# Steps $1$ and $3$ of the proof of the Second Existence Theorem for weak solutions of elliptic equations

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!

## 1 Answer

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

• 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. – George Jul 4 '19 at 12:32
• 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. – whpowell96 Jul 5 '19 at 21:39
• I think you want to say "observe that $u = \frac{Kg}{\gamma}$ by $(14)$ and $(18)$" and not "letting $u = g$", right? – George Jul 9 '19 at 15:16