# 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 $\dim \mathrm{im}(T+S)<\infty$.

## 2 thoughts on “Finite codimension”

1. 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 invertible and compact, 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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2. 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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