consider $r>0 , p>1$ and $K \subset B(x_0 , 2r) \subset R^n$ . $K$ compact. Define the sets :

$$A = \{ u \in C^{\infty}_{0} (B(x_0 , 2r)); \textit{ such that } \ u=1 \textit{ in a open neighborhood of K} \}$$

$$ B = \{ u \in H^{1,p}_{0} (B(x_0 , 2r)) \cap C (B(x_0 , 2r)) : u=1 \ on \ K , \ 0 \leq u \leq 1\}$$

I want to show that

$$ \displaystyle\inf_{u \in A} \displaystyle\int_{B(x_0 , 2r)} |\nabla u|^p \ dx= \displaystyle\inf_{u \in B} \displaystyle\int_{B(x_0 , 2r)} |\nabla u|^p \ dx$$

For this i am trying to aproximate (in the sense of the norm of $H^{1,p}$) the function of B , by function of $A$.

Somenone can give me a hint ?

thanks in advance


Let me define $$\tilde B = B = \{ u \in H^{1,p}_{0} (B(x_0 , 2r)) \cap C (B(x_0 , 2r)) : u=1 \ on \ K\}.$$ Then by Stampacchia's lemma, it is easy to see that the $\inf$ over $\tilde B$ equals the $\inf$ over $B$. Since $A \subset \tilde B$, it remains to show $\inf_A \le \inf_{\tilde B}$. Therefore, the idea is to take $f \in \tilde B$ and approximate it with $g_\varepsilon \in A$, such that $$\int |\nabla f|^p dx +\varepsilon \ge \int |\nabla g_\varepsilon|^p dx.$$ This shows $$\inf_{\tilde B}(\ldots) + \varepsilon \ge \inf_{A}(\ldots).$$ Since this holds for all $\varepsilon > 0$, we have $$\inf_{\tilde B}(\ldots) \ge \inf_{A}(\ldots).$$

It remains to give an idea for the approximation of $f \in \tilde B$ with $g_\varepsilon \in A$. Since your functional is continuous w.r.t. the norm in $H^{1,p}$, we just need to ensure that we can construct a $g_\varepsilon$ with $$\| f - g_\varepsilon\|_{1,p} \le \varepsilon$$ for all $\varepsilon > 0$. Therefore, take $\delta > 1$ and observe that $\delta \, f \ge \delta$ on $K$. Hence, the set $X = \{x : \delta \, f(x) \ge 1\}$ is a neighbourhood of $K$. Now, convolve $\delta\,f$ with a smooth kernel $\varphi$ (which is close to the dirac delta). This yields a smooth function $g = \varphi \star \delta \, f$. Since $\delta\,f \ge 1$ on $X$ and since $\varphi$ has small support and $\int \varphi = 1$, we get $g \ge 1$ on a neighbourhood of $K$. Finally, you get $$\|f - g\|_{1,p} \to 0$$ as $\delta \to 1$ and $\varphi$ approaches the dirac delta.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.