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Let $p\in [1,\infty [$. I want to prove that $$\forall u\in W^{1,p}(\Omega ): \int_\Omega u=0, \ \ \|u\|_{L^p}\leq C\|\nabla u\|_{L^p}.$$

In the proof the say : suppose it's not true. Then, $$\exists (u_n)\subset W^{1,p}(\Omega ): \int_{\Omega }u_n=0,\|u_n\|_{L^p}=1\quad \text{and}\quad \|\nabla u_n\|_{L^p}\to 0,$$ but I don't understand why such a sequence exist.

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You want to prove that: $$ \exists C \ \forall u \in W^{1,p} \Big(\int u = 0 \Rightarrow \|u\|_{L^p} \le C\|\nabla u\|_{L^p}\Big). $$ You want to prove that by contradiction. The negation of the previous statement is:

$$ \forall C \ \exists u_C \in W^{1,p} \Big(\int u_C = 0 \ \land \|u_C\|_{L^p} > C\|\nabla u_C\|_{L^p}\Big). $$ Therefore if you assume that, you can take $C = n$ for every $n = 1, 2, \dots$ and you can also assume $\|u_n\|_{L^p} =1$ (otherwise you can renormalize $u_n$ ).

Therefore picking $C = n$ and after renormalising $u_n$ you have: $$ 1 > n \| \nabla u_n \|_{L^p}. $$ thus $$ \| \nabla u_n \|_{L^p} < \frac{1}{n}. $$

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  • $\begingroup$ Emphasize that since the inequality strict $u_C$ it cannot be vanish. That is why you can consider its norm to be 1. $\endgroup$ – Guy Fsone Jan 19 '17 at 10:14
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    $\begingroup$ thanks. But why $\|\nabla u_n\|\to 0$ ? $\endgroup$ – MathBeginner Jan 19 '17 at 10:14
  • $\begingroup$ @MathBeginner I made an edit. I hope it's more clear now. $\endgroup$ – Onil90 Jan 19 '17 at 10:19
  • $\begingroup$ @MathBeginner In the proofs by contradiction you always have to make clear what is the negation of your statement. Sometimes writing things in a rigorous way and using some basic logic can be very helpful. $\endgroup$ – Onil90 Jan 19 '17 at 10:22
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Think of the inequality as

$$ \frac{ \Vert u \Vert_p }{ \Vert \nabla u \Vert_p } \le C. $$

If there is no $C$ such that this holds, it means the ratio takes on arbitrarily large values on the specified domain $\{ u : \int u = 0 \};$ i.e. for all $n$ we can find $f_n$ such that $\Vert f_n \Vert_p \ge n \Vert \nabla f_n \Vert_p$. Normalizing these functions by defining $u_n = f_n / \Vert f_n \Vert_p$ we get a sequence satisfying $\Vert u_n \Vert_p = 1$ and $\Vert \nabla u_n \Vert_p \le \frac 1n\to0.$

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