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It is rather easy question but I'm already struggling with this problem for a long time. I'm trying to estimate the value

$$\inf\int|\nabla g|^p\,dx$$

where $\mathbf{inf}$ is taken over all continuous functions $g \in L^1_p(\mathbb R)$ such that $g(0)=1$ and $g(1)=0$, $p>1$. I guess Poincaré inequality should be used here.

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$g$ should be only continuous? To make sense $\nabla g$, $g \in W^{1,p}(0,1)?$ –  user29999 Nov 29 '12 at 15:02
    
@user29999 OK, $g$ continuous and $g \in L^1_p(\mathbb R)$ –  nikita2 Nov 29 '12 at 15:04
    
Sorry, but I dont know the notation $ L^{1}_{p}(\mathbb{R})$. –  user29999 Nov 29 '12 at 15:07
    
$L^1_p(\mathbb R)$ is the set of locally summable function and $\nabla g \in L_p(\mathbb R)$ –  nikita2 Nov 29 '12 at 15:09

1 Answer 1

up vote 1 down vote accepted

If you only want locally summable functions $g$, the infimum is equal to 1 and is attained for $$ g(x) = 1 \, (x < 0), \, g(x) = 1-x \, (0 \le x \le 1), \, g(x) = 0 (x > 1) \, .$$ The reason is that on the interval $(0,1)$ the Euler-Lagrange equation is equivalent to $g''(x) = 0$ (so the minimizer should be linear there) and outside that interval clearly constant functions minimize the integrand.

If you want $g \in L^1(\mathbb{R})$, the infimum does not exist. To see this, you can define $g_n(x) = g(x) \, (x \ge 0)$ where $g$ is as before and $g_n(x) = \max(0,1 + x/n)$ for $x < 0$. Then $$ \int_\mathbb{R} |\nabla g_n|^p = 1 + n^{1-p} $$ so we have a minimizing sequence and the infimum should be 1. But that's clearly impossible, there is no integrable function with this property.

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Perfect! Now I see that the key is the Euler-Lagrange equation. And I wanted only $\nabla g \in L_p(\mathbb R)$. –  nikita2 Nov 29 '12 at 15:41

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