# Closure of the space of Fredholm operators

Let $$X,Y$$ be two Banach spaces. A bounded operator $$A$$ is Fredholm if $$\ker A$$ and $$\mathrm{coker} A$$ are finite dimensional. Denote by $$Fred(X,Y) \subset \mathcal{L}(X,Y)$$ the space of Fredholm operators endowed with the subspace topology (hence the operator norm).

It is well known that $$Fred(X,Y)$$ is an open subset of $$\mathcal{L}(X,Y)$$.

Q. What is the closure of $$Fred(X,Y)$$ in $$\mathcal{L}(X,Y)$$? ($$X,Y$$ infinite dimensional)

I would be happy to have an answer even for the case $$X=Y$$ separable Hilbert space.

• For the closure of invertible operators in a Hilbert space, see Bouldin, Proc AMS, and reference therein: ams.org/journals/proc/1990-108-03/S0002-9939-1990-1000147-9/…
– YCor
Jul 14, 2021 at 16:07
• In particular, for a separable Hilbert space, a bounded operator $T$ is in the norm closure of invertible elements if and only either it has non-closed range, or the cardinals $\dim(\mathrm{Ker}(T))$ and $\dim(\mathrm{Ker}(T^*))$ are equal.
– YCor
Jul 15, 2021 at 7:31
• If $X$ is a separable space for which every bounded self-operator is scalar+compact (this exists by Argyros-Haydon), then the norm closure of the open subset invertible operators is the whole space of operators. This applies more generally in every $X$ for which every bounded operator has spectrum of empty interior (but I don't know of any further example).
– YCor
Jul 15, 2021 at 11:13

$$\DeclareMathOperator\Ker{Ker}$$

Let $$\mathcal{H}$$ be a separable Hilbert space. Then a bounded operator $$T$$ is not in the norm closure of Fredholm operators iff either $$T$$ or $$T^*$$ has a finite-dimensional kernel and has a closed image of infinite codimension.

(Note that in particular, the only operators that are in the closure of Fredholm operators but not in the closure of invertible operators, are Fredholm operators of nonzero index.)

Proof: Let $$T$$ not be in the norm closure of Fredholm operators. So it is not in the norm closure of invertible operators. By the Bouldin result mentioned previously, $$T$$ has a closed image and $$\dim(\Ker(T))$$ and $$\dim(\Ker(T^*))$$ differ. If both are finite, $$T$$ is Fredhlom. So one is finite and the other is infinite.

Conversely, suppose that $$T$$ satisfies the condition. Up to adjoint, we can suppose that $$T$$ has closed range, has finite-dimensional kernel and infinite-dimensional cokernel. So there exists a block-decomposition under which we can write $$T$$ as $$T=\begin{pmatrix} A_0 & 0 \\ 0 & 0\end{pmatrix},$$ where the right-hand column is finite-dimensional, $$A_0$$ is invertible, and both rows are infinite-dimensional. Let $$T'$$ be norm-close to $$T$$: write it as $$T'=\begin{pmatrix} A & b \\ C & d\end{pmatrix},$$ so $$A$$ being close to $$A_0$$, it is invertible. Suppose by contradiction that $$T'$$ is Fredholm. So there exists a bounded operator $$S$$ such that $$TS-I$$ has finite rank. Write $$S=\begin{pmatrix} E & F \\ g & h\end{pmatrix}$$ in the transpose block decomposition. (Small letters emphasize the fact that they have either finitely many rows or columns.) Then $$T'S=\begin{pmatrix} AE+bg & AF+bh \\ CE+dg & CF+dh\end{pmatrix},$$ this time in the symmetric block decomposition corresponding to the row decomposition of $$T$$ (hence with both blocks infinite-dimensional. Hence $$AF+bh$$ has finite rank. Since $$h$$ has finite rank, we deduce that $$AF$$ has finite rank. Since $$A$$ is invertible, we deduce that $$F$$ has finite rank. But then $$CF+dh$$ has finite rank, which contradicts that $$CF+dh-I$$ has finite rank.

• Thank you, I am convinced by your proof but the assumption does not look straightforward to me. How did you go from Bouldin's theorem, which gives conditions in terms of the spectral projector of $|T|, |T^*|$, to restate it in terms of closedness of the range and dimension of kernel/cokernel? Jul 15, 2021 at 9:22
• I guess that, by Bouldin's result, the characterization extends to arbitrary Hilbert spaces as follows (using Bouldin's notation): $T$ belongs to the closure of compact operators iff $\textrm{essnull} T$ and $\textrm{essnull} T^*$ are equal or both finite. Jul 15, 2021 at 9:22
• @WarlockofFiretopMountain see the last few lines in the first page (=page 721) of the linked (1990 Proc AMS) paper.
– YCor
Jul 15, 2021 at 10:51
• @MikaeldelaSalle you probably mean Fredholm rather than compact.
– YCor
Jul 15, 2021 at 10:51
• @YCor Sure, this is what I meant. Jul 15, 2021 at 19:17