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Let $B$ be a Banach space, $H,K$ be closed subspaces and let $K$ be finite dimensional.

Suppose $B = H\oplus K$ and $T:B\to B$ is a bounded linear operator.

How do I show that $T(B)$ is closed $\iff$ $T(H)$ is closed ?

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1 Answer 1

up vote 8 down vote accepted

The internal sum of a closed subspace and a finite-dimensional subspace is closed, so if $T(H)$ is closed then $T(B) = T(H) + T(K)$ is closed. (See the solution to exercise 41 here or the comments below for a proof).

For the other direction, we can assume that $T$ is injective, otherwise we consider the factorization of $T$ over $B/\ker T = (H / (H \cap \ker T)) \oplus (K / (K \cap \ker T))$ (noting that $T$ and its factorization have the same image).

Suppose that $T(B)$ is closed. Then $T(H)$ is a subspace of $T(B)$ of finite codimension, so it has an algebraic complement $Z$, so $T(B) = T(H) \oplus Z$ as vector spaces. Since $Z$ is finite-dimensional, $Z$ is closed, so the linear operator $S \colon H \oplus Z \to T(B)$ defined by $S(h,z) = T(h) + z$ is a continuous linear bijection between Banach spaces, hence it is a homeomorphism by the open mapping theorem. Since $H$ is closed in $H \oplus Z$, we have that $T(H) = S(H,0)$ is closed in $T(B)$.

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"The internal sum of a closed subspace and a finite-dimensional subspace is closed" How to prove this? –  lee Jan 9 '13 at 23:44
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Let $C$ be a closed subspace and let $F$ be a finite-dimensional subspace. Consider the quotient map $\pi\colon X \to X/C$. Then $X/C$ is a normed space since $C$ is closed and the projection $\pi \colon X \to X/C$ is continuous. Since $\pi(F) \subset X/C$ is finite-dimensional, it is closed. It follows that $C + F = \pi^{-1}(\pi(F))$ is closed as a pre-image of a closed set under a continuous map. –  Martin Jan 10 '13 at 6:43
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Incidentally, if you had bothered to Google the exact sentence of your follow-up question, you'd have found this exercise sheet containing a full solution as a first hit... –  Martin Jan 10 '13 at 10:04
    
Thanks! I get the idea! –  lee Jan 11 '13 at 4:16
    
Very good! You're welcome. // I would recommend to be a bit more patient before you downvote an answer in which there are some details you would like to have explained further. For instance, you can check if someone has visited the site since you requested information by clicking on their user name and visiting their profile. You could have seen that I wasn't given a chance to answer your query (remember people here are from all over the world!). Many people react rather negatively to downvotes and, after all, you are asking for help... Anyway, I'm glad this has been cleared up. –  Martin Jan 11 '13 at 5:14

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