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Let be $\Omega\subset \mathbb{R}^n $ an open bounded set of class $C^1$ and let be $\Omega'\subset \subset \Omega ''\subset \subset \Omega$.

I'm trying to construct a function $\varphi\in C^\infty (\Omega)$such that:

  • $0\leq\varphi\leq1$
  • $\varphi\equiv 1$ in $\Omega'$
  • $\varphi\equiv 0$ in $\Omega \setminus \Omega''$
  • $||\nabla \varphi ||_{L^{\infty}(\Omega, \mathbb{R}^n)}\leq \dfrac{2}{{\rm dist}(\Omega',\partial \Omega'')}$

My attempt: My idea is to use the distance function ${\rm dist }$ and the mollifiers. Let be $\alpha={\rm dist}(\Omega',\partial \Omega'')$.

I have in mind this function:

$$\varphi(x)=\big [\dfrac {2}{\alpha} {\rm dist}(x,\tilde{\Omega}^c)\wedge 1\big ]*\rho_\epsilon$$

where $\tilde{\Omega}=\big \{ x\in \Omega \;| \;{\rm dist}(x,\Omega')<\dfrac{\alpha}{2} \big \}$, $\rho_\epsilon$ is the standard mollifier (I use the convolution in order to have a $C^\infty$ function) and $\wedge$ is the minimum operator.


Is my example correct?

My main concern is about the last point. I've constructed this function thinking of the domains as balls and for them the distance function is linear with respect to the radius.

Does my function satisfy the last request?

Edit What can I say about the gradient of the function $x \mapsto {\rm dist }(x,\tilde{\Omega}^c)$? Can I say that it is linear or can I do some estimations?

Thanks for the help!

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    $\begingroup$ Do the double inclusions mean e.g. that $A \subset \subset B$ is for the closure of $A$ to be open in $B$? $\endgroup$ – coffeemath Aug 14 '16 at 16:02
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    $\begingroup$ @coffeemath $A$ is strongly included in $B$ ( the closure of $A$ is compact and $\bar{A}\subset B$) $\endgroup$ – Gggl. Aug 14 '16 at 16:05
  • $\begingroup$ Do you know what the gradient of the distance function is? (I don't which is why I'm asking instead of telling you.) Maybe that could be used to get a bound on the $L^{\infty}$ norm which shows that the example is correct. Or you could derive a bound for the gradient, and then use that bound to get another bound for the $L^{\infty}$. In any case, assuming you can find some results showing that mollifying the function won't change your requests, it seems like it would be simplest to start with the unmollified case first. $\endgroup$ – Chill2Macht Aug 15 '16 at 0:44
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    $\begingroup$ @William No, I don't know. But I think that, since I don't know the exact shape of the domain, there is no simple expression for the gradient. Your comment has pointed out my main problem, I think. Thanks! $\endgroup$ – Gggl. Aug 15 '16 at 7:36
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Since you are actively thinking about this exercise, I will answer your question giving you just some hints. Tell me if you want to see more details.

  • Your example is almost correct. In order to guarantee that $\varphi\equiv 1$ on $\Omega'$, you should mollify a function which equals $1$ on a neighbourhood of $\overline{\Omega'}$ (not just on $\Omega'$). To accomplish this, it suffices to correct the definition of $\tilde\Omega$ a little: just take $\tilde{\Omega}=\big \{x\in \Omega\mid{\rm dist}(x,\Omega')<\frac{2}{3}\alpha\big\}$ instead.
  • Call $\psi(x):=\frac{2}{\alpha}{\rm dist}(x,\tilde{\Omega}^c)\wedge 1$ the function you are mollifying and think $\varphi$ and $\psi$ as functions from $\mathbb{R}^n$ to $\mathbb{R}$, extending them by zero outside $\Omega$ (notice that we still have $\varphi=\psi*\rho_\epsilon$ on $\mathbb{R}^n$). $\psi$ is a Lipschitz function: can you estimate its Lipschitz constant?
  • What happens to the Lipschitz constant of a Lipschitz function $f:\mathbb{R}^n\to\mathbb{R}$ when you mollify it? (this one is the crucial step)
  • How is $\|\nabla\varphi\|_{L^\infty(\mathbb{R}^n,\mathbb{R}^n)}$ related to the Lipschitz constant of $\varphi$? Conclude.

As a side note, although $\psi$ could be non-differentiable somewhere, one can still give a meaning to the 'gradient' of $\psi$. There are many ways to do that (Sobolev spaces, Clarke's subdifferential, Dini derivatives, ...): you will surely meet some of them if you pursue your studies in analysis.

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  • $\begingroup$ Thank you very much for your answer!!! I denote with $[f]$ the Lipschitz constant of the Lipschitz function $f$ defined on $\mathbb{R}^n$. I know that the function distance is $1$-Lip so $[\psi]=\dfrac{2}{ \alpha}$. And since $||\nabla \varphi||\leq [\varphi]\leq [\psi]$ I can conclude. Is it right? Thank you! $\endgroup$ – Gggl. Aug 17 '16 at 12:54
  • $\begingroup$ Exactly! Well done $\endgroup$ – Mizar Aug 17 '16 at 14:45

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