Evans - p298 - Lax Milgram

Let $A:H\to H$ be a linear operator on the Hilbert space $H$.

He shows that for some $\alpha,\beta>0$ that: $$\beta \|u\|\leq \|Au\|\leq \alpha \|u\|$$

Why does this immediately give me that $A$ is one to one, and that the range of $A$ is closed in $H$?

Is this what was in mind for one-to-one?

Say $Au=Av$ then $Au-Av=0=A(u-v)$ where $\beta \|u-v\| \leq 0 \leq \alpha\|u-v\|$, but $\beta >0$ and a norm is positive, contradiction, hence $Au\ne Av$.

I have no idea how to show that this would make $A$ a closed operator.

  • $\begingroup$ this only can be true if $\beta\neq 0$, otherwise for finite-dimensional Hilbert spaces if $\beta=0$ then zero is a singular value of $A$ and hence an eigenvalue of $A$. $\endgroup$ – Masacroso Jul 13 '17 at 9:39
  • $\begingroup$ @Masacroso Actually I wrote the wrote thing there. Thanks though, it was meant to be $\alpha,\beta>0$. $\endgroup$ – F.White Jul 13 '17 at 9:46
  • $\begingroup$ my previous comment was innacurate (it was too late when I wrote it). The correct way to express it is saying that $$\beta=\sup\{\lambda\in[0,\infty):\lambda\|v\|\le\|Tv\|\,\forall v\in V\}$$ because zero is ever a lower bound of $\|Tv\|$. $\endgroup$ – Masacroso Jul 13 '17 at 20:08

Yes, this is what is meant by one-to-one. Well assume you have a sequence $\{y_n\}$ in $R(A)$ which converges to $y \in H$. Since $A$ is injective there is a sequence $\{x_n\}$ in $H$ with $Ax_n=y_n$. Now $||x_n-x_m|| \le ||Ax_n-Ax_m|| /\beta \to 0$, since $\beta>0$ is fixed and $\{y_n\}$ is Cauchy. Hence $\{x_n\}$ is Cauchy and therefore $x_n \to x$ fo some $x$ because of completeness. By boundedness (=continuity) of $A$ it follows that $y = \lim y_n=\lim A x_n = Ax$, hence $y \in R(A)$.


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