Finite codimension

Let E,F be Banach (more generally Fréchet) spaces and let T,S:E\to F be continuous linear maps. Suppose that T is surjective and S is compact. Prove that T+S has closed image and \dim \mathrm{im}(T+S)<\infty.

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Finite codimension

One thought on “Finite codimension

  1. alejosalvatore says:

    The proof for Banach spaces: Write Y=E^*, X=F^*, A=T^* and K=-S^*. X and Y are Banach spaces, A:X\to Y is bounded below, K:X\to Y is compact and (T+S)^*=A-K. We shall show that A-K has closed range and \dim N(A-K)<\infty. By the closed range theorem this imples that T+S has closed range, and then \dim (F/R(T+S))^*=\dim R(T+S)^\perp = \dim N(A-K)<\infty. Thus \dim (F/R(T+S))<\infty.

    We prove first that H=N(A-K) is finite-dimensional. To see this, note that A|_H=K|_H, so A|_H is compact and bounded below. Then A|_H has closed range and A|_H:H\to A(H) is compact and invertible by the Open Mapping Theorem, hence \dim (H)<\infty.

    Since H is finite-dimensional, it has a closed complement V in X. To prove that A-K has closed range it suffices to show that (A-K)|_V is bounded below since clearly R(A-K)=R((A-K)|_V) and bounded below operators have closed range. Suppose that (A-K)|_V is not bounded below, then there exists a sequence (v_n)_n of unit vectors in V such that (A-K)v_n\to 0. Since K is compact, by passing to a subsequence we may assume that Kv_n\to y for some y\in Y. Then Av_n\to y, which implies that y\in R(A|_V) since R(A|_V) is closed. Thus y=Av for some v\in V, and A(v_n-v)\to 0. This implies that v_n\to v as A is bounded below. In particular, \|v\|=\lim_{n\to \infty} \|v_n\|=1. However, (A-K)v=\lim_{n\to \infty} (A-K)v_n=0, so v\in H, and we reach contradiction since H\cap V=0. Therefore (A-K)|_V is bounded below, and R(A-K) is closed.

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