This post here states that if $E$ is an infinite dimensional space and if $T$ is an injective, bounded,non surjective opertor with closed range in $E$, then $T$ cannot be approximated in operator norm by invertible bounded linear operators on $E$. Can anyone tell why?

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    $\begingroup$ We have $\|Tx\| \ge c \|x\|$ for some $c > 0$ and all $x \in E$ . Assume that there are invertible operators $T_n$ such that $T_n \to T$ wrt the operator norm. For all sufficiently large $n$ we then have $\|T_n x\| \ge c/2 \cdot \|x\|$ for all $x$, so $\|T_n^{-1}\| \le 2/c$. Thus, $(T_n^{-1})$ is a Cauchy sequence and hence convergent to an operator $S$. Now, one readily checks that $ST = TS = I$, so $T$ is invertible and hence surjective. $\endgroup$ Sep 30, 2018 at 12:47
  • $\begingroup$ Why is $(T_n^{-1})$ Cauchy? $\endgroup$ Sep 30, 2018 at 14:45
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    $\begingroup$ @M.González: We have $T_n^{-1} - T_m^{-1} = T_n^{-1} (T_m - T_n) T_m^{-1}$. $\endgroup$ Sep 30, 2018 at 15:31

1 Answer 1


If $E$ is a Banach space and $T:E\to E$ is an injective bounded operator with closed range $R(T)$, then there exists a number $\delta>0$ such that any bounded operator $S:E\to E$ with $\|S-T\|<\delta$ is also injective, has closed range and $\dim \frac{E}{R(S)} =\dim\frac{E}{R(T)}$.

The fact that $S$ is injective with closed range is easy: $T$ injective with closed range implies that $T$ is bounded below; i.e., there exists $C>0$ such that $\|Tx\|\geq C\|x\|$ for every $x\in X$. Thus if $\|S-T\|<C$, then $S$ is bounded below, hence injective with closed range.

To prove that $R(S)$ and $R(T)$ have the same codimension in $E$ is a bit more difficult. It can be seen as a consequence of the fact that the subset of semi-Fredholm operators $SF(E)$ is an open subset of the set of all bounded operators on $E$, and the index is constant in each connected component of $SF(E)$. See this link.


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