Does there exist a surjective compact operator $T:l^{\infty} \to l^{\infty}$ ?

Even though this might be tagged as a repeat question, i still have some doubts that i would like to clarify.

My attempt:

$l^{\infty}$ is a Banach space and T being a compact operator is bounded. Thus by the Open mapping theorem $T^{-1}$ is bounded.

Further since $T$ is compact and $T^{-1}$ is bounded therefore $T^{-1}T =I$ is compact in $l^{\infty}$.

That would then be equivalent to saying that every closed unit ball in $l^{\infty}$ is compact which is impossible as $l^{\infty}$ is infinite dimensional.

So, no surjective compact operator exists on $l^{\infty}$.

Even though i have tried to write an argument for this question I am not very sure about whether this is correct or not.

First, can i assume the injectivity of $T$ for the application of the Open mapping theorem? If not,then the inverse may not always exist,so in that case how will i argue the non existence of a surjective compact operator?

Second,I have seen similar questions posted here, wherein,$T^{-1}T =I$ is compact in $l^{\infty}$ implies that every closed unit ball in $l^{\infty}$ is comopact. I don't quite see how that follows ?


1 Answer 1


There is no such operator. Your proof works when $T$ is injective, but you are right that it has a gap when it isn't since in that case "$T^{-1}$" is not a function but a one-to-many mapping, and none of our results about bounded operators apply to such mappings.

Instead, use the original statement of the open mapping theorem: a surjective linear operator is open, i.e. it maps open sets to open sets. So if we let $B$ be the open unit ball of $\ell^\infty$ and $\bar{B}$ the closed unit ball, then we have that $T(B)$ is open and $T(\bar{B})$ is precompact. Therefore $T(B)$ is a nonempty, open, precompact open set, so it contains an open ball whose closure is compact. As you know, this is not possible in an infinite dimensional normed space.

  • $\begingroup$ Thanks, though what is your view about the second question assuming that $T^{-1}$ exists ? $\endgroup$
    – johny
    Sep 5, 2015 at 6:14

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