Prove that a linear and continuous operator admits inverse in Hilbert space

Question

Let $(H,(\cdot,\cdot))$ an Hilbert space and $A:H\rightarrow H$ a linear and continuous operator such that there exists $\alpha >0$ such that $$(Au,u)\geq \alpha \|u\|^2 \text{ for each } u\in H.$$

Prove that $A$ is invertible and $\|A^{-1}\|\leq {1\over \alpha}$.

I've already proved that $A$ is injective, regarding surgectivity, my professor suggested first to prove that $Im(A)$ is closed in $H$ and then that $Im(A)$ is dense.

I've thought to take a sequence $(v_h)_h\subset Im(A)$ such that there exists $v\in H$ such that $$v_h\rightarrow v \text{ in } H.$$

But now I can't prove that $v\in Im(A)$, and also I really would appreciate if someone could give me some hints to solve the other points.

Answer 1

Notice that $v_h=Au_h$ for some (actually a unique) $u_h$. Using the assumption with $u:=u_k-u_j$, we obtain after an application of Cauchy-Schwarz inequality that $(u_k)_{k\geqslant 1}$ is Cauchy.

For density, notice that if $v$ is orthogonal to the range of $A$, then $(Av,v)=0$.

For the norm inequality, use $(A(A^{-1}u),A^{-1}u)\geqslant\alpha (A^{-1}u,A^{-1}(u))$ and Cauchy-Schwarz inequality.