# Obtaining orthogonality from a variational inequality in $L^2$

I'm working on the following problem:

Let $$\Omega$$ be a bounded domain in $$\mathbb{R}^n$$ and let $$H$$ be a closed linear subspace of $$L^2(\Omega)$$. Let $$\gamma : \mathbb R \to \mathbb R$$ be a continuous increasing function satisfying $$|\gamma(t)| \leq A + B|t|$$ for some constants $$A,B$$. Let $$F \in L^2(\Omega)$$.

Show that there exists a unique $$u \in H$$ such that $$u + \gamma \circ u -F \in H^{\perp}$$.

(Hint: solve the variational inequality $$(u + \gamma \circ u -F,v-u) \geq 0 \quad \forall v \in H$$.)

Question: if $$u \in H$$ is a solution of the inequality, why does this imply $$u + \gamma \circ u -F \in H^{\perp}$$?

We assume that $$u \in H$$ is such that $$(u + \gamma \circ u -F,v-u) \geq 0$$ for every $$v \in H$$. In particular, this means that for $$w \in H$$ we can take $$v = u \pm w$$ to obtain $$(u + \gamma \circ u -F,\pm w) \geq 0$$ which in turn implies that $$(u + \gamma \circ u -F, w) = 0$$. Since $$w$$ was arbitrary, this shows that $$u + \gamma \circ u -F \in H^\perp$$.