Consider a manifold $M \subset \mathbb{R}^2$ with boundary $\partial M$. Then, consider a zero-form field, $\phi^{(0)}$, defined on the whole of $M$ (including the boundary). Then, the following holds: \begin{equation} (d\phi^{(0)},d\phi^{(0)})_M - (\phi^{(0)},d^\star d\phi^{(0)})_M = (\phi^{(0)},d\phi^{(0)})_{\partial M} = \int_{\partial M} \phi^{(0)} \wedge \star d\phi^{(0)}. \end{equation}
Then, can I use Cauchy-Schwarz inequality on the boundary term to write the following? \begin{equation} |(\phi^{(0)},d\phi^{(0)})_{\partial M}| \leq (\phi^{(0)},\phi^{(0)})_{\partial M}^{1/2}\,\,\,\,\,(d\phi^{(0)},d\phi^{(0)})_{\partial M}^{1/2}. \end{equation}
If yes, how would I go about evaluating these norms on the right hand side? For instance, considering the first norm, \begin{equation} ||\phi^{(0)}||_{\partial M}^2 = (\phi^{(0)},\phi^{(0)})_{\partial M} = \int_{\partial M} \phi^{(0)} \wedge \star \phi^{(0)}, \end{equation} but since $\star \phi^{(0)} = \phi^{(0)}dx \wedge dy$, am I integrating a 2-form on a curve? Or does $\star \phi^{(0)}$ need to be evaluated differently? If so, could some one please help me write down an explicit formula assuming $dx$ and $dy$ as the co-vector basis?
Any help would be greatly appreciated! (This is NOT homework.)
Side note: I have a 1D interval, $I := [a,b]$, and a map $F$ such that \begin{equation} F: I \rightarrow \partial M. \end{equation} Then I want to evaluate the following. \begin{equation} ||\phi^{(0)}||_{\partial M}^2 = \int_{F(I)} \phi^{(0)} \wedge \star \phi^{(0)} = \int_{I} F^*(\phi^{(0)}) \wedge F^*\star (\phi^{(0)}). \end{equation}
EDIT: Thanks, Ted. I'm sorry, I forgot to mention. The manifold is a smooth region in $\mathbb{R}^2$.
Figured it out. What I wrote above is incorrect. $\int_{\partial M} \phi^{(0)} \wedge \star d\phi^{(0)} \neq (\phi^{(0)},d\phi^{(0)})_{\partial M}.$ It doesn't even make sense to write it like that because it implies an inner product between a 0 and a 1-form. Instead, what it should be is $(\phi^{(0)},(d\phi^{(0)})_n)_{\partial M}$ where $(d\phi^{(0)})_n$ is the normal component of the differential form (which is a 0-form on $\partial M$) and then the inner product makes sense.