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.
 A: 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).
