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I'm studying by myself PDE by the Evans' book and I'm trying understand the Second Existence Theorem for weak solutions of elliptic equations and I'm trying understand the steps $1$ and $3$ of the proof (this topic has an image with the theorem and the proof), but there are two points that I didn't understand:

1) Why $(13)$ is true? I think it is used the Lax-Milgram Theorem, but I don't sure because $u$ would be in $U$ and not in $H^1_0(U)$, right? Because $g \in L^2(U)$ and $u$ must be an element of the domain of $g$ by the Lax-Milgram Theorem.

2) How obtain $|| Kg ||_{H^1_0(U)} \leq C ||g||_{L^2(U)}$?

Thanks in advance!

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  1. You are correct that you can verify this via Lax-Milgram lemma. I think you have confused $g$ and the linear functional $H^1_0(U)\to \mathbb{R}$ given by $v\mapsto (g,v)$. The domain of that operator is $H^1_0(U)$, not $U$. $U$ is the domain of the functions $u,v,g,f$.

  2. It tells you to refer back to a previous section, which I believe introduces the trace operator, which relates a function to its boundary values and establishes the bound you mention

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  • $\begingroup$ 1. This makes more sense now, thanks! 2. Can you give more details please? There is the Trace Theorem here, but I can't see how this is related to my doubt. $\endgroup$ – George Jul 4 '19 at 12:32
  • $\begingroup$ Sorry, I misread the theorem the first time I answered. The statement above the inequality you are referring to is proving that $K$ is bounded. Dividing both sides of that statement by $\|u\|^2_{H_0^1(U)}$ and then letting $u=g$ and dividing by $\beta$ gives you the desired inequality. $\endgroup$ – whpowell96 Jul 5 '19 at 21:39
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    $\begingroup$ I think you want to say "observe that $u = \frac{Kg}{\gamma}$ by $(14)$ and $(18)$" and not "letting $u = g$", right? $\endgroup$ – George Jul 9 '19 at 15:16

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