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