Estimate the integral of a function on the boundary using the integral of its gradient

Question

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.

Answer 1

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 $C$.