MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $D \subset \Bbb C$ with smooth boundary. Let $f \in C^{1,1}(D)$. Let $\varphi$ be the supremum of all members in the set

$$\lbrace g \in C^{\infty}(\overline{D})| g \ is \ subharmonic \ and \ g \leq f\rbrace$$

It is known that $\varphi$ is subharmonic in $D$ and harmonic on $E =\lbrace \varphi < f\rbrace$, the complement of the contact set. It is also known that $\varphi$ is C^{1,1}.

Suppose now that $\lbrace f_t \rbrace$, $t \in(0,1)$ is a smooth family of smooth obstacles. Will the variation of the family $\lbrace \varphi_t \rbrace$ be at least $C^1$?

If this is too much to ask, then let $\tilde E \subset (0,1)\times D$, be the open? domain for which each $t-$slice $E_t$ is the set where $\varphi_t$ will be harmonic. Will the variation of $\lbrace \varphi_t\rbrace$ be $C^1$ on $\tilde E$?

Any reference or chunk of information will be appreciated.

I don't know almost anything about obstacle problems (but you know this by now :D). There seems to be no obvious reference out there. Can someone recommend something to a grad student who is faimiliar with the Gilbarg and Trudinger stuff but not much more?

share|cite|improve this question
Interesting question! It seems to me that if we start with a convex obstacle (like $f = |x|^2$) and smoothly deform it to an obstacle with "2 humps" something would be irregular; the set where $\phi$ agrees with $f$ will start out as a sphere and pinch off into 2 spheres. – Connor Mooney Dec 22 '12 at 18:31
Indeed. Nice observation. My intention with the first question was to lead into the second one which seems more involved. I think in general only $C^{0,1}$ variation is possible. As the boundary of the contact set varies it is hard to believe that along this set full $C^1$ variation is possible(as your example also suggests). – The Common Crane Dec 22 '12 at 22:13
up vote 2 down vote accepted

I am not sure what you mean by $C^1$ variation, but it doesn't seem to be true. Here is an example.

Let $f_0(x) = |x|^4 - |x|^2$, thus $f$ has a local maximum at the origin, then it turns around and goes to infinity as $|x| \to \infty$. The corresponding $\varphi_0$ coincides with $f_0$ outside of a disk and it is constant inside the disk.

Let $f_t$ be $\min(f_0(x), 10|x|^4 - t)$, smoothing out the edge. That is $f_t$ is the same as $f_0$ outside of a small disk centered at the origin, but it has a smooth peak pointing down at the origin which is being lowered at speed one in $t$.

For $t$ small enough, $\varphi_t = \varphi_0$, since one can see directly that $\varphi_0 \leq f_t$. Let $T$ be the suppremum of these values of $t$. We will have $f_T(0) = \varphi_T(0) = \varphi_0(0)$. For $t>T$, the origin will be inside the contact set. We will get $\varphi_t(0) = f_t(0) = -t$ for $t>T$. So $\varphi_t(0)$ is not a $C^1$ function of $t$, since it has a corner at $t=T$.

As you say $C^{0,1}$ variation is possible, meaning that $\|\varphi_t - \varphi_0\|_{L^\infty(D)} \leq \|f_t - f_0\|_{L^\infty(D)}$.

You didn't mention anything about boundary conditions, so I assumed the implied boundary condition of your definition which is $\varphi_t = f_t$ on $\partial D$.

The typical references for the obtacle problem are either the lecture notes of Caffarelli in the Fermi lectures, or a paper called "The obstacle problem revisited" which can be found online. They contain almost the same material. I believe there is something more about the stability of the free boundary in particular in the PhD thesis of Ivan Blank.

Your statement of the obstacle problem is upside down compared with the usual setting.

share|cite|improve this answer
It may also be worth mentioning that the Wikipedia page on the obstacle problem is not bad. – Luis Silvestre Dec 24 '12 at 2:34
Thank you for your careful answer. – The Common Crane Jan 2 '13 at 22:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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