# Lax-Milgram problem

I am trying to solve this problem:

Let $H$ a Hilbert space, $A:H\times H\rightarrow\mathbb{R}$ a bilinear form, bounded and $H$-elliptic, and $F\in H^{\prime}$ ($H^{\prime}$ = dual space). Besides, let $\{H_n\}_{n\in\mathbb{N}}$ a sequence of subspaces of finite dimension of $H$, and for each $n\in\mathbb{N}$ consider a bounded bilinear form $A_n:H_n\times H_n\rightarrow\mathbb{R}$ such that the sequence $\{A_n\}_{n\in\mathbb{N}}$ is uniform elliptic. This is, there is $\tilde{\alpha} > 0$, independent of $n$ m such that $A_n(v_n,v_n) \geq \tilde{\alpha}\|v_n\|_H^2$, $\forall\ v_n\in H_n$, $\forall\ n\in\mathbb{N}$.

1. Show that there's unique $u\in H$ and $u_n\in H_n$ such that $$A(u,v)\ =\ F(v),\quad \forall\ v\in H$$ and $$A_n(u_n,v_n)\ =\ F(v_n),\quad \forall\ v_n\in H_n.$$
2. Prove that there's $C>0$, independent of $n\in\mathbb{N}$, such that $$\|u - u_n\|_H\ \leq\ C\inf_{v_n\in H_n}\left\{\|u - v_n\|_H + \sup_{\mbox{w_n\in H_n \atop w_n\neq 0}}\frac{\left|A(v_n,w_n) - A_n(v_n, w_n)\right|}{\|w_n\|_H}\right\}.$$

I think I solved it 1., by direct applications of Lax-Milgram Lemma, but I really don't know how to solve 2. I tried to use the subordinate forms: $$A(w,v)\ =\ \langle\mathbb{A}(w),v\rangle, \quad \forall (w,v)\in H\times H,$$ $$A_n(w_n,v_n)\ =\ \langle\mathbb{A}_n(w_n),v_n\rangle, \quad \forall (w_n,v_n)\in H_n\times H_n,$$ and the best approximation theory, but I couldn't do it.

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Is there something like $\cup_n H_n$ dense in $H$? Also, is there any relation between $A$ and $A_n$? –  TZakrevskiy Aug 7 '13 at 17:36
No, I wrote the problem exactly. –  FASCH Aug 7 '13 at 17:42
The only relation, if I am correct, is $\mathbb{A}(u) = \mathcal{R}(F)$ and, $\langle\mathbb{A}_n(u_n), v_n\rangle = \langle\mathcal{R}(F), v_n\rangle$, for all $v_n\in H_n$. –  FASCH Aug 7 '13 at 17:45

For all $v_n \in H_n$ we have \begin{align*} A(u-u_n, u-u_n) &= A(u-u_n, u - v_n) + A(u, v_n - u_n)-A(u_n, v_n - u_n) \\ &=A(u-u_n, u - v_n) + A_n(u_n, v_n - u_n)-A(u_n, v_n - u_n)\end{align*} This implies the estimate \begin{align*} \alpha \, \|u - u_n\|^2 &\le \|A\| \, \|u-u_n\| \, \|u -v_n\|\\ &\qquad + \|\mathbb A_nu_n-\mathbb A u_n\| \, \|u_n - v_n\|\\ &\le \|A\| \, \|u-u_n\| \, \|u -v_n\|\\ &\qquad + \|\mathbb A_nu_n-\mathbb A u_n\| \, \|u - u_n\|\\ &\qquad + \|\mathbb A_nu_n-\mathbb A u_n\| \, \|u - v_n\|\\ \end{align*} Here, $\alpha$ is the coercivity constant of $A$. Now, we use Young's inequality on the second term of the right-hand side and obtain \begin{align*} \frac\alpha2 \, \|u - u_n\|^2 &\le \|A\| \, \|u-u_n\| \, \|u -v_n\|\\ &\qquad + \|\mathbb A_nu_n-\mathbb A u_n\| \, \|u - v_n\|\\ &\qquad \frac1{2\,\alpha}\,\|\mathbb A_nu_n-\mathbb A u_n\|^2. \end{align*} Now, take Young's inequality on the first two terms on the right-hand side to obtain $$c \, \|u - u_n\|^2 \le \|u - v_n\|^2 + \|\mathbb A_nu_n-\mathbb A u_n\|^2$$ with some $c > 0$ (depending only on $\alpha$ and $\|A\|$). By taking the $\inf$ over all $v_n \in H_n$, using the equivalency of the $2$- and the $1$-norm in $\mathbb R^2$ and taking the square root, we get the claim. Note that this does not require the uniform coercivity of $A_n$.
How, taking the inf over all $v_n\in H_n$, you can change $\|\mathbb A_nu_n-\mathbb A u_n\|^2$ for $\|\mathbb A_nv_n-\mathbb A v_n\|^2$ ? Because, in other hand, I don't see how get the claim. –  FASCH Aug 7 '13 at 20:35