Show that for a sufficiently smooth boundary of $\omega$ and any $\epsilon > 0$, there exists $C$ such that

$$\int_{\partial \omega} u^2\ ds \leq C\int_{\omega} u^2\ dx + \epsilon \int_{\omega} |\bigtriangledown{u}|^2dx$$

My attempt: I choose a smooth vector field $\sigma$ on $\omega$ s.t $\sigma =$ outward unit normal on $\partial {\omega}$. Then $$\int_{\partial \omega} u^2\ ds\ =\ \int_{\partial \omega} u^2\sigma\cdot v\ ds = \int_{\omega}{\text{div}(\sigma u^2)\ ds} $$ (by divergence theorem) $$= \int_{\omega} (\text{div}(\sigma)\cdot u^2) + (\sigma)\cdot \text{div}(u^2) =(\int_{\omega} 0 + 2\sigma\cdot u\ \text{div}(u))\ ds$$ $$ \leq 2\int_{\omega} |v|\ |\text{div}(v)|dx \tag{1} $$

(since $|\sigma|=1$ as $\sigma$ is a unit normal vector, and $u\cdot \text{div}(u) = |u|\ |\text{div}(u)|\cos(u, \text{div}(u))\leq |u|\ |\text{div}(u)|$)

By Young's inequality, we have: $\forall\ \epsilon > 0$, exists $C_{\epsilon} = \frac{1}{\epsilon} > 0$ such that $$\ 2|u|\ |\text{div}(u)|\leq C_{\epsilon} v^2 + \epsilon |\text{div}(u)|^2\tag{2}$$

From $(1)$ and $(2)$, we obtain $\int_{\partial \omega} u^2\ ds \leq C_{\epsilon} \int_{\omega} u^2\ dx + \epsilon \int_{\omega} |\text{div} (u)|^2\ dx$ for any $\epsilon > 0$ and some $C_{\epsilon}$ (Q.E.D)

Question: Can anyone please verify if my proof above is correct? Any help would greatly be appreciated in case it was incorrect.


The general structure is correct, but some steps are not.

  1. Divergence is not normally applied to scalar functions; they have a gradient instead: $$\operatorname{div}(\sigma u^2) = (\operatorname{div}\sigma)u^2 + \sigma\cdot \nabla(u^2)$$

  2. The property $|\sigma|=1$ is known to hold only on the boundary. You don't really know what $\sigma$ is inside, other than it's smooth on closed domain and therefore is bounded. I guess you could reason that $\sigma$ can be chosen so that $|\sigma|\le 1$, but why bother, if we can simply estimate
    $$ |\sigma\cdot \nabla(u^2)| \le 2M|u||\nabla u|,\quad M=\sup_\omega|\sigma| $$ This $M$ is easily dealt with later; just use $\epsilon M^{-1}$ in Young's inequality.

  3. I don't think that $\int_\omega (\operatorname{div}\sigma)u^2 =0 $ is true in general. Why would this expression, with a general function $u^2$ in it, magically integrate to zero? Instead, estimate it by $M'\int_\omega u^2$ where $M'=\sup_\omega |\operatorname{div}\sigma|$, and absorb this $M'$ in the constant $C$.

  • $\begingroup$ Thank you so much for your help!! It's very helpful to learn where my mistakes are. For 3, the reason was because I thought $\text{div} \sigma = 0$, but I think that was incorrect. However, is the existence of $M'$ because $\sigma$ is smooth on $\omega$? $\endgroup$ – user177196 Sep 7 '15 at 5:32
  • 1
    $\begingroup$ A continuous function on a compact set is bounded. $\endgroup$ – user147263 Sep 7 '15 at 5:49
  • $\begingroup$ I'm so dumb:( Thank you for your patience. $\endgroup$ – user177196 Sep 7 '15 at 6:03

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