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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.

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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).

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  • $\begingroup$ Why do you claim that $(Au_{n})$ is Cauchy? $\endgroup$
    – zorro47
    Commented Mar 29, 2017 at 14:39
  • $\begingroup$ Because $Au_n = v_n $ (remeber $v_n$ is in the range of $A$) for every $n$ and $v_n$ is convergent, hence Cauchy. $\endgroup$
    – GaC
    Commented Mar 29, 2017 at 14:41
  • $\begingroup$ 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. $\endgroup$
    – zorro47
    Commented Mar 29, 2017 at 14:50
  • $\begingroup$ 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. $\endgroup$
    – zorro47
    Commented Mar 29, 2017 at 16:54
  • $\begingroup$ 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. $\endgroup$
    – GaC
    Commented Mar 30, 2017 at 13:07

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