An self-adjoint operator between infinite-dimensional Hilbert spaces $T:H\rightarrow K$ is said to be compact if $T(B)$ is precompact (i.e. $\bar{T(B)}$ is compact),$\forall B\subset H$ bounded.

By saying $T$ is invertible, I mean there exists $T^{-1}:K\rightarrow H$ such that the composition of operators $T\circ T^{-1}=id_{rangeT}; T^{-1}\circ T=id_{domainT}$. $id_\bullet$ is the identity operator of the space. The operator $T,T^{-1}$ does not have to be surjective, that is $T(H)$ can be a strict subset of $K$, so is $T^{-1}$.

(1)Is there a characterization (sufficient and necessary ,what I have in mind is some spectral condition on $T$) for those invertible operators whose inverse $T^{-1}$ is also compact?

(2)What if $H,K$ are both Banach spaces(with infinite dimensions)? Are the charcterization in (1) still correct?

(3)What if $T$ is not self-adjoint? Are the charcterization in (1) still correct?

(4)What if I only require the left-inverse $T^{-1}_{left}$ (i.e. $T^{-1}_{left}\circ $T$=id_{H}$ but not necessary right-inverse) to be compact?

(5)Is there an example such that $T$ is invertible and compact, but $T^{-1}$ is not a compact operator?


  • $\begingroup$ In your first sentence I suppose you mean "infinite-dimensional"? For finite-dimensional spaces all operators are compact. $\endgroup$ Mar 20, 2017 at 19:34
  • $\begingroup$ @NateEldredge Yes, corrected. For finite dimensions, they are no more than matrices. $\endgroup$
    – Henry.L
    Mar 20, 2017 at 19:36
  • 1
    $\begingroup$ It depends on exactly what you mean by "invertible". Can you clarify? If it includes "surjective" then the answer is "there are no invertible compact operators". $\endgroup$ Mar 20, 2017 at 19:36
  • $\begingroup$ @NateEldredge I added definition, and also propose a new subproblem.$T$ does not have to be surjective. Thanks for your attention! $\endgroup$
    – Henry.L
    Mar 20, 2017 at 19:42
  • $\begingroup$ I'm wondering the same kinds of things. If you have a better answer now, I would really apretiatte to hear about it $\endgroup$
    – The Ju
    Oct 11, 2021 at 18:11

1 Answer 1


No such operators exist. If $T : H \to K$ where $H,K$ are Banach spaces and $K$ is infinite-dimensional, then $T$ cannot be both compact and surjective. Proof: if it were, then by the open mapping theorem it is an open map. So if $B$ is the open unit ball of $H$ then $T(B)$ is open as well as precompact, which is impossible by the Riesz lemma (i.e. $K$ is not locally compact).

For (1), even if you do not assume $T$ compact, you are asking for $T^{-1}$ to be compact and surjective. For (4), you are asking for the left inverse to be compact and surjective.

Likewise, if $H$ is infinite-dimensional it is not possible for $T$ to be compact and have a bounded left inverse $S$. (In this case the issue is that $T(B)$ is precompact, hence so is $S(T(B)) = B$.)

  • $\begingroup$ Thanks for the quick a nice response ! I guess what I really want to formulate is $id_{domain T}$, if invertibility is defined in that way, how will the answer change? Sorry for the confusion and I have updated the OP again. $\endgroup$
    – Henry.L
    Mar 20, 2017 at 20:04
  • $\begingroup$ Ah, you want to allow unbounded operators now? That wasn't clear. I think the relevant term to search is "compact resolvent". $\endgroup$ Mar 20, 2017 at 20:05
  • $\begingroup$ Thanks for the pointer, I was not very familiar with the associated terms in functional analysis, thanks much! $\endgroup$
    – Henry.L
    Mar 20, 2017 at 20:08

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