Consider the Sobolev space $W^{k,p}(\Omega)$ for $k\in \mathbb N$, $p\in [1,\infty]$ and some open domain $\Omega\subset \mathbb R^n$ $^\*$. Then it is known that $W^{k,p}(\Omega)$ is an ordered Banach space, and indeed a *lattice*-ordered Banach space if $k=1$, but not a Banach lattice because the norm is not monotone on the positive cone $W^{k,p}(\Omega)\_+$. Furthermore, it is known that in the reflexive range $W^{k,p}(\Omega)$ is isomorphic, but in general **not** *lattice* isomorphic to an $L^p$ space. It is also clear that the positive cone of $W^{k,p}(\Omega)$ is a closed subset, hence at least in the reflexive range each element of $W^{k,p}(\Omega)$ can be projected onto $W^{k,p}(\Omega)\_+$.


That said: 



> 
> Does anybody have an idea of how said orthogonal projection looks like?
> 
> 
> 


$^\*$ I would be happy already with an answer in the case of $n=1$, $k=1$, $p=2$, $\Omega$ bounded.



projection of sobolev spaces onto cones
I will consider the case $k = 1$ and $p = 2$ (some arguments may generalize to $k \in \mathbb{N}$).
Let us use the norm $\|u\|^2 = \|u\|^2 + \|\nabla u\|^2$ in $H^1(\Omega)$ (both are $L^2$-norms).
The associated scalar product is denoted by $(\cdot,\cdot)$.
We denote by $K = \{v \in H^1(\Omega) : v \ge 0\}$ the positive cone in $H^1(\Omega)$.
Then, the projection of $u \in H^1(\Omega)$ onto $K$ is given by the minimizer $v$ of $\frac12 \, \| u - v \|^2$ over $K$.


Now, one can show, that there is a Lagrange multiplier $\lambda \in (H^1(\Omega))'$.
This multiplier lies in the polar cone $K^\circ$, that is
$$
 \langle \lambda, w \rangle \le 0 \quad\mbox{for all } w \in K
$$
and is orthogonal to $v$, i.e., $\langle \lambda , v \rangle = 0$.
Moreover, we have the equation
$$
(v - y, w) + \langle \lambda, w \rangle = 0 \quad\mbox{for all } w \in H^1(\Omega),
$$
which couples all involved quantities. This equation is the weak formulation of a PDE.


Note that these optimality conditions are nothing more than an equivalent reformulation of the projection inequality
$$
 (v - y, w - v) \ge 0 \quad\mbox{for all } w \in K.
$$


The inclusion $\lambda \in K^\circ$ can be interpreted as non-positivity of $\lambda$.
Moreover, it should be possible to show (at least in the case $n = 1$, you are interested in), that $\lambda$ is represented by some negative measure (along the lines that negative distributions are essentially measures).


Then, the complementarity $\langle \lambda , v \rangle = 0$ means, that $\lambda$ is only strictly negative, where $v = 0$ (i.e. where the projection is active).


As you can see,
the calculation of the projection requires the solution of a nonlinear (even non-differentiable) PDE.
This PDE is just the optimality condition (i.e., the projection inequality) of minimizing the distance in the Dirichlet energy.
Hence, there is no easy formula.


Finally, let me comment on the case $u \in H^2(\Omega)$.
That is, we are projecting a more regular element (but still w.r.t. the $H^1$-norm).
Then one can show (see, e.g., the book by Kinderlehrer and Stampacchia - I can provide an exact reference, if needed),
that (under some regularity of $\Omega$)
again $v \in H^2(\Omega)$ and $\lambda \in L^2(\Omega)$.
Due to this regularity, on can interpret the above relations pointwise and obtain
$\lambda(x) \ge 0$ and $v (x) \, \lambda(x) = 0$ for almost every $x \in \Omega$.
Finally, the PDE can be written as
$$
\max\Big( {-\Delta(u-v)} + (u-v), \; v \Big) = 0 \quad\mbox{almost everywhere in }\Omega.
$$
Again, one can see, that there is no easy (explicit) formula for the projection.


I found this 'high regularity case' interesting, because its similar to the case of projecting a $H^1$-function w.r.t. the $L^2$-norm, where the projection is again $H^1$.


