# Lax - Milgram. Problem with the proof.

I'm learning Lax - Milgram theorem (from L.C. Lawrance book) and I have a problem with understanding the proof.

Assume $B\colon H\times X \rightarrow \mathbb{R}$ ($H$ - Hilbert space) is bilinear and

$|B(u,v)|\leq \alpha\|u\| \|v\|$ for $u,v \in H$,

$\beta\|u\|^{2}\leq B(u,u)$ for $u\in H$.

Let $f\colon H\rightarrow \mathbb{R}$ be bounded linear functional on $H$. Then there exists unique element $u\in H$ such that $B(u,v)=(f,v)$ for all $v\in H$.

So I follow the standard pattern and introduce an operator $A:H\rightarrow H$ as in Evans book. I easily obtain that $\beta \|u\|\leq\|Au\|$. I make a use of the latter inequality to prove that $A$ is injective. In the book however there is a statement that the inequality implies that Range(A) is closed. That is the part I don't get.

I assume that $A$ is linear, if we suppose that $A$ is also continuous (or bounded, equivalently) we can show that $\mathcal{R}(A)$ is closed.

Indeed, let $(v_n)_n \in \mathcal{R}(A)$ be a sequence such that $\lim_{n} v_n = v \in H$, we want to show that $v \in \mathcal{R}(A)$, i.e $A \overline{v} = v$ for some $\overline{v} \in H$. Since $(v_n)_n \in \mathcal{R}(A)$, for every $n$ there exists $u_n \in H$ such that $Au_n = v_n$; since $v_n$ is convergent, $(Au_n)_n$ is a Cauchy sequence in $H$ and the following inequality holds for every $m,n \in \mathbb{N}$: $$\beta\| u_m - u_n \| \le \| A(u_m-u_n) \|= \| Au_m - Au_n \| \xrightarrow[m,n ]{} 0$$

whence also $(u_n)_n$ is a Cauchy sequence. By the completeness of $H$, $(u_n)_n$ converges, say to $u \in H$. Finally,

$$v=\lim_{n} v_n = \lim_{n} Au_n \underbrace{=}_{\text{continuity}} A(\lim_{n} u_n )= Au.$$

Hence $v \in \mathcal{R}(A)$ and $\mathcal{R}(A)$ is closed (note that in an Hilbert space topological closedness is equivalent to sequential closedness).

• Why do you claim that $(Au_{n})$ is Cauchy? – zorro47 Mar 29 '17 at 14:39
• Because $Au_n = v_n$ (remeber $v_n$ is in the range of $A$) for every $n$ and $v_n$ is convergent, hence Cauchy. – GaC Mar 29 '17 at 14:41
• I was thinking of proving that $(u_{n})$ is convergent straight from the fact that $v_{n}=Au_{n}$ is convergent but it can not be done. Now, I understand the aim of using Cauchy condition and the inequality I mentioned before. – zorro47 Mar 29 '17 at 14:50
• There is one more doubt. We have that $\mathcal{R}(A)$ is closed, hence $A$ is surjective. Now, from the Riesz representation theorem there exists unique $w$ such that $f(v)=(w,v)$ for every $v$. Since $A^{-1}$ exists, so there is $u_{0}$ such that $Au_{0}=w$. $u_{0}$ is unique. On the oher hand, $B(u_{0},v)=(Au_{0},v)=(w,v)=f(v)$, so there is no need to show uniqueness, still it is a part of Evans proof. – zorro47 Mar 29 '17 at 16:54
• Uh, maybe I should look at the complete proof to answer your last question. I'm going to the library next week and take a look at Evans' proof. – GaC Mar 30 '17 at 13:07