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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:

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

Then:

  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)

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

Five. $E\left(\phi\right)-E\left(\phi_{0}\right)

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"?2012-02-10
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    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".2012-02-10

1 Answers 1

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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.

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    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.2012-02-14
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    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.2012-03-04
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    Correction: in the new newest version, $\|\ \|_\mathfrak F$ becomes $\|\ \|_\infty$ on $\Omega$. Same remark as before.2012-03-04
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    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?2012-03-04
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    Correction:I use iteration to solve the minimum, not mininal. Sorry for that.2012-03-04
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    The absolute value of *what*? The norm is supposed to measure a distance between two *functions*, the continuity of $E$ is related to a norm on *functions*. If you do not specify a norm on the space $\mathfrak F$ of *functions* that $E$ is defined on (space $\mathfrak F$ that you still did not specify), the assertion that $E$ is continuous has no meaning.2012-03-04
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    Hi, I mean $|x|$is the absolute value of $x$. $\|H(\phi)-H(\phi_{0})\|_{2}=|H(\phi)-H(\phi_{0})|=H(\phi)-H(\phi_{0}) \,\,\,\,(H(\phi)>H(\phi_{0}))\\ H(\phi_{0})-H(\phi) \,\,\,\,(H(\phi)<=H(\phi_{0}))$ since $H(\phi)\in R$, yes, it is $\|\|_2$ The norm is $\|\|_2$. This norm is normal in the real world and I can image it so I did not specify it.2012-03-04
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    Once again: call $E_3(\phi)$ the $\nu$ term in $E(\phi)$. The continuity of the functional $E$ would imply that $|E_3(\phi_1)-E_3(\phi_2)|$ is small for every functions $\phi_1$ and $\phi_2$ (in the, still unspecified, domain of $E$) such that $\|\phi_1-\phi_2\|_\mathfrak F$ is small. You say in your comment that this holds with $\|\ \|_\mathfrak F=\|\ \|_2$, and (in contradiction with this choice of norm) your post pretends to prove this holds for $\|\ \|_\mathfrak F=\|\ \|_\infty$. Alas, both results seem wrong, for the reason I already explained once in my answer and twice in comments ../..2012-03-04
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    ../.. So, yes, there is a problem, which must be faced if you want to solve the question.2012-03-04
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    the space of functions that $E$ is defined on proximate signed distance functions $\phi R^{2}\rightarrow R$ and I try to calculte $E(\phi)-E(\phi_{0})$. Since $E(\phi)$ would be a real number, i.e. $E(\phi)\in R$, by the way $E(\phi)>0$ according to equation above. So since $R$ is one dimension, I think $|x|=\|\|_{2}=\|\|_{\infty}$. Thank you very much for explanation three times.2012-03-04
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    First, $E(\phi)$ simply does not exist when $\phi$ is not differentiable. So, no, the space of functions $\mathfrak F$ **cannot be** the space of all functions from $\mathbb R^2$ to $\mathbb R$. Second, the norm you need to even define your question is a norm **on a space of functions** and you indicate a norm on $\mathbb R$. Sorry, all this does not add up...2012-03-04
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    Hi, @Didier Piau. I am sorry I have not learnt any example to solve this problem before. I just try to calculate intuitively. All definition I think should follow the most usual rules, like Euclidean distance. But this makes sense in the reality and I have computer programming. I can show you it can work, but I want more theoretical certification so I post this question.2012-03-05