# Existance of Solution in Neumann Variational Inequality

I was working on the following question and got stuck (Chap 2, #10 Kinderleher and Stampaccia)

Let $$\Omega$$ be a bounded domain with a smooth boundary $$\partial \Omega$$ and suppose given $$f \in H^{-1}(\Omega)$$ and $$g \in L^2(\Omega)$$. For $$\lambda \in \mathbb{R}$$ study the existence and uniqueness of solutions of the variational inequality:

$$u \in K : \int_\Omega[u_{x_i}(v-u)_{x_i} + \lambda(v-u)]\,dx \geq \langle f,v-u\rangle + \int_{\partial \Omega} g(v-u)\,dx \quad \forall v \in K$$

Where $$K = \{v \in H^1(\Omega):v \geq 0\text{ on }\partial \Omega\}$$. (Here I've used summation convention and partial derivatives are to be understood in the weak sense)

I think I have a lot of the pieces:

1. The LHS, written as $$a(u,v) = \int_\Omega[u_{x_i}(v)_{x_i} + \lambda(vu)]\,dx$$ is coercive on $$H^1(\Omega)$$ when $$\lambda >0$$
2. K is convex in $$H^1$$

So I think there will exist a weak solution to the Neumann Problem: $$-\Delta u + \lambda u = f \\ \frac{\partial u}{\partial n}=g$$ where $$g \geq 0$$.

I'm having trouble dealing with $$f \in H^{-1}$$ though. This $$H^{-1}$$ is dual to $$H^1_0$$, so the dual pairing on the RHS of the inequality only works when $$v-u \in H^1_0$$ But taking two elements arbitrary $$v,u \in K$$ doesn't necessarily give that $$u-v$$ is going to zero at boundary. This seems inconsistent and I'm having trouble resolving it.