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.