I have a question about Poincare-Wirtinger inequality for $H^{1}(D)$.
Let $D$ is an open subset of $\mathbb{R}^d$. We define $H^{1}(D)$ by \begin{equation*} H^{1}(D)=\{f \in L^{2}(D,m): \frac{\partial f}{\partial x_i} \in L^{2}(D,m),\ 1\le i \le d\}, \end{equation*} where $\partial f/\partial x_i$ is the distributional derivative of $f$ and $m$ is the Lebesgue measure on $D$. $H^{1}(D)$ becomes a Hilbert space with the usual Sobolev norm $\|\cdot\|_{H^{1}(D)}$.
In the following, we write $\nabla f$ for $(\partial f/\partial x_1,\ldots \partial f/\partial x_d)$. We also write $\|f\|$ for $(\int_{D}f^2\,dx)^{1/2}$
[Theorem 1] Let $D$ be a connected open subset of $\mathbb{R}^d$ with finite volume. We assume the canonical embedding $H^{1}(D) \subset L^{2}(D,m)$ is compact. Then, there exists a positive constant $C>0$ such that for any $f \in H^{1}(D)$ \begin{equation*} \|f-f_{D}\|\le C \|\nabla f\|, \end{equation*} where $f_{D}:=m(D)^{-1}\int_{D}f\,dm$
[Proof of Theorem 1]
Define $X \subset H^{1}(D)$ by $\{f \in H^{1}(D) : f_{D}=0 \}$. Then, $X$ is a closed subspace of $H^{1}(D)$. It suffices to prove there exists a $C>0$ such that \begin{equation*} (*)\quad \|f\|\le C \|\nabla f\| \ \text{ for any }f \in X \end{equation*} We prove $(*)$ by contradiction. We assume there exist a $\{f_n\} \subset X$ such that $\|f_n\|>n \|\nabla f_n\|$. We may assume $\|f_n\|_{H^{1}(D)}=1$. As $f_n$ is a bounded sequence in $X$, there exists a subsequence $f_{n_k}$ of $f_n$ and a $f \in X$ such that $f_{n_k} \to f$ weakly in $X$. $f_{n_k}$ is also denoted by $f_n$. By assumption, $f_n \to f$ in $L^{2}(D,m)$. Since $$1=\|f_n\|_{H^{1}(D)}\text{ and }\|f_n\|>n \|\nabla f_n\|$$ holds, $\|f_n\|\to \underline{1=\|f\|}$ and $\|\nabla f_n\| \to 0$. Since $f_n \to f$ in $L^{2}(D,dx)$ and $\|\nabla f_n\| \to 0$, \begin{equation*} \int_{D}(\nabla f_n,\nabla g)+\int_{D}f_n g\,dx\to \int_{D}fg\,dx. \end{equation*} for any $g \in X$. On the other hand, since $f_n$ convenes to $f$ weakly in $X$, \begin{equation*} \int_{D}(\nabla f_n,\nabla g)dx+\int_{D}f_n g\,dx\to \int_{D} (\nabla f,\nabla g)\,dx+\int_{D}fg\,dx. \end{equation*} Hence $\int_{D}(\nabla f,\nabla g)dx=0$ for any $g \in X$, which implies $\nabla f=0$ and $\Delta f=0$ in weak sense. By Wyle's Lemma $f$ is smooth on $D$. Hence, $\nabla f=0$ in classical sense. Since $D$ is connected, $f$ is a some constant. Since $ f \in X$, we have $f=0$, which contradicts to $\underline{1=\|f\|}$.
My question and remark
From the above proof, we do not know concrete estimates of C. If you know an another proof, please let me know.