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I'm studying about statistical learning theory and I bumped into the following statement in my study material:

Claim:

If a linear normed space $E_1$ contains bounded noncompact sets, then the inverse operator $A^{-1}$ for an absolutely continuous operator $A:E_1\rightarrow E_2$ ($E_2$ also linear normed space) need not be continuous.

Proof:

Consider a bounded noncompact set in linear normed space $E_1$. Select in this set an infinite sequence:

$$f_1, f_2, ...,\;(||f_j||_{E_1} \leq c \in \mathbb{R})\;\;\;\;\;\;\;\;\;\;\;\;(1)$$ such that no subsequence of it is convergent. An infinite sequence $$Af_1, Af_2, ...,\;\;\;\;\;\;\;\;\;\;\;\;(2)$$ from which convergent subsequence may be selected (since $A$ is absolutely continuous) corresponds in $E_2$ to this sequence. If the operator $A^{-1}$ were continuous, then a convergent subsequence

$$f_{k}, f_{k+1}, ...,\;\;\;\;\;\;\;\;\;\;\;\;(3)$$

could be extracted from $(1)$ in space $E_1$. This however contradicts the choice of sequence $(1)$.

The above was somewhat clear to me, but I wasn't 100% sure, so I wanted to try prove this claim myself and then verify here if I understood it or not.

UPDATE: I have added my own attempt as a possible answer.

My question was: is my proof valid? Or did I misunderstand something?

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  • $\begingroup$ Looks good to me. You should say that $A$ is assumed to be injective (which allows its inverse to exist). $\endgroup$ – DanielWainfleet Feb 4 '19 at 1:32
  • $\begingroup$ @DanielWainfleet thank you very much for your help =) $\endgroup$ – jjepsuomi Feb 4 '19 at 10:18
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    $\begingroup$ Other names for absolutely continuous operator are completely continuous operator and compact operator. $\endgroup$ – DanielWainfleet Feb 4 '19 at 15:35
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Claim:

If a linear normed space $E_1$ contains bounded noncompact sets, then the inverse operator $A^{−1}$ for an absolutely continuous one-to-one operator $A:E_1→E_2$ ($E_2$ also linear normed space) need not be continuous.

Proof:

Denote $\rho_k := ||f_k-f_{k+1}||_{E_1}$ and $r_k := ||Af_k-Af_{k+1}||_{E_2}$. Since $A$ is absolutely continuous, for arbitrary $\delta >0$ we can always find $\varepsilon>0$ for which:

$$\sum_{k} r_k <\delta \;\;\;\text{when}\;\;\; \sum_{k}\rho_k <\varepsilon,$$

for some subsequence $f_k, f_{k+1}, ...$ of $(1)$. That is, $r_k \to 0$ and $A f_k \to A f_0$, so $(2)$ converges to $Af_0$. In other words, due to absolutely continuity of $A$ we can always select a subsequence from $(1)$ which has a convergent image.

Lets now assume that we have a continuous inverse operator $A^{-1}$. This means that for arbitrary $k$ and $\delta_1>0 $, we can always find a $\varepsilon_1>0$ for which:

$$ \rho_k <\delta_1 \;\;\;\text{when}\;\;\;r_k <\varepsilon_1, $$

which means that $(1)$ has a convergent subsequence. But this is a contradiction since $(1)$ is an infinite divergent sequence $\blacksquare$

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