This is from PDE Evans, 2nd edition, pages 327, 328, and 330. I have a question regarding one piece of the proof.
The theorem concerned is THEOREM 1 (Interior $H^2$ regularity) which is stated on page 327.
On page 328, the proof begins by saying
Fix any open set $V \subset \subset U$, and choose an open set $W$ such that $V \subset \subset W \subset \subset U$. Then select a smooth function $\zeta$ satisfying \begin{cases}\zeta \equiv 1 \text{ on }V, \zeta \equiv 0 \text{ on }\mathbb{R}^n – W \\ 0 \le \zeta \le 1. \end{cases}
(Note that $U \subset \mathbb{R}^n$ is a bounded, open set.)
Now my question on page 329 concerns applying Cauchy's inequality with $\epsilon$. The inequality states that
$$ab \le \frac{a^2}2+\frac{b^2}2 \qquad(a,b \in \mathbb{R}).$$
If we have (from the textbook)
$$|A_2| \le C \int_U \zeta |D_k^h Du| |D_k^h u| + \zeta |D_k^h Du| |Du| + \zeta |D_k^h u| |Du| \, dx$$for some appropriate constant $C$
how is Cauchy's inequality with $\epsilon$ used to obtain the bound of
$$|A_2| \le \epsilon \int_U \zeta^2 |D_k^h Du|^2 \, dx + \frac{C}{\epsilon} \int_W |D_k^h u|^2 + |Du|^2 \, dx$$
I am confused particularly because Cauchy's inequality deals with a product, like $ab$. However, $|A_2|$ is not a product; rather it is a sum of three product terms, all of which are integrated with respect to $x$.
Please also see my next question.
Best Answer
$|A_2|$ is not a product, but it can be estimated by one. Note that $$ \zeta |D_k^h Du| |D_k^h u| + \zeta |D_k^h Du| |Du| + \zeta |D_k^h u| |Du| \leq (\zeta|D_k^h Du|+\zeta|D_k^h u|)(|D_k^h u|+|Du|). $$ Then one can apply Cauchy's theorem (or its $\epsilon$-version $ab=(a\sqrt\epsilon)(b/\sqrt\epsilon)\leq\frac\epsilon2a^2+\frac1{2\epsilon}b^2$) and $(a+b)^2\leq C(a^2+b^2)$ to get $$ |A_2| \leq \epsilon \int_W \zeta^2 (|D_k^h Du|^2 + |D_k^h u|^2) \, dx + \frac{C}{\epsilon} \int_W (|D_k^h u|^2 + |Du|^2) \, dx. $$ Now you can just absorb the second term in the first integral to the first term of the second integral. (Since $\zeta$ is supported in $W$, you only really need to integrate over $W$. Expanding $W$ to $U$ after all is done is a trivial step.)