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I am reading Functional Analysis, Sobolev Spaces and Partial Differential Equations, by Haim brezis, and I am a bit confused about the proof. The theorem is stated as:

Let $\Omega \subset \mathbb{R}^n$ be an open set of class $C^2$ with $\partial \Omega$ bounded. Let $f \in L^2(\Omega)$ and $u \in H_0^1(\Omega)$ satisfies the weak formulation (Laplacian): $$\int_\Omega \nabla u \nabla \varphi + \int_{\Omega}u \varphi=\int_{\Omega}f\varphi \quad \forall \varphi \in H_0^1(\Omega)$$ Then $u \in H^2(\Omega)$ and $||u||_{H^2} \leq C ||f||_{L^2}$.

I don't understand the part where Brezis uses tangential translations. First, he says: Brezis301

I understand this well, we can "translate" the domain $\mathbb{R}^n_+= \{(x_1,...,x_n) \ | \ x_n>0\}$ and it is clear that $\mathbb{R}^n_+ + h = \mathbb{R}^n_+$ for $h \in \mathbb{R}^{n-1} \times \{0\}$. Then he uses a trick to show the result for all the derivatives except for $\frac{\partial \varphi}{\partial x_n x_n}$. This is also clear since $h \in \mathbb{R}^{n-1} \times \{0\}$.

enter image description here

With the domain $Q_+=\{(x_1,...,x_n) \ | \ (\sum_{i=1}^{n-1}x_i^2)^{1/2} <1 \ and \ 0 < x_n < 1 \}$ and uses $ h \in Q_0= \{(x_1,...,x_n) \ | \ x_n=0 \ and \ \sum_{i=1}^{n-1}x_i^2)^{1/2} <1\}$ and argues for $h$ "small enough".

Then he also splits the case $\frac{\partial \varphi}{\partial x_n x_n}$. Why? If you can choose $h$ "small" so that for $w \in H_0^1(Q_+)$ $w(x+h) \in H_0^1(Q_+)$, Why do you need to pick it in $Q_0$ and then split for $\frac{\partial \varphi}{\partial x_n x_n}$? Why not choose $h$ in the boundary of $Q_+$ (Wich is a bounded "cilinder") and then argue it is "small enough" ...?

Note: This is what follows:

enter image description here

As you can see, it only shows the inequality for $(k,l) \neq (n,n)$:

enter image description here

enter image description here

Update: I paste here a drawing of the situation that helped me understand the proof, thanks to Sandwich:

enter image description here

Brown is the covering of the boundary of $\Omega$.

Red is the open set we are dealing with.

Green is the support of $\theta_i $, $\Rightarrow$ the support of $\theta_i u$.

Grey is $\Omega \cap U_i$.

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  • $\begingroup$ It's not easy to answer without a book, since the part you quoted is not the part you have a question about. With Google Books, I couldn't even find on what page $Q_+$ appears. $\endgroup$ – user147263 May 23 '16 at 2:22
  • $\begingroup$ @sandwich You are right, I have added the next part. The set $Q_+$ is specified bellow. In the case $\mathbb{R}^n_+$ it is clear why he picks $h \in \mathbb{R}^{n-1} \times \{0\}$, as the domain does not change. But if you can pick $h$ "small enough" so that $w \in H_0^1(Q_+)$, which is bounded, Why do you need to do it for $h \in Q_0$ (i.e. lying on the plane $\{x_n=0\}$)? $\endgroup$ – D1X May 23 '16 at 10:46
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The footnote 31 on page 305 explains the logic of the choice of $h$: the support of $w$ is at some distance from the top and side surface of the cylinder $ Q_+$, but it need not be separated from the bottom surface, which is $x_N=0$. This is how the chosen partition of unity works: in order to cover the domain by finitely many sets, we need these sets to touch the boundary. I sketched the situation below, where the blue square represents $Q_+$, the green square inside of it is the support of $w$, and the other similar squares are supports of other parts of the partition of unity.

partition

So, translating $w$ left and right keeps it within the blue square, but vertical translation is constrained. Since the argument involves both $D_h$ and $D_{-h}$, it only works for $h$ parallel to the boundary.

The author eventually gets around this by using the PDE, which relates the second $x_N$ derivative to the other second partials.

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  • $\begingroup$ Why does the support of $w$ "touch" $x_n=0$? $\endgroup$ – D1X May 23 '16 at 18:45
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    $\begingroup$ Why shouldn't it? The support of $u$ does. (You know that $H_0^1$ doesn't mean compact support, right?) $\endgroup$ – user147263 May 23 '16 at 18:46
  • $\begingroup$ Wow. I think only a book can answer this question. $\endgroup$ – ibnAbu May 23 '16 at 19:16
  • $\begingroup$ @sandwich Okay, now I understand it. Partition of the Unity $\{\theta_1,..., \theta_k\}$ gives us open sets $U_1,...,U_k$ covering the Boundary (The case we are studying now). $\{\theta_1,..., \theta_k\}$ are compactly supported in $U_i$ (So is $\theta_i u$), but the support contains "a portion" of the boundary of $\Omega$. As the boundary is $C^2$, we use $H$ to transfer $\Omega \cap U_i$ to $Q_+$ and $U_i \cap \partial \Omega$ is mapped to $Q_0$, and the support of $\theta_i u$ can touch $\partial \Omega \cap U_i$, but not the other parts of the boundary of $\Omega \cap U_i$ . $\endgroup$ – D1X May 24 '16 at 12:17
  • $\begingroup$ @sandwich I have added another drawing of $\Omega$ sumarizing the situation of the support of $\theta_i u$ in $\Omega$, explaining my previous comment. Could you check it? Thank you very much. $\endgroup$ – D1X May 24 '16 at 12:21

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