> [!theorem] > > Let $1 \le p < \infty$, $W^{1, p}_0(\real^d)$ be the closure of $C_c^\infty(\real^d)$ in the [[Sobolev Space|Sobolev space]], and $f \in W^{1, p}_0(\real^d)$ be compactly supported. If $\supp{f} \subset [-R, R]^d$, > $ > \norm{f}_{L^p} \le 2R\norm{Df}_{L^p} > $ > *Proof*. Let $q$ be the [[Hölder's Inequality|Hölder]] conjugate of $p$. First suppose that $f \in C_c^\infty(\real^d)$. Let $K = \supp{f}$ be its support and $R \ge 0$ such that $K \subset [-R, R]^d$, then for each $x \in \real^d$, > $ > \begin{align*} > \abs{f(x)} &\le \int_{\real} \abs{Df(t, x_2, \cdots, x_d)}dt \\ > \abs{f(x)} &\le (2R)^{1/q}\braks{\int_{\real} \abs{Df(t, x_2, \cdots, x_d)}^pdt}^{1/p} \\ > \int_{\real^d}\abs{f}^p&\le (2R)^{1 + p/q}\int_{\real} \abs{Df}^pdt = (2R)^p\norm{Df}_{L^p}^p > \end{align*} > $ > [!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 $\norm{f}_{L^p} = 1$.