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The functional $E$ is defined on some real valued functions $\phi$ defined on $\Omega\subseteq\mathbb R^2$ ($R^{2}\rightarrow R$)and $E(\phi)$ is a real number. $\phi$ is differentiable. My equation is as follows: $E\left(\phi\right)=\int_{\Omega}\left|\left(g\left(\vec{x}\right)-A\right)\right|^{2}\left(1-H(\phi\left(\vec{x}\right))\right)d\vec{x} +\mu\int_{\Omega}H(\phi\left(\vec{x})\right)d\vec{x} +\nu\int_{\Omega}\delta\left(\phi\left(\vec{x}\right)\right)\left|\nabla\phi\left(\vec{x}\right)\right|d\vec{x}$

where $g$ is a known function $R^{2}\rightarrow R$, and A is a know scalar, $H$ is a Heaviside function,i.e. $H\left(\phi\right)=\frac{1}{2}+\frac{1}{\pi}\arctan\left(\phi\right)$ (I choose this kind of approximation to smear out $H$) and $\delta$ is a Delta function(I also smear out $\delta=\frac{\partial{H}}{\partial{\phi}}$), I want to know whether $E$ is continuous. I specify the norm is $\|\|_2$. Since $E(\phi)\in R$, $\|E(\phi)-E(\phi_{0})\|_{2}=|E(\phi)-E(\phi_{0})|$ And I am not sure whether my derivation is right.

Is this right???

For any $\varepsilon>0$, given:

$\underset{x\epsilon\Omega}{max}|\phi\left(x\right)-\phi_{0}\left(x\right)|<\frac{\varepsilon}{\int|\frac{\left(\mu-N\left(x\right)\right)}{\pi}|\, dx}$


  1. $E\left(\phi\right)-E\left(\phi_{0}\right)=\int_{\Omega}\left|\left(g\left(\vec{x}\right)-A\right)\right|^{2}\left(H(\phi_{0}\left(\vec{x}\right))-H(\phi\left(\vec{x}\right))\right)d\vec{x} +\mu\int_{\Omega}\left(H(\phi\left(\vec{x})\right)-H(\phi_{0}\left(\vec{x}\right))\right)d\vec{x} +\nu\int_{\Omega}\left(\delta\left(\phi\left(\vec{x}\right)\right)\left|\nabla\phi\left(\vec{x}\right)\right|-\delta\left(\phi_{0}\left(\vec{x}\right)\right)\left|\nabla\phi_{0}\left(\vec{x}\right)\right|\right)d\vec{x}$

  2. $E\left(\phi\right)-E\left(\phi_{0}\right)=\int_{\Omega}\left(\left|\left(g\left(\vec{x}\right)-A\right)\right|^{2}-\mu\right)\left(H(\phi_{0}\left(\vec{x}\right))-H(\phi\left(\vec{x}\right))\right)d\vec{x} +\nu\int_{\Omega}\left(\delta\left(\phi\left(\vec{x}\right)\right)\left|\nabla\phi\left(\vec{x}\right)\right|-\delta\left(\phi_{0}\left(\vec{x}\right)\right)\left|\nabla\phi_{0}\left(\vec{x}\right)\right|\right)d\vec{x}$

  3. $E\left(\phi\right)-E\left(\phi_{0}\right)<\int_{\Omega}\left(\left|\left(g\left(\vec{x}\right)-A\right)\right|^{2}-\mu\right)max\left(H(\phi_{0})-H(\phi)\right)d\vec{x} +\nu\int_{\Omega}\left(\delta\left(\phi\left(\vec{x}\right)\right)\left|\nabla\phi\left(\vec{x}\right)\right|-\delta\left(\phi_{0}\left(\vec{x}\right)\right)\left|\nabla\phi_{0}\left(\vec{x}\right)\right|\right)d\vec{x}$

And let $H$ is signed distance function. So$\left|\nabla\phi_{0}\left(\vec{x}\right)\right|\approx1$ for all $\vec{x}\in\Omega$. Then equation 3 becomes:

Four. $E\left(\phi\right)-E\left(\phi_{0}\right)<max\left(H(\phi_{0})-H(\phi)\right)\int_{\Omega}\left(\left|\left(g\left(\vec{x}\right)-A\right)\right|^{2}-\mu\right)d\vec{x} +\nu\int_{\Omega}\left(\delta\left(\phi\left(\vec{x}\right)\right)-\delta\left(\phi_{0}\left(\vec{x}\right)\right)\right)d\vec{x}$

$$max(\delta(\phi)-\delta(\phi_{0})) = (\delta(\phi(\vec{x^{*}}))-\delta(\phi_{0}(\vec{x^{*}})))$$ $$\vec{x^{*}}\in{\left\{x|max(\delta(\phi(\vec{x}))-\delta(\phi_{0}\vec{x}))for\,\, all\, x\in\Omega\right\}}$$

Five. $E\left(\phi\right)-E\left(\phi_{0}\right)<max\left(H(\phi_{0})-H(\phi)\right)\int_{\Omega}\left(\left|\left(g\left(\vec{x}\right)-A\right)\right|^{2}-\mu\right)d\vec{x} +\nu max(\delta(\phi)-\delta(\phi_{0}))\int_{\Omega}d\vec{x}$

Make $max(\delta(\phi)-\delta(\phi_{0}))=\frac{\varepsilon}{2\nu\int_{\Omega}d\vec{x}}$ and\,\, $max(H(\phi)-H(\phi_{0}))=\frac{\varepsilon}{2\int_{\Omega}\left(\left|\left(g\left(\vec{x}\right)-A\right)\right|^{2}-\mu\right)d\vec{x}}$

$E\left(\phi\right)-E\left(\phi_{0}\right)<\varepsilon$ So $E$ is continuous. I am not sure whether I can derive like this since I am confused little about $\phi$. $\phi$ is a function $R^2\rightarrow R$ and so $H$ is $R\rightarrow R$.

I really appreciate if you could tell my whether this is right or how can I to deriive it.

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First: continuity depends on the notion of convergence for the functions $\phi$ (more precisely, on the topology on the space of functions $\phi$). Which one do you use? Second: What is $H$ precisely? Is is a function of the real line? Is it the discontinuous Heaviside or a differentiable approximation? Third: Is this related to "active contours without edges"? – Dirk Feb 10 '12 at 8:11
Thank you, Dirk. H is defined as above question. This function is a energy function which I want to minimize. Yes, this is related to "active contours without edges". – user18481 Feb 10 '12 at 9:19

Since the question leaves much to be explained, let us try to do some reconstruction (note that this mainly expands on the first part of @Dirk's comment). The functional $E$ is defined on some real valued functions $\phi$ defined on $\Omega\subseteq\mathbb R^2$ and $E(\phi)$ is a real number. For $E(\phi)$ to exist, it is necessary that $\phi$ is measurable, and even differentiable since the last integral involves the gradient of $\phi$.

To summarize, one considers a functional $E:\mathfrak F\to\mathbb R$, $\phi\mapsto E(\phi)$, defined on some unspecified subspace $\mathfrak F$ of $C^1(\Omega,,\mathbb R)$. The continuity of $E$ at $\phi$ in $\mathfrak F$ would be the statement that $E(\psi)$ is close to $E(\phi)$ for every $\psi$ in $\mathfrak F$ close to $\phi$. Closeness in such settings is often (but not always) quantified by a norm, say $\|\ \|_{\mathfrak F}$, and the continuity of $E$ at $\phi$ in $\mathfrak F$ would then read as follows: $$ \forall\varepsilon\gt0,\quad\exists\alpha\gt0,\quad\forall\psi\in\mathfrak F,\quad \|\phi-\psi\|_{\mathfrak F}\leqslant\alpha\implies|E(\psi)-E(\phi)|\leqslant\varepsilon. $$ Hence, what is also missing at this stage is a definition of $\|\ \|_{\mathfrak F}$. One can guess that $\phi$ or integrals of $\phi$ could be involved, but also $\nabla\phi$ or some of its integrals. In any case, to specify the space $\mathfrak F$ and the norm $\|\ \|_{\mathfrak F}$ is mandatory for the question to make sense.

share|cite|improve this answer
Thank you so much for your answer. The assumption is totally right. I will reedit my question based on your answer after I fix the problem of my exploreer. – user18481 Feb 14 '12 at 2:43
You seem to have made up your mind about the norm and chosen $\|\ \|_\mathfrak F=\|\ \|_2$ (the only available definition of a norm anywhere in your post, so far). This choice is hopeless, as explained in my post, since one needs a control on the gradient as well. – Did Mar 4 '12 at 9:53
Correction: in the new newest version, $\|\ \|_\mathfrak F$ becomes $\|\ \|_\infty$ on $\Omega$. Same remark as before. – Did Mar 4 '12 at 11:33
thank you again for your comment. No, actually, I use absolute value as the norm. I want to know whether $E$ is continuous since I want to calculate its minimal and I really did. I use iteration to solve the mininal. I mean I calculate the minianal iteratively. I update $\phi$ each time to try to maintain a signed distance function. If I see this kind of update is tiny, can I know whether $E$ is continuous proximately? – user18481 Mar 4 '12 at 11:36
Correction:I use iteration to solve the minimum, not mininal. Sorry for that. – user18481 Mar 4 '12 at 11:51

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