I have a bit of an an odd question with this poincare-type inequality. I have been able to come up with something, given below, for $k=0,l=1$ but am unsure what to do and where to go when $p=\infty$. I wonder if you could possibly give me a hand.
Denote by $D^{k}v(\boldsymbol{x})$, $k\in\mathbb{N}\cup\left\{ 0\right\} $, the total derivative of order k of a function $v(\boldsymbol{x})$, e.g., for $k=1$ the gradient of $v(\boldsymbol{x})$. Let $\Omega$ be convex and be included into a ball of radius R. Let $l\in\mathbb{N}\cup\left\{ 0\right\} $ with $k\leq l$ and let $p\in R$ with $p\in[1,\infty)$. Assume that $v\in W^{l,p}(\Omega)$ satisfies the following:
\begin{align*} {{\int_{\Omega}\partial_{\boldsymbol{\alpha}}v(\boldsymbol{x})\ d\boldsymbol{x}=0\,\text{for all }\left|\boldsymbol{\alpha}\right|\leq l-1,}}\\ \end{align*}
Then the following estimate holds:
\begin{align*} \left\Vert D^{k}v\right\Vert _{L^{p}\left(\Omega\right)} & \leq CR^{l-k}\left\Vert D^{l}v\right\Vert \\ \end{align*}
Where the constant $C$ does not depend on $\Omega$ and on $v\left(\boldsymbol{x}\right)$.
What I have thus far:
Since $\Omega$ is assumed to be convex, the integral mean value theorem can be written as follows:
\begin{align*} {v(\boldsymbol{x})-v(\boldsymbol{y})=\int_{0}^{1}\nabla v(t\boldsymbol{x}+(1-t)\boldsymbol{y})\cdot(\boldsymbol{x}-\boldsymbol{y})\;dt,\quad\boldsymbol{x},\boldsymbol{y}\in\Omega}\\ \end{align*}
integrate resp.$\boldsymbol{y}$:
\begin{align*} {v(\boldsymbol{x})\int_{\Omega}~d\boldsymbol{y}-\int_{\Omega}v(\boldsymbol{y})~d\boldsymbol{y}=\int_{\Omega}\int_{0}^{1}\nabla v(t\boldsymbol{x}+(1-t)\boldsymbol{y})\cdot(\boldsymbol{x}-\boldsymbol{y})~dt~d\boldsymbol{y}}\\ \end{align*}
From the assumption that the second integral on the left-hand side vanishes we get:
\begin{align*} {v(\boldsymbol{x})=\frac{1}{|\Omega|}\int_{\Omega}\int_{0}^{1}\nabla v(t\boldsymbol{x}+(1-t)\boldsymbol{y})\cdot(\boldsymbol{x}-\boldsymbol{y})\ dt\ d\boldsymbol{y}}\\ \end{align*}
Take absolute values on both sides and apply Cauchy Schwarz and using the estimate $\left\Vert \boldsymbol{x}-\boldsymbol{y}\right\Vert _{2}\leq2R$:
\begin{align*} {|v(\boldsymbol{x})|=\frac{1}{|\Omega|}\left|\int_{\Omega}\int_{0}^{1}\nabla v(t\boldsymbol{x}+(1-t)\boldsymbol{y})\cdot(\boldsymbol{x}-\boldsymbol{y})\ dt\ d\boldsymbol{y}\right|}\\ \end{align*}
\begin{align*}{\leq\ \frac{1}{|\Omega|}\int_{\Omega}\int_{0}^{1}|\nabla v(t\boldsymbol{x}+(1-t)\boldsymbol{y})\cdot(\boldsymbol{x}-\boldsymbol{y})|\ \ dt\ d\boldsymbol{y}}\\ \end{align*}
\begin{align*} {\leq\,\frac{2R}{|\Omega|}\int_{\Omega}\int_{0}^{1}\|\nabla v(t\boldsymbol{x}+(1-t)\boldsymbol{y})\|_{2}\,\,\,dt\,\,d\boldsymbol{y}}\\ \end{align*}
From this point, I start to get a little lost. From applying Fubin, Integral Mean Value theorem and other theorems I arrive at the following:
\begin{align*} {\int_{\Omega}|v(\boldsymbol{x})|^{p}~~d\boldsymbol{x}\leq2CR^{p}\left\Vert \nabla v\right\Vert _{L^{p}\left(\Omega\right)}^{p}}\\ \end{align*}
and from there I have no idea which is why I ask . Thank you in advance for reading this rather rambly post by me, I appreciat and solutions or help rendered greatly!