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.

  • $\begingroup$ I see you have created (regularity). When you create a tag, it is good to create also tag-wiki or at least tag-excerpt to indicate what you intend the tag to be used for. In this particular case, regularity has several meanings in mathematics. If you wish to discuss this further, feel free to drop a line in chat. $\endgroup$ – Martin Sleziak Aug 20 '14 at 6:15
  • $\begingroup$ @MartinSleziak I created a tag excerpt as you suggested. However, you said regularity has several meanings in math. My description of regularity focused on PDEs, so I may be biased. Please add your edits if you can or need. $\endgroup$ – Cookie Aug 20 '14 at 12:55
  • $\begingroup$ Thanks for creating tag-excerpt. I have posted a question on meta about this tag. Hopefully, some people knowledgeable about PDEs will tell us their opinion on utility of this tag. $\endgroup$ – Martin Sleziak Aug 20 '14 at 13:18

$|A_2|$ 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.)


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