# Is this functional convex?

Maybe this is a strange or un-professional question. I want to know whether my equation is convex.

My equation is as follows: $E\left(\phi\right)=\int_{\Omega}\left|\left(g\left(\overline{x}\right)-A\right)\right|^{2}\left(1-H(\phi\left(\overline{x}\right))\right)d\overline{x} +\mu\int_{\Omega}H(\phi\left(\overline{x})\right)d\overline{x} +\nu\int_{\Omega}\delta\left(\phi\left(\overline{x}\right)\right)\left|\nabla\phi\left(\overline{x}\right)\right|d\overline{x}$

where $g$ is a known function $R^{2}\rightarrow R$, and A is a know scalar, $H$ is a Heaviside function,e.g. $H\left(\phi\right)=\frac{1}{2}+\frac{1}{\pi}\arctan\left(\phi\right)$ and $\delta$ is a Delta function, I want to find the $\phi,R^{2}\rightarrow R$ which minimize $E$. So I want to know whether $E$ is convex. So if I find its extremum, then I find its extrernes.

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I am not sure if that will be called Convex functional Analysis. – user21436 Feb 5 '12 at 8:53
At this level of generality, all I can suggest is that you use the definition of "convex". In most cases I've encountered, it would be fairly easy to decide. We might be able to help if you gave us some specific details about the particular problem. – Robert Israel Feb 5 '12 at 9:26
Right now, this question is on its 14th edit, and looks completely different from its original form. Please open a new question if you have something new to ask, instead of changing the question entirely, as it makes the existing comments and answers seem irrelevant. See this meta thread for more info. – Rahul Feb 10 '12 at 5:00
Thank you so much for your advise. I will turn the original question back. – user18481 Feb 10 '12 at 6:24

First question you should ask yourself is whether your functional is even continuous over the function space you are interested in. For example, using the reduction by Robert Israel above, over any $L^p$ space your functional is not even continuous. Which means that you are not guaranteed to be able to approach its infimum using a minimizing sequence. But maybe your functional is upper semi continuous (this would in part depend on you convention for the Heaviside function)? At the very least you know that your functional is bounded below by 0, so you can't have the minimum running off to infinity.

The fact that your functional is not continuous also means that you can not use the calculus of variations as you described in the second half of your question. To do so it is necessary that your functional is once continuously differentiable (if you just want to use the Euler-Lagrange equations to find a critical point) and twice continuously differentiable (if you want to also use the second variation to check whether the critical point is a max, min, or saddle point).

One possibility could be potentially to construct the actual convex hull of your functional. Whether that is approachable depends strongly on how complicated your initial functional is. If your original functional has only one minimum point (this is a uniqueness statement that requires proving), then the convex hull will be a continuous functional with the same minimum point, which can be slightly easier to analyse.

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There are hundreds of books on the subject, a small list is available at this Wikipedia link. For general Convex Optimisation a standard reference seems to be the book(s) by Hiriart-Urruty and Lemaréchal. If you are interested in the convex hull argument that I mentioned in the final paragraph, you may want to look at Part III of Dacorogna's Direct methods in Calculus of Variations – Willie Wong Feb 8 '12 at 14:01
Thank you so much. – user18481 Feb 8 '12 at 21:57

Consider the case $g(x)=A$, $\nu = 0$, $\mu=1$, so your functional reduces to $E(\phi) = \int_{\Omega} H(\phi(x))\ dx$. I assume $\Omega$ has finite measure. For constant functions $\phi$, $E(\phi)$ is a step function: $0$ if $\phi < 0$ and $m(\Omega)$ if $\phi > 0$. That is not convex, so your functional is not convex.

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After your simplification, this functional is a non convex. What theory could I use to analyse a functional? I think this functional is a little complex, too many variable. And actually, $g\left(x\right)$ and $A$ is uncertain, $g\left(x\right)$ means a image. The image can change to another image. Are there any condition which if I meet then the functional is convex? – user18481 Feb 8 '12 at 8:45