I'm currently reading a book about inverse scattering theory and in this book there is a section about ill-posed problems and there's a proof I'm not completely sure I understand. There might be need of this definition if it's going to make sense
$\textbf{Definition}$ Let $X$ and $Y$ be normed spaces and let $A:X\rightarrow Y$ be an injective bounded linear operator. Then a family of bounded linear operators $R_{\alpha}:Y\rightarrow X,\ \alpha >0$, with the property of pointwise convergence
$$\underset{\alpha\rightarrow 0}{\text{lim}}\ R_{\alpha}A\psi=\psi$$
for all $\psi \in X$ is called a regularization scheme for the operator $A$. The parameter $\alpha$ is called the regularization parameter.
This also means that $R_{\alpha}f\rightarrow A^{-1}f,\ \alpha\rightarrow 0$, for all $f\in A(X)$, where $A(X)$ is the range of $X$.
The theorem which proof I don't really get is the following
$\textbf{Theorem}$ Let $X$ and $Y$ be normed spaces, let $A:X\rightarrow Y$ be a compact linear operator, and let $\text{dim}\ X=\infty$. Then for a regularization scheme the operators $R_{\alpha}$ cannot be uniformly bounded with respect to $\alpha$ and the operators $R_{\alpha}A$ cannot be norm convergent as $\alpha\rightarrow 0$.
Here comes the proof. The first statement I do understand so I'm going to leave that out. The prood of the 2nd statement (norm convergence) goes like this:
Assume that we have norm convergence. Then there exists $\alpha > 0$ such that $||R_{\alpha}A-I||<1/2$. Now for all $f\in A(X)$ we can estimate
$$||A^{-1}f||\leq ||A^{-1}f-R_{\alpha}AA^{-1}f||+||R_{\alpha}f||\leq \frac{1}{2}||A^{-1}f||+||R_{\alpha}||\ ||f||,$$
whence $||A^{-1}f||\leq 2||R_{\alpha}||\ ||f||$ follows. Therefore, $A^{-1}:A(X)\rightarrow X$ is bounded and we have the same contradiction as in the first statement (contradiction is that $\text{dim}\ X=\infty$).
I don't really uderstand the inequalities. I understand that $||R_{\alpha}A-I||<1/2$ for some $\alpha$ since $R_{\alpha}A\rightarrow A^{-1}A=I$ as $\alpha\rightarrow 0$ hence if $\alpha$ is fixed we can get within a certain tolerance from $I$. Furthermore, my idea is that we can look at the inequality
$$||A^{-1}f-R_{\alpha}AA^{-1}f||=||A^{-1}f-R_{\alpha}f||,$$
Since $A^{-1}A=I$, but I can't quite seem to get further from here.
Any hints are highly appreciated!