I want to show the following Poincaré inequality for an open, bounded and convex region $\Omega \subset \mathbb{R}^d$,
$$ \int\limits_{\Omega} \vert u\vert^p \, dx \leq CD^p\int\limits_{\Omega} \vert \nabla u\vert^p \, dx $$
where $D$ is the diameter of $\Omega$ and $u \in C^1(\Omega)$, having the mean value property, i.e. $(u)_{\Omega} = \frac{1}{\alpha(\Omega)}\int\limits_{\Omega} u\, dx = 0$.
I do know the poincaré inequality for functions $u \in C^1_c(\Omega)$, $u \in W^{1,2}_0(\Omega)$ and $ u \in W^{1, p}(\Omega)$. Particularly, the last one already uses $(u)_{\Omega}$ and hence for $(u)_{\Omega} = 0$ the desired inequality follows readily for $W^{1, p}(\Omega)$. However, how does it follow that $u \in C^1(\Omega)$ then? I think $W^{1, p}(\Omega) \subset L^2(\Omega)$ but how to get from here?
Moreover, I got a hint that said to prove
$$\vert u(x) \vert \leq D\int\limits_{\Omega}\int\limits_{0}^{1} \vert \nabla (mx + (1 - m)x^*)\vert \, dm\,dx^*$$
in order to prove the inequality directly. It is according to FTC
$$ \vert u(x) - u(x^*) \vert = \Big\vert \int\limits_{0}^{1} \frac{\partial}{\partial m} u(mx + (1 - m)x^*)dm\Big\vert $$
Now integrating with respect to $x^*$ over $\Omega$ gives
$$\vert u(x) \vert \leq \vert \Omega \vert \int\limits_{\Omega}\int\limits_{0}^{1} \vert \nabla u(mx + (1 - m)x^*)\cdot(x - x^*)\,ds\vert $$
But how do I proceed here and reach the desired inequality?
EDIT: this is much longer than I anticipated. This proof gives a direct derivation of the inequality. There is a much shorter proof arguing by contradiction that doesn't describe the constant as accurately.
It helps to use polar coordinates. If $x \in \Omega$ and $\omega \in S^{n-1}$, the ray $x + r \omega$, $r > 0$, eventually crosses the boundary of $\Omega$. The quantity $$d(x,\omega) = \sup\{r > 0 : x + r \omega \in \Omega\}$$ can be thought of as the distance from $x$ to $\partial \Omega$ in the direction of $\omega$.
Fix $x \in \Omega$. Any $y \in \Omega$ other than $x$ itself may be uniquely written as $y = x + r\omega$ with $r > 0$ and $\omega \in S^{n-1}$, and the integral of a function $f$ over $\Omega$ may be expressed in polar coordinates as $$\int_\Omega f(y) \, dy = \int_{S^{n-1}} \int_0^{d(x,\omega)} f(x + r\omega) r^{n-1} \, drd\omega.$$ Since the integral of $u$ on $\Omega$ vanishes, this leads to $$|\Omega| u(x) = \int_\Omega u(x) - u(y) \, dy = \int_{S^{n-1}} \int_0^{d(x,\omega)} (u(x) - u(x + r\omega) ) r^{n-1} \, dr d\omega.$$
Focus on the integrand. Define a function $t$ by $\phi(t) = u(x) - u(x+t\omega)$. Then $\phi'$ is given by $\phi'(t) = Du(x + t\omega) \cdot \omega$ and the fundamental theorem of calculus gives you $$u(x) - u(x+r\omega) = \phi(0) - \phi(r) = - \int_0^r \phi'(t) \, dt = -\int_0^r Du(x + t\omega) \cdot \omega \, dt.$$ Since $|\omega| = 1$ for each $\omega \in S^{n-1}$ and $r <d(x,\omega)$, this leads to the estimate $$ | u(x) - u(x+r\omega)| \le \int_0^r |Du(x + t\omega)| \, dt \le \int_0^{d(x,\omega)} |Du(x + t\omega)| \, dt$$ where the last integral makes sense since $x + t\omega \in \Omega$ for all $0 \le t < d(x,\omega)$. We may use the triangle inequality, Tonelli's theorem, and the fact that $d(x,\omega) \le \mathrm{diam\,} \Omega$ for any $x$ and $\omega$ to obtain \begin{align*}|\Omega| |u(x)| &\le \int_{S^{n-1}} \int_0^{d(x,\omega)} \int_0^{d(x,\omega)} |Du(x + t\omega)| \,dt r^{n-1} \, dr d\omega \\ &\le \int_{S^{n-1}} \int_0^{\mathrm{diam\,}\Omega} \int_0^{d(x,\omega)} |Du(x + t\omega)| \,dt r^{n-1} \, dr d\omega \\ &=\int_0^{\mathrm{diam\,}\Omega} \int_{S^{n-1}} \int_0^{d(x,\omega)} |Du(x + t\omega)| \,dt d\omega r^{n-1} \, dr \\ &= \frac{(\mathrm{diam\,}\Omega)^n}{n} \int_{S^{n-1}} \int_0^{d(x,\omega)} |Du(x + t\omega)| \,dt d\omega. \end{align*} Next go back to rectangular coordinates: \begin{align*}\int_{S^{n-1}} \int_0^{d(x,\omega)} |Du(x + t\omega)| \,dt d\omega &= \int_{S^{n-1}} \int_0^{d(x,\omega)} \frac{|Du(x + t\omega)|}{t^{n-1}} t^{n-1}\,dt d\omega \\ & = \int_{S^{n-1}} \int_0^{d(x,\omega)} \frac{|Du(x + t\omega)|}{|x - (x + t\omega)|^{n-1}} t^{n-1}\,dt d\omega \\ & \int_\Omega \frac{|Du(y)|}{|x-y|^{n-1}} \, dy.\end{align*} This yields the pointwise estimate $$|u(x)| \le \frac{(\mathrm{diam\,}\Omega)^n}{n|\Omega|} \int_\Omega \frac{|Du(y)|}{|x-y|^{n-1}} \, dy.$$
Let $\alpha_n$ denote the volume of the $n$-ball. We use polar coordinates and the fact that the surface area of the $n-1$ sphere is $n \alpha_n$ to find $$ \int_{\Omega} \frac 1{|x-y|^{n-1}} \, dy = \int_{S^{n-1}} \int_0^{d(x,\omega)} \frac{1}{r^{n-1}} r^{n-1} \, dr d\omega \le \int_{S^{n-1}} d(x,\omega) d\omega \le (\mathrm{diam\,}\Omega) n\alpha_n$$
Holder's inequality with indices $p,p'$ applied to the pointwise estimate gives you $$\int_\Omega \frac{|Du(y)|}{|x-y|^{n-1}} \, dy \le \left( \int_\Omega \frac{|Du(y)|^p}{|x-y|^{n-1}} \, dy \right)^{1/p} \left( \int_{\Omega} \frac 1{|x-y|^{n-1}} \, dy \right)^{1/p'}$$ so that $$|u(x)|^p \le \frac{(\mathrm{diam\,}\Omega)^{np}}{n^p|\Omega|^p} (\mathrm{diam\,}\Omega)^{p/p'} n^{p/p'}\alpha_n^{p/p'} \int_\Omega \frac{|Du(y)|^p}{|x-y|^{n-1}} \, dy .$$
Finally integrate over $\Omega$ with respect to $x$. Tonelli's theorem gives you $$\int_\Omega \int_\Omega \frac{|Du(y)|^p}{|x-y|^{n-1}} \, dy \, dx = \int_\Omega |Du(y)|^p \int_\Omega \frac{1}{|x-y|^{n-1}} \, dx \, dy \le (\mathrm{diam\,}\Omega) n\alpha_n \int_\Omega |Du(y)|^p \, dy.$$ Thus (after some simplifying) $$\int_\Omega |u(x)|^p \, dx \le \frac{(\mathrm{diam\,}\Omega)^{np}}{|\Omega|^p} (\mathrm{diam\,}\Omega)^p \alpha_n^p \int_\Omega |Du(y)|^p \, dy$$ giving you the desired inequality with $$C = \left( \frac{(\mathrm{diam\,}\Omega)^{n}\alpha_n}{|\Omega|} \right)^p.$$