I have a simple question but so far google has failed me.
Suppose $A$ is a bounded compact and injective linear operator. Suppose $m$ is some function in a space $M$ and in the range of $A$ and so there exists a function $f$ so that $Af=m$. Suppose that $R_{\epsilon}$ is some regularized inverse of $A$ with regularization parameter $\epsilon$ for example this could be the Tikhonov regularized inverse or the spectral-cut off inverse. What I want to know is whether $||(AR_{\epsilon}-I)m||$, where $I$ is the identity operator, converges to zero uniformally in the space of $m$ (as $\epsilon$ goes to zero with $\epsilon$ not dependent on $m$)? Standard results about say Tikhonov regularization imply convergence for a given $m$ but the key here is that I need uniform convergence. If this isn't true for the regularization schemes I mentioned are there ones for which it is true? What if I restrict the space $M$ so that all functions in that space are say, uniformaly Holder continuous?
Any help hugely appreciated.
This is not possible. You want to show that for all $\sigma>0$ there is $\epsilon_0>0$ such that $$ \| (AR_\epsilon-I)m\|_M \le \sigma $$ for all $m\in M$ and $\epsilon\in (0,\epsilon_0)$. This inequality depends on the norm of $m$, hence it is only satisfied if $AR_\epsilon-I=0$, which is impossible.
A better version would be: for all $\sigma>0$ there is $\epsilon_0>0$ such that $$ \| (AR_\epsilon-I)m\|_M \le \sigma \|m\|_M $$ for all $m\in M$ and $\epsilon\in (0,\epsilon_0)$. This is equivalent to $AR_\epsilon \to I$ in the operator norm. Since $A$ and $AR_\epsilon$ is compact, this is only possible if $I$ is compact, which is equivalent to finite-dimensionality of $M$.