Hi please view the attachment. I am interested in how Lemma 1.11 is used in the proof of Theorem 2.10.
Based on the statement of Lemma 1.11 it seems that in order to use Lemma 1.11 in we require $\sum_{i=1}^{n}g^{i}(x)x_{i} \geq 0~~~~\forall c^{n} \in \mathbb{R}^{n} \text{ with } |c^{n}| = R_{0}$, with the conclusion being that there exists a solution $c^{n}$ of (2.5) such that $|c^{n}| \leq R_{0}$. Instead in the proof it is used that for all $c^{n}$ such that $\| u_{n} \| = R_{0}$ there exists a solution $u_{n}$ where $\| u_{n} \| \leq R_{0}$. I'm having difficulty seeing how the Lemma 1.11 is applied in this way? It seems to use $c^{n}$ and $u_{n}$ interchangeably.
Note I have only attached the beginning of the proof. Also, the definition of demicontinuous is that $A: (V,norm) \rightarrow (V^{*}, weak)$ continuous. While the assumed hemicontinutiy on $A: V \rightarrow V^{*}$ is just used to show the demicontinutiy.
Any assistance would be greatly appreciated.
Best Answer
The proof maybe is written in a confusing way. I will try to rephrase it a little to make it clearer. If you let repeat me a bit of what is happening in the proof:
We want to show that for all $n \in \mathbb{N}$ there is a solution $u_n \in X_n:=\{e_1,\dots,e_n\}$ to $$ \forall\,k \in \{1,\dots,n\}\colon \langle A u_n - b, e_k\rangle_X = 0. $$ This is possible by using Lemma 1.11 as we will show in a moment.
Before we come to this, we introduce a map that will allow us to identify the space $\mathbb{R}^n$ with $X_n$. This is simply done by the map $$ \mathbb{R}^n \rightarrow X_n~~~~~~~ \\ ~~~~~~c \mapsto \sum_{k=1}^n c^k e_k. $$
So, instead of the above problem we now try to solve the following problem for $c=(c_1,\dots,c_n)\in\mathbb{R}^n$: $$ \forall\,k \in \{1,\dots,n\}\colon \left\langle A (\textstyle\sum_{i=1}^n c^i e_k)- b, e_k\right \rangle_X = 0. $$ Note that this problem is equivalent to the one stated in the beginning because our map $\mathbb{R}^n\rightarrow X_n$ defined above is bijective (it's even a linear isomorphism).
Now, we again reformulate this problem. In order to use Lemma 1.11 we need a map $g\colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ with appropriate properties. We get this by defining $$ g\colon \mathbb{R}^n \rightarrow \mathbb{R}^n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~c \mapsto \left( \left\langle A (\textstyle\sum_{i=1}^n c^i e_k)- b, e_k\right \rangle_X \right)_{k\in\{1,\dots,n\}}. $$ (Note that $g$ is also dependent on $n$ but we omit giving an index $n$ to the function $g$ out of convenience.) If we are able to find a $c_n\in\mathbb{R}^n$ such that $g(c_n) = 0$ we have a solution to the Galerkin problem above and ultimately solve the Galerkin problem stated in the beginning by setting $u_n:= \sum_{k=1}^n c_n^k e_k$.
By the coercivity of $A$, which was one of the assumptions of the theorem, we have that there exists $R_0 \in \mathbb{R}$ such that $\langle Au - b, u\rangle_X > 0$ for all $u \in X$ fulfilling $\Vert u\Vert \geq R_0$. (See the proof given in the original post; is this plausible?)
In particular, this means that for all $c \in \mathbb{R}^n$ with $\Vert\sum_{k=1}^n c^k e_k\Vert \geq R_0$ we have $$ 0 < \left\langle A\textstyle\sum_{k=1}^n c^k e_k - b, \textstyle\sum_{k=1}^n c^k e_k\right\rangle_X = \sum_{k=1}^n \left\langle A\textstyle\sum_{k=1}^n c^k e_k - b,e_k\right\rangle_X c^k = \sum_{k=1}^n g^k(c)c^k, $$ which together with the remark below implies condition (1.11) needed for Lemma 1.11. (Note that we showed this using "$c$" instead of "$x$" and that we even were able to show this condition for all $\Vert c\Vert \geq R$ and not only $\Vert c\Vert = R$.)
From this we get that Lemma 1.11 is applicable and thus that an appropriate solution $c_n \in \mathbb{R}^n$ to the Galerkin problem exists (and in particular an $u_n$ for the problem we stated initially).
Remark: In order to get condition (1.11) from the above property we need to show that there exists a $R \in \mathbb{R}$ such that $\Vert \sum_{k=1}^n c^k e_k \Vert \geq R_0$ for all $c\in\mathbb{R}^n$ with $\Vert c\Vert \geq R$. This follows quickly if you use that all norms on finite-dimensional spaces are equivalent.