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This question has been post on MO a week ago. I move it here to get more luck.

Let $\Omega\subset \mathbb R^N$ be open bounded, smooth boundary. Let $u_1$, $u_2\in H^{1}(\Omega)$ such that $T[u_1]=T[u_2]=T[\omega]$ where $T$ stands for the trace operator and $\omega\in H^1(\Omega)$ is a fixed function.

Define $$ F(u):=\inf_{v\in\mathcal V}\left\{\int_\Omega |\nabla u|^2v^2dx + \int_\Omega |\nabla v|^2+(1-v)^2dx \right\}, $$ where $\mathcal V:=\{v\in H^1(\Omega),\,0\leq v\leq 1\}$.

Question: does there exist a path $a(t): [0,1]\to H^1(\Omega)$ between $u_1$ and $u_2$ satisfies the following conditions?

  1. $a(0)=u_1$, $a(1)=u_2$
  2. $T[a(t)]=T[\omega]$ for all $t\in (0,1)$
  3. $a(t)$ is continuous in $L^2$ sense, i.e., if $t\to t_0$, then $a(t)\to a(t_0)$ in $L^2$.
  4. $F(a(t))\leq \max\{F(u_1),F(u_2)\}$, for all $t\in (0,1)$.

Any help, hint, or reference would be really welcome!


Update: based on @Jason's answer, I wrote, for arbitrary $v\in\mathcal V$, $$ F(a(t))\leq G(a(t),v)\leq tG(u_1,v)+(1-t)G(u_2,v). $$ Let's denote by $v_1$ and $v_2$ that $F(u_1)=G(u_1,v_1)$ and $F(u_2)=G(u_2,v_2)$, such $v_1$ and $v_2$ exists and unique by the properties of $H^1$ function.

I understand that $v$ on the right hand side is arbitrary so we may replace it with either $v_1$ or $v_2$ and we have $$ F(a(t))\leq tG(u_1,v_1)+(1-t)G(u_2,v_1)=t F(u_1)+(1-t)G(u_2,v_1)\tag 1 $$ or $$ F(a(t))\leq tG(u_1,v_2)+(1-t)F(u_2) $$ But we may can not go further from here. Take, from example, $(1)$. We can not switch $v_1$ by $v_2$ here since if we do, we will change $F(u_1)$ as well.

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My first instinct at looking at this problem was to take a convex combination of $u_1$ and $u_2$, and this turns out to be the right thing to do. So let

$$a(t)=(1-t)u_1+tu_2.$$

Properties 1,2 and 3 all follow fairly immediately, so it remains to check property 4. First, it will be helpful to write

$$G(u,v)=\int_\Omega\bigg(|\nabla u|^2v^2+|\nabla v|^2+(1-v)^2\bigg)\,\mathrm dx$$

and note that it is sufficient to prove $G(a(t),v)\le\max\{G(u_1,v),G(u_2,v)\}$ for all $v\in\mathcal V$. So fix $v\in\mathcal V$ and observe that $F$ and $G$ are both now fairly irrelevant since we need only prove that the function $t\mapsto|(1-t)\nabla u_1+t\nabla u_2|^2$ does not take a maximum in $(0,1)$. It is fairly straightforward to show that the given map is convex (indeed it is smooth, so look at the second derivative) which implies the result.

A few little extra things to think about:

  • Property 3 is redundant given that $a$ is continuous into $H^1$;
  • Your 'relaxed' version of property 4 is actually equivalent to it;
  • One can show that $F(u)$ is attained as a minimum. This isn't particularly useful to this problem, but perhaps may be helpful in the wider context of where this question was found.
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  • $\begingroup$ Thx for your answer! However, if you click the link I provided, you will see an already tried to use convex combination, but it does not work, or at least we can't prove it works... $\endgroup$ – spatially Dec 18 '15 at 7:16
  • $\begingroup$ But I did prove it works! There's an answer there that provides a proof, but you don't even need all of the details of that proof. Are there any steps of my proof you don't agree with? $\endgroup$ – Jason Dec 18 '15 at 15:31
  • $\begingroup$ I wrote some details according to your answer, as well as my concern and problem. Can you have a look? Thank you! $\endgroup$ – spatially Dec 18 '15 at 18:26

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