How can I prove this special version of the Poincaré inequality?

Question

I am reading a paper, Is $L^2$ Physics-Informed Loss Always Suitable for Training Physics-Informed Neural Network?, which uses an "application" of Poincaré's inequality. The form I know and can find everywhere online is $$\lVert u\rVert_p ≤ C\lVert Du\rVert_p$$ where $\lVert.\rVert_p$ is the $L^p$ norm on some compact domain, but the version they use is $$\lVert u\rVert_{2,p} ≤ C\lVert D^2u\rVert_p$$ where $\lVert.\rVert_{2,p}$ is the $W^{2,p}$ Sobolev norm on a compact domain and $D^2u$ is the vector $D^2u_i=\frac{\partial^2u}{\partial x_i^2}$, and I can't figure out how they arrive at that. The assumption that $u$ dissapears on the boundary is still there. I've searched various literature like Jost's PDEs book, Evans' PDE book, and Adams' Sobolev Spaces, and can't find the answer anywhere.

Answer 1

The following is a very non-generalizable result specific to the second order Poincare inequality.

Thm: Let $\Omega$ be open, bounded, and connected. Further assume it has Lipschitz boundary. Then for every $p\in [1,\infty)$ there exists a constant $C$ such that for any $u\in C^0(\bar{\Omega}) \cap C^2(\Omega)$ with $u|_{\partial\Omega = 0}$, $$ \|u\|_{L^p} + \|D u\|_{L^p} \leq C \|D^2 u \|_{L^p}$$ (Here $Du$ is the gradient of the function $u$, and $D^2 u$ the Hessian matrix.)

Proof: Classical Poincare inequality provides (per the boundary value assumption) that $\|u\|_{L^p} \lesssim \|Du\|_{L^p}$. Next observe that $\|Du\|_{L^p} \lesssim \sum_{i = 1}^n \|\partial_i u\|_{L^p}$, and the boundary value assumption on $u$ implies $\int_{\Omega} \partial_i u = 0$ for any $i$. And therefore we may apply the Poincare-Wirtinger inequality (the version subtracting the mean) to conclude that $\|\partial_i u\|_{L^p} \lesssim \|D (\partial_i u)\|_{L^p}$. Sum over $i$ and we have the desired result.

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Note that I am interpreting $D^2$ differently then how you stated it in the question. In fact, in the paper you cited, the final sentence of section A.2 supports my interpretation (and not yours).

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By density this also extends to $u\in W^{1,p}_0(\Omega) \cap W^{2,p}(\Omega)$.

Answer 2

It is noted in Willie Wong's answer that the definition of the operator $D^2$ in the OP is different from the one in the paper cited by the OP. Namely, in the paper $D^2$ involves both the mixed and unmixed partial derivatives, whereas in the OP $D^2$ involves only the unmixed partial derivatives.

However, the original question just as stated in the OP still makes sense, and answering that original question involves the additional task of bounding the $p$-norms of the mixed partial derivatives by the sum of the $p$-norms of the unmixed partial derivatives.

This was done in the previous version of this answer only for $p=2$, but now such a bound is available for all $p\in(1,\infty)$.

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To simplify the writing, suppose that the dimension is $2$. Then, without loss of generality, $u$ is supported on $S:=[0,1]^2$. Let $D_j$ denote the operator of the partial differentiation wrt the $j$th argument. Then, for real $p\ge1$ and $(x,y)\in S$, by Jensen's inequality, $$|u(x,y)|^p\le\Big(\int_0^x (x-s)\,|(D_1^2u)(s,y)|\,ds\Big)^p \le\int_0^1|(D_1^2u)(s,y)|^p\,ds,$$ $$|D_1u(x,y)|^p\le\Big(\int_0^x|(D_1^2u)(s,y)|\,ds\Big)^p \le\int_0^1|(D_1^2u)(s,y)|^p\,ds,$$ so that $$\|u\|_p\le\|D_1^2u\|_p,$$ $$\|D_1u\|_p\le\|D_1^2u\|_p,$$ and similarly $$\|D_2u\|_p\le\|D_2^2u\|_p.$$

It remains to note that, thanks to this recent answer,
$$\|D_1D_2u\|_p\le C_p(\|D_1^2u\|_p+\|D_2^2u\|_p)$$ for each $p\in(1,\infty)$ and some real constant $C_p$ depending only on $p$.