> [!theorem] > > Let $U \subset \real^d$ be bounded, connected, and [[Open Set|open]] with $\partial U \in C^1$, $1 \le p < \infty$, and $W^{1, p}(U)$ be the [[Sobolev Space|Sobolev space]], then there exists $C_{d, p, U} \ge 0$ such that > $ > \norm{f - \frac{1}{m(U)}\int f}_{L^p} \le C\norm{Du}_{L^p} > $ > for all $f \in W^{1, p}(U)$. > > *Proof*. Suppose that the above does not hold. Let $\seq{f_n} \subset W^{1, p}(U)$ such that $\int f_n = 0$, $\norm{f_n}_{L^p} = 1$, and $\norm{Df_n} \le 1/n$ for all $n \in \nat$, then $\seq{f_n}$ is bounded in $W^{1, p}(U)$. > > By the [[Sobolev Embedding Theorem|Sobolev Embedding Theorem (I)]], there exists $\seq{n_k} \subset \nat$ and $f \in L^p(U)$ such that $f_{n_k} \to f$ in $L^p(U)$, $\int f = 0$, and $\norm{f}_{L^p} = 1$. Moreover, $Df_{n_k} \to 0$ in $L^p$, so $Df = 0$ almost everywhere. Thus $f$ is constant a.e. on connected components of $U$. Given that $U$ is connected, $f = 0$ a.e., which contradicts the fact that $\norm{f}_{L^p} = 1$.