Let $X$ and $Y$ be some infinite dimensional Banach spaces. Let $T:X\longrightarrow Y$ be some compact linear operator. It is easy to understand that $T$ cannot be surjective: the Open Mapping Theorem due to Banach, states that surjective operators should be open, and it follows that the image of the unit ball $\mathbb{B}_{1,X}$ should contain an open ball $\{||y||<r\}$ for some $r>0$, so its closure can't be compact. Is it true that the image of $T$ cannot contain any closed subspace of infinite dimension?


2 Answers 2


Let $F\subset T(X)$ be a closed (in $Y$) subspace. Then $E = T^{-1}(F)$ is a closed subspace of $X$, and $T\lvert_E \colon E \to F$ is a compact surjective operator. Since $F$ is closed in $Y$, the open mapping theorem implies that $F$ is finite-dimensional.

So: the image of a compact operator cannot contain an infinite-dimensional Banach space.

  • $\begingroup$ Sorry, but why did $F$ being closed implied by the O.M.T. that $F$ is finite dimensional? $\endgroup$
    – user770533
    Jan 20 at 19:06

A compact operator has a closed range iff it has a finite dimensional range. Without loss of generality,we can assume that the range of A is closed,otherwise we can consider the restriction of A to E where E=T−1(F),In all cases the theorem assures that F is finite dimensional. to prove the theorem,consider the canonical map associated with A and the fact that the closedness of the range of the operator imply the invertibility of canonical map and hence the finite dimensionality of the range (the identity is compact).


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