u - u_B| \, d\mathcal{L}^n \le \liminf_{k \to \infty} \int_B \big| u_k - (u_k)_B \big| \, d\mathcal{L}^n \le C(n) \, r \, \lim_{k \to \infty} |Du_k|(B) = C(n) \, r \, |Du|(B). \end{align*} This is the BV Poincaré inequality with constant $C(n)$ from the smooth case.[/step]

[guided]We pass to the limit on both sides of the inequality from Step 2. *Right-hand side: $|Du_k|(B) \to |Du|(B)$.* This is the strict convergence guaranteed by [BV Smooth Approximation](/theorems/3131). Multiplying by the constant $C(n) \, r$: \begin{align*} \lim_{k \to \infty} C(n) \, r \, |Du_k|(B) = C(n) \, r \, |Du|(B). \end{align*} *Left-hand side: convergence of means.* Since $u_k \to u$ in $L^1(B)$, the means \begin{align*} (u_k)_B = \frac{1}{\mathcal{L}^n(B)} \int_B u_k \, d\mathcal{L}^n \end{align*} converge to $u_B = \mathcal{L}^n(B)^{-1} \int_B u \, d\mathcal{L}^n$. Quantitatively, \begin{align*} |(u_k)_B - u_B| = \frac{1}{\mathcal{L}^n(B)} \left| \int_B (u_k - u) \, d\mathcal{L}^n \right| \le \frac{1}{\mathcal{L}^n(B)} \int_B |u_k - u| \, d\mathcal{L}^n = \frac{\|u_k - u\|_{L^1(B)}}{\mathcal{L}^n(B)} \to 0. \end{align*} *Lower semicontinuity of the LHS.* We show $\liminf_k \int_B |u_k - (u_k)_B| \, d\mathcal{L}^n \ge \int_B |u - u_B| \, d\mathcal{L}^n$. By the triangle inequality applied pointwise, \begin{align*} |u_k(x) - (u_k)_B| \ge |u(x) - u_B| - |u_k(x) - u(x)| - |(u_k)_B - u_B|. \end{align*} Integrating over $x \in B$, \begin{align*} \int_B |u_k - (u_k)_B| \, d\mathcal{L}^n \ge \int_B |u - u_B| \, d\mathcal{L}^n - \|u_k - u\|_{L^1(B)} - \mathcal{L}^n(B) \, |(u_k)_B - u_B|. \end{align*} Both error terms $\|u_k - u\|_{L^1(B)}$ and $\mathcal{L}^n(B) \, |(u_k)_B - u_B|$ tend to $0$ as $k \to \infty$, the first by hypothesis and the second by the means convergence above. Taking $\liminf_k$ on both sides: \begin{align*} \liminf_{k \to \infty} \int_B |u_k - (u_k)_B| \, d\mathcal{L}^n \ge \int_B |u - u_B| \, d\mathcal{L}^n. \end{align*} (Equivalently, by the dominated/triangle reverse argument, $\int_B |u_k - (u_k)_B|\,d\mathcal{L}^n \to \int_B |u - u_B|\,d\mathcal{L}^n$, but only the lower-bound direction is needed.) *Combination.* From Step 2, \begin{align*} \int_B |u_k - (u_k)_B| \, d\mathcal{L}^n \le C(n) \, r \, |Du_k|(B). \end{align*} Taking $\liminf_k$ of the LHS and $\lim_k$ of the RHS: \begin{align*} \int_B |u - u_B| \, d\mathcal{L}^n \le \liminf_{k \to \infty} \int_B |u_k - (u_k)_B| \, d\mathcal{L}^n \le \lim_{k \to \infty} C(n) \, r \, |Du_k|(B) = C(n) \, r \, |Du|(B). \end{align*} This is the desired BV Poincaré inequality with the same dimensional constant $C(n)$ as in the smooth case from Step 1. The proof is complete.[/guided]