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This question is a direct continuation of my previous question. However, this one is requesting only for a relatively simple explanation.

Note: $W \subset U \subset \mathbb{R}^n$.

On page 330 of PDE Evans, 2nd edition, it says:

Now Theorem 3(i) in §5.82 implies \begin{align} \int_u |v|^2 \, dx &\le C \int_U |D(\zeta^2 D_k^h u)|^2 \, dx \\ &\le C \int_W |D_k^h U|^2 + \zeta^2 |D_k^h Du|^2 \, dx \\ &\le C \int_U |Du|^2 + \zeta^2 |D_k^h Du|^2 \, dx \end{align}

For the second inequality, I have a feeling they used the triangle inequality with exponent $p$ of $(a+b)^p \le 2^p(a^p+b^p)$, for all $a,b \in \mathbb{R}$ and $1 \le p < \infty$. However, how can we suddenly change from integrating over the region $U$ to over the subregion $W$, upon applying the triangle inequality with exponent $p$?

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From looking at your previous question it seems like $\zeta$ is nonnegative and supported in W, so that's why you can do it, it doesn't change anything.

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Since $\zeta$ is supported in $W$ (assuming the same notation as in your other question), also $D(\zeta^2 D_k^h u)$ is supported in $W$.

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