Let $\Omega \subset \mathbb{R}^d$ be open and bounded. Consider the Laplace-Operator given by $$ \Delta u=\sum_{j=1}^d \partial_j^2 u \quad \text { for all } u \in W^{2,2}(\Omega) . $$ (i) Show that for every $u \in C_c^{\infty}(\Omega)$ $$ \int_{\Omega}|\Delta u(x)|^2 d x=\sum_{i, j=1}^d \int_{\Omega}\left|\partial_i \partial_j u(x)\right|^2 d x $$
Answer:
Consider a function $u$ in $C_c^{\infty}(\Omega)$. The Laplace operator $\Delta u$ is defined as the sum of the second partial derivatives of $u$ with respect to each variable, i.e., $\Delta u=\sum_{j=1}^d \partial_j^2 u$
We start by examining the left-hand side of the integral identity, $\int_{\Omega}|\Delta u(x)|^2 d x$. Substituting in the definition of the Laplace operator, this is $\int_{\Omega}\left|\sum_{j=1}^d \partial_j^2 u(x)\right|^2 d x$.
Expanding the square of the sum leads to $\int_{\Omega} \sum_{i=1}^d \sum_{j=1}^d \partial_i^2 u(x) \partial_j^2 u(x) d x$. This includes both the square terms $\left(\partial_i^2 u(x)\right)^2$ and the cross-terms $\partial_i^2 u(x) \partial_j^2 u(x)$ for $i \neq j$.
Applying Fubini's Theorem, we can interchange the order of summation and integration: $\sum_{i=1}^d \sum_{j=1}^d \int_{\Omega} \partial_i^2 u(x) \partial_j^2 u(x) d x$.
Integration by parts is applied to each term in the double sum. For the square terms $i=j$, this leads directly to $\int_{\Omega}\left|\partial_i^2 u(x)\right|^2 d x$ as boundary terms vanish due to $u$ 's compact support. For the cross-terms $i \neq j$, additional justification is needed to handle these terms, such as referencing a theorem or providing a detailed argument.
Summing these integrals, including both square and cross-terms, results in $\sum_{i, j=1}^d \int_{\Omega}\left|\partial_i \partial_j u(x)\right|^2 d x$. This sum represents the right-hand side of the integral identity, completing the proof.
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