proof on Poincare's inequality. This might be a silly question. 
So basically, I have proved the Poincare's inequality for $p=1$ case. That is, for $u\in W^{1,1}(\Omega)$, I have $||u-\bar{u}||_{L^1}\leq C||\nabla u||_{L^1}$. Here $\bar{u}$ is the average of $u$ on $\Omega$.
Now I need to get the general $p$ case, i.e., for $u\in W^{1,p}(\Omega)$, there is $||u-\bar{u}||_{L^p}\leq C||\nabla u||_{L^p}$. My professor in class argued very briefly. He said just look at $g(z)=u(z)-\bar{u}$ and $h(z)=|g(z)|^p$. Apply the pre-established $p=1$ Poincare's inequality, and it's done. But now I don't see why.
 A: What your professor meant to say takes just one line which looks like that
$$
\|g\|_{L^p}^p=\int\limits_{\Omega}|h(z)-\overline{h}|\,dz\leqslant
C\!\!\int\limits_{\Omega}|\nabla h(z)|\,dz\leqslant 
Cp\!\!\int\limits_{\Omega}|\nabla g(z)|^{p-1}|\nabla g(z)|\,dz\leqslant
\|g\|_{L^p}^{p-1}\|\nabla g\|_{L^p}
$$
where the Hölder's inequality is applied with the conjugate exponents $p$ and
$p'=p/(p-1)$.  But to follow this way one should have chosen   $h(z)-\overline{h}=|g(z)|^p\,$ which immediately implies $g\equiv 0$,  i.e.,  $u\equiv\overline{u}=const$.  Taking just  $h(z)=|g(z)|^p$  doesn't work.  A miscalculation can be corrected if only one takes $\,h(z)=g(z)|g(z)|^{p-1}$ in order to  start with the Poincaré-like inequality
$$
\int\limits_{\Omega}|h(z)|\,dz\leqslant
C\!\!\int\limits_{\Omega}|\nabla h(z)|\,dz\quad\forall\, h\in W^{1,1}(\Omega)\,\colon 
\int\limits_{\Omega} h(z)|h(z)|^{-1/p'}\,dz=0. \tag{$\ast$}
$$
Proving $(\ast)$ by inverse argument amounts to a trivial exercise due to the obvious fact that $h(z)|h(z)|^{-1/p'}=g(z)=u(z)-\overline{u}$,  and hence
$$
\int\limits_{\Omega} h(z)|h(z)|^{-1/p'}\,dz=0 \iff 
\int\limits_{\Omega} g(z)\,dz=\int\limits_{\Omega} \bigl(u(z)-\overline{u}\bigr)\,dz=0.
$$
But it is rather doubtful whether any direct proof of $(\ast)$ be known as yet. 
