The following is an exercise from Halmos book "A Hilbert space problem book" :
Exercise: If $H$ and $K$ are Hilbert spaces, and if $A$ is a bounded linear transformation that maps $H$ one to one and onto $K$, then $A$ is invertible.
He gives the following solution for this:
I do not know why he consider the operator $0$ in the role of $A^*$. Clearly $0$ is not bijective, so why does he state this part? Please help me. Thanks.
For operator $A^*=0$ the condition $\Vert A^*g\Vert=1$ is always false. By the fundamental rules of logic False implies anything and in particular that $\Vert g\Vert\leq 1/\delta$.
One could say for example that $\Vert A^*g\Vert=1\implies\pi=10^5$ and would be absolutely correct.