Let $X$ be a Banach space. It is well known that an operator $T\in B(X)$ is Fredholm if and only if $\pi(T)$ is invertible in the Calkin algebra $B(X)/K(X)$.

Now suppose that the invertible elements in $B(X)/K(X)$ are dense. Can we conclude that Fredholm operators are dense in $B(X)$?

Note that there exist spaces for which every operator $T\in B(X)$ is Fredholm with index 0.


Yes, if the invertible elements in $B(X)/K(X)$ are dense then the Fredholm operators are dense.

Let $x \in B(X)$ and consider $\pi(x) \in B(X)/K(X)$. By density of the invertible elements $I$ in $B(X)/K(X)$, there exists a (Cauchy) sequence $y_n \in I$ such that $y_n \rightarrow \pi(x)$ as $n \rightarrow \infty$. Using an argument similar to the one that shows that the quotient of a Banach space by a closed subspace is a Banach, we can obtain a Cauchy sequence $x_n \in B(X)$ such that $\pi(x_n) = y_n$. Note that the $x_n \in B(X)$ are Fredholm since $\pi(x_n) \in I$. Completeness of $B(X)$ then implies that $x_n$ converges to some $x_\infty \in B(X)$ which must satisfy $\pi(x_\infty) = \pi(x)$. Thus, there exists some Fredholm operator $k \in K(X)$ such that $x_\infty + k =x$. It follows that the sequence $x_n + k$ is a sequence of Fredholm operators that converges to $x$.

Note that this proof doesn't use anything special about Fredholm operators but rather follows simply from properties of Banach spaces and their quotients.

Also, the converse statement is straightforward to show.

  • $\begingroup$ There are spaces $X$ for which the Calkin algebra is isomorphic to the algebra of continuous functions on the Cantor set. Take some non-trivial projection $p$ in that algebra and lift it to $B(X)$. How can we approximate it by Fredholm operators? $\endgroup$ – user512365 Apr 18 '17 at 22:56
  • $\begingroup$ @RobertBarg: in response to why we can take the $x_n$ to be Cauchy, see for example the argument in the proof of theorem 3.1 on p. 7 of people.math.gatech.edu/~heil/6338/summer08/section6a.pdf Regarding your possible counter-example, are the invertibles dense in that Calkin algebra? $\endgroup$ – Maxwell Stolarski Apr 18 '17 at 23:24
  • $\begingroup$ Yes, if a space is zero dimensional, then invertibles are dense in $C(X)$. $\endgroup$ – user512365 Apr 19 '17 at 8:47
  • $\begingroup$ I think it is okay --- every projection $p$ can be approximated by invertible hence by image of a Fradholm operator. $\endgroup$ – user512365 Apr 19 '17 at 9:15

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