# Prove: the inverse of an absolutely continuous operator $A:E_1 \to E_2$ need not be continuous.

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.

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

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

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 $$\blacksquare$$