On a compact subset $\Omega \subset \mathbb{R}^n$, the Poincaré inequality states
$$\|u\|_{L^p(\Omega)} \leq C \|\nabla u \|_{L^p(\Omega)}. \tag{1}$$
When we generalize to a compact Riemannian manifold $M$ with metric $g$ the inequality is given as:
$$\|u – \overline{u}\|_{L^p(M)} \leq C \|\nabla u\|_{L^p(M)}$$
where $$\overline{u} = \frac{1}{\text{Vol}_{(M,g)}} \int_M u \,d\text{Vol}_g. \tag{2}$$
I have two questions regarding this inequality.
- Why are we able to omit the average value of $u$ on $\mathbb{R}^n$ as in (1) but not in general as in (2)?
- Is the gradient in (2) interpreted as the Riemannian gradient with respect to $g$, i.e. $\nabla u = (\partial_i u) g^{ij}\partial_j $? In some texts (for example, Jost's Riemannian Geometry and Geometric Analysis) he writes
Let $M$ be a compact Riemannian manifold. If $f \in H^{1,2}(M)$ satisfies $\int_M f = 0$, then $$\|f\|_{L^2(M)} \leq \text{const Vol}(M)^{\frac1n}\|Df\|_{L^2(M)}.$$
It is not too clear what is $D$ in his notation, but judging by how it is used earlier in the text for $\mathbb{R}^n$ I assume he means a weak derivative. If so, I am confused on how the right-hand side of (2) should be interpreted. Should it be the norm of the Riemannian gradient (which depends on the metric $g$) or the weak derivative (which doesn't depend on the metric $g$)?
Best Answer
It might help me a bit more if I knew where you're getting the statements of the theorems from - is it from Jost?
(1)
Looking at wikipedia, you'll notice that your inequality (1) only holds the Sobolev space of traceless functions $W^{1,p}_0$, which means the functions "go to zero at the boundary of $\Omega$" (quotes because the functions are officially only defined up to measure 0, so you need a different way of saying they are zero on the boundary). A more precise statement is that $W^{1,p}_0$ is the closure of $C^\infty_0$ of smooth functions which actually are zero on the boundary.
On the other hand you can also put the average value back into the inequality and not worry about the zero trace condition. (This is on wikipedia too. I'm also assuming $\Omega$ has nice enough boundary in both cases.)
You can come up with some counter example sequences of functions which break both cases. E.g. $u_n(x) = u(x) + n$ makes the integral on the left side of (1) large, but doesn't change the gradient.
(2)
If he says $f\in H^{1,2}(M)$ that means the weak derivative $Df$ exists.
Don't both norms depend on $g$? I would imagine $|\nabla f|_g = |\mathrm d f|_g = |D f|_g$.