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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 continuous 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 $t\in (0,1)$.

Moreover, if $F(u_1)\leq F(a(t))\leq F(u_2)$ would be great, but it is not necessary... (here we assume $F(u_1)\leq F(u_2)$)

I tried the affine connection, i.e., $a(t)=tu_2+(1-t) u_2$, but it does not seems to work....

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

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    $\begingroup$ Why doesn't the affine path work? Seems to work to me... Moreover $\nabla\alpha_t=t\nabla u_2+(1-t)\nabla u_1$ and the convexity of the norm squared even gives you 4. and the additional condition you want. $\endgroup$
    – Teri
    Dec 4, 2015 at 18:09
  • $\begingroup$ @Teri: here the post is updated. $\endgroup$
    – JumpJump
    Dec 4, 2015 at 19:38
  • $\begingroup$ @Teri: I edit the problem a bit more. Now it is definitely a non-trivial problem... $\endgroup$
    – JumpJump
    Dec 5, 2015 at 18:28
  • $\begingroup$ Note that the compactness properties of $H^1(\Omega)$ imply that the infimum is attained. Using this one can still prove that $F$ is a convex functional, i.e. $F(tu_2+(1-t)u_1)\le tF(u_2)+(1-t)F(u_1)$. Property 4. and the additional condition always follow from this convexity. $\endgroup$
    – Teri
    Dec 6, 2015 at 20:50
  • $\begingroup$ @Teri Can you be more specific and write your comments as an answer? Thank you! $\endgroup$
    – JumpJump
    Dec 10, 2015 at 22:05

1 Answer 1

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I've to admit that my original argument had a (stupid) mistake. However the argument below should work.

First let me relabel: call the two functions $u_1,u_2$ by $u_0,u_1$ so as to be compatible with $u_t:=tu_1+(1-t)u_0$.

Note that given any $u\in H^1(\Omega)$, the infimum in $$F(u)=\inf_{v\in \mathcal V}\{\int_\Omega|\nabla u|^2v^2 dx+\int_\Omega(|\nabla v|^2+(1-v)^2)dx\}$$ is obtained: take a minimizing sequence for $F(u)$ and, using the Rellich-Kondrachov theorem, find $v\in \mathcal V$ and a subsequence $v_k$ so that $v_k\to v$ in $L^2$, $\nabla v_k\rightharpoonup \nabla v$ weakly in $L^2$, and $v_k\to v$ pointwise almost everywhere. Then \begin{align*} F(u)\le &\int_\Omega|\nabla u|^2v^2 dx+\int_\Omega(|\nabla v|^2+(1-v)^2)dx\\ \le &\liminf_{k\to\infty}\left[\int_\Omega|\nabla u|^2v_k^2 dx+\int_\Omega(|\nabla v_k|^2+(1-v_k)^2)dx\right]=F(u). \end{align*}

Now let $w_j$ be a function in $\mathcal V$ for which the infimum $F(u_j)$ is attained, $j=0,1$, and let $w_t=tw_1+(1-t)w_0$. By the edit below $$t\mapsto w_t^2|\nabla u_t|^2$$ is convex so we may estimate \begin{align*} F(u_t)\le & \int_\Omega |\nabla u_t|^2w_t^2 dx+\int_\Omega(|\nabla w_t|^2+(1-w_t)^2)dx\\ \le & t\int_\Omega|\nabla u_1|^2w_1^2dx+(1-t)\int_\Omega |\nabla u_0|^2w_0^2dx\\ &\ \ + t\int_\Omega(|\nabla w_1|^2+(1-w_1)^2)dx+(1-t)\int_\Omega(|\nabla w_0|^2+(1-w_0)^2)dx\\ =&tF(u_1)+(1-t)F(u_0). \end{align*} Thus $F$ is convex and this gives you the properties you want.

EDIT: Denote $$f(t)=f_x(t)=|\nabla u_t(x)|^2, g(t)=g_x(t)=w_t(x)^2.$$ Both are convex functions, but this alone doesn't guarantee the convexity of their product. Instead we use the fact that they are essentially quadratic.

Let us show that $(fg)''(t)\ge 0$. Compute \begin{align*} f'(t)=&2\nabla(u_1-u_0)\cdot \nabla u_t\\ g'(t)=&2(w_1-w_0)w_t\\ f''(t)=& 2|\nabla(u_1-u_0)|^2\\ g''(t)=& 2(w_1-w_0)^2. \end{align*} Next check that \begin{align*} (fg)''(t)&=f''(t)g(t)+2f'(t)g'(t)+g''(t)f(t)\\ =&2[|\nabla(u_1-u_0)|^2w_ t^2+4w_t\nabla(u_1-u_0)\cdot (w_1-w_0)\nabla u_t+(w_1-w_0)^2|\nabla u_t|^2]\\ \ge& 2(|\nabla(u_1-u_0)|w_t-|w_1-w_0||\nabla u_t|)^2-4|w_1-w_0|w_t|\nabla(u_1-u_0)||\nabla u_t|\\ \ge & 0. \end{align*}

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  • $\begingroup$ you mean that $w_t^2$ and $|\nabla u_t|^2$ is nonnegative and nondecreasing so that the product is convex? Since generally the product of nonnegative convex functions may not convex. However, we have $w_t^2$ is convex but $\nabla u_t$ may not be positive always. $\endgroup$
    – JumpJump
    Dec 11, 2015 at 20:30
  • $\begingroup$ I mean, why the function $|\nabla u_t|^2w_t^2$ is convex? We usually need $|\nabla u_t|^2$ and $w_t^2$ have nonnegative or nonpositive derivative at the same time. But I don't think we have in this case. $\endgroup$
    – JumpJump
    Dec 11, 2015 at 22:38
  • $\begingroup$ @tankonetoone You're right, the convexity of the product needs a separate argument, it doesn't follow automatically. I've made an edit to my answer to deal with this. $\endgroup$
    – Teri
    Dec 14, 2015 at 19:31
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    $\begingroup$ Sorry if this question is naive. But how you obtain the last inequality? You are basically saying that $(|a|-|b|)^2-2|ab|\geq 0$, which is not true in generally. For example, take $a=b$. $\endgroup$
    – JumpJump
    Dec 14, 2015 at 20:29
  • $\begingroup$ @tankonetoone You're right, again. I'm busy with a lot of stuff currently (hence the slow replies) but I'll try to see if anything can be done about the question. It seems that the convexity of $w_t^2|\nabla u_t|^2$ does not necessarily hold.. $\endgroup$
    – Teri
    Dec 19, 2015 at 14:52

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