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.

  • $\begingroup$ It's not totally clear what you mean by projection. Of course, you can take the element in the positive cone with the smallest distance (which is unique for $p \in (1,\infty)$). But I don't think that this is your question. In the case $k = 1$, $p = 2$, this projection is given by the obstacle problem $\endgroup$
    – gerw
    Mar 25, 2013 at 13:43
  • $\begingroup$ gerw, thanks for your comment. sure, i can take the vector with smallest distance. the problem is that i'd like to have an explicit, analytic expression of how this element of minimal distance look like. let me be more explicit: if we were in $L^2$, and not in $H^1$, then for all $f\in L^2$ its orthogonal projection onto the cone of positive-valued $L^2$-functions would be simply $f^+$, the positive part of $f$: that is, the pointwise maximum between $f(x)$ and $0$ (a.e.). but i have no idea how an analogous formula could look like, if one considers the $W^{1,2}$-distance instead. $\endgroup$ Mar 25, 2013 at 20:20
  • $\begingroup$ btw, i know too little of calculus of variations to appreciate your reference to the obstacle problem. i guess that what you call the "obstacle problem" (or a solution thereof) is simply the abstract formula which in functional analysis is used to characterize the orthogonal projection. but of course this characterization is in general purely abstract, that is, you come nowhere close to an explicit expression. $\endgroup$ Mar 25, 2013 at 20:23
  • $\begingroup$ Now I know what you are interested in. I will put it in a separate answer. $\endgroup$
    – gerw
    Mar 26, 2013 at 7:17

1 Answer 1


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 $H^1$.

  • $\begingroup$ thanks. i do not yet see why you write "hence, there is no easy formula" - several special solutions of complicated nonlinear pdes are known, and of course that minimum is the inf of an lsc functional (perhaps even smoother), but i do agree that "morally" this will be hard. $\endgroup$ Mar 26, 2013 at 9:16
  • $\begingroup$ More or less, this is meant in the sense of "morally" hard. However, many people work on / with the obstacle problem (also in terms of numerical approximation). Therefore, I think there is no explicit formula for arbitrary $u$. $\endgroup$
    – gerw
    Mar 26, 2013 at 9:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.