How is the inner product in $H^{-1/2}$ defined? Since $H^{1/2}$ is a Hilbert space, $H^{-1/2}$ must also be a Hilbert space by the isomorphism of Riesz representation theorem. How is the inner product defined there?
We know there is a nice Cauchy-Schwarz $\| f\cdot g\|_{L^2(\Gamma )}\leq \| f\|_{L^2(\Gamma )}\| g\|_{L^2(\Gamma )}$. Is there a counterpart of this in $H^{-1/2}$? If $f,g\in L^2(\Gamma )$, is it true that $\| f\cdot g\|_{H^{-1/2}(\Gamma )}\leq \| f\|_{H^{-1/2}(\Gamma )}\| g\|_{H^{-1/2}(\Gamma )}$?
Or is $\| f\cdot g\|_{H^{-1/2}(\Gamma )}\leq \| f^2\|_{H^{-1/2}(\Gamma )}\| g^2\|_{H^{-1/2}(\Gamma )}$ the best we can do?
 A: If I understood you first question well, there is some ways to define inner product in $H^{-1/2}$. For example by using Riesz theorem, if we indetify $F,G\in H^{-1/2}$ with $f,g\in H^{1/2}$, then we can define the inner product $(\cdot,\cdot)$ in $H^{-1/2}$ by $$(F,G)=((f,g))$$
where $((\cdot,\cdot))$ is the inner product in $H^{1/2}$.
Another way is: define in $H^{-1/2}$ the norm $$\|F\|_{H^{-1/2}}=\sup_{f\in H^{1/2},\ \|f\|_{H^{1/2}}=1}\langle F,f\rangle$$
You can verify that $\|\cdot\|_{H^{-1/2}}$ satisfies the paralelogram law and hence it is possible to define a inner product with it (do you know how to do it?). Note that, if we assume Riesz theorem here, we have the first definition.
For your second question, first note that your statement is wrong, as pointed out by @40votes. The right statement is: if $(\cdot,\cdot)$ denotes the inner product in $H^{-1/2}$ then for all $f,g\in L^2$ we have that $$|(f,g)|\leq \|f\|_{H^{-1/2}}\|g\|_{H^{-1/2}}$$ 
This is true because every inner product satisfies Cauchy-Schwarz inequality and also $L^2\subset H^{-1/2}$.
