Let be Banach (more generally Fréchet) spaces and let be continuous linear maps. Suppose that is surjective and is compact. Prove that has closed image and .

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# Finite codimension

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2 thoughts on “Finite codimension”

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Let be Banach (more generally Fréchet) spaces and let be continuous linear maps. Suppose that is surjective and is compact. Prove that has closed image and .

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The proof for Banach spaces: Write , , and . and are Banach spaces, is bounded below, is compact and . We shall show that has closed range and . By the closed range theorem this imples that has closed range, and then . Thus .

We prove first that is finite-dimensional. To see this, note that , so is compact and bounded below. Then has closed range and is invertible and compact, hence .

Since is finite-dimensional, it has a closed complement in . To prove that has closed range it suffices to show that is bounded below since clearly and bounded below operators have closed range. Suppose that is not bounded below, then there exists a sequence of unit vectors in such that . Since is compact, by passing to a subsequence we may assume that for some . Then , which implies that since is closed. Thus for some , and . This implies that as is bounded below. In particular, . However, , so , and we reach contradiction since . Therefore is bounded below, and is closed.

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The proof for Banach spaces: Write , , and . and are Banach spaces, is bounded below, is compact and . We shall show that has closed range and . By the closed range theorem this imples that has closed range, and then . Thus .

We prove first that is finite-dimensional. To see this, note that , so is compact and bounded below. Then has closed range and is compact and invertible by the Open Mapping Theorem, hence .

Since is finite-dimensional, it has a closed complement in . To prove that has closed range it suffices to show that is bounded below since clearly and bounded below operators have closed range. Suppose that is not bounded below, then there exists a sequence of unit vectors in such that . Since is compact, by passing to a subsequence we may assume that for some . Then , which implies that since is closed. Thus for some , and . This implies that as is bounded below. In particular, . However, , so , and we reach contradiction since . Therefore is bounded below, and is closed.

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