This implication is true ? I believe that is ..

Consider $E$ a compact subset of $R^n$. $1< q < p$ . Supoose that for all $\Omega \subset R^n$ ($\Omega$ open )occurs

$$\displaystyle\inf_{\varphi \in C^{\infty}_{0}(\Omega) , \varphi (E \cap \Omega) \geq 1 } \displaystyle\int_{\Omega} |\nabla \varphi|^{p/q} = 0$$

then $p/q \leq n.$

I am studying a theorem and if this affirmation is true , I understand the theorem .

Someone can help me prove this affirmation ( if the affirmation is true ) ?

I have no idea how to prove ..

thanks in advance =)

  • $\begingroup$ What does $\varphi(E \cap \Omega) \ge 1$ mean? $\endgroup$ – Umberto P. Jul 27 '13 at 5:59
  • $\begingroup$ I would guess $\varphi \ge 1$ on the set $E \cap \Omega$. $\endgroup$ – gerw Jul 27 '13 at 10:31

Define $r = p/q$. Then, your statement is equivalent to "$E$ is a set of $r$-capacity zero", i.e., a $r$-polar set.

In the case where $E$ is the empty set, this holds trivially for all possible values of $r$.

If $E$ is not the empty set, than it contains at least one point. Hence, this point is $r$-polar. But this implies $r \le n$, since for $r > n$, all points have positive $r$-capacity.

Note that you get an even better bound, if you know that $E$ contains "larger" sets. E.g. if a $n-1$-dimensional (smooth) manifold is contained in $E$, then $r < 2$, since $n-1$-dimensional manifolds have positive $2$-capacity.

After writing this answer, I have just seen, that you required $1 < p < q$, which gives $p/q < 1$ immediately. Surely, this is a typo.

  • $\begingroup$ you're right , the correct is $1<q<p$. sorry for my mistake $\endgroup$ – math student Jul 27 '13 at 18:12
  • $\begingroup$ I agree with your solution. beatiful solution! . thanks ! $\endgroup$ – math student Jul 27 '13 at 18:30

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