Let $X$ be a Banach space and $N,R$ two subspace of $X$. Suppose further that $X=N\oplus R$ (i.e., $N\cap R=\{0\}$ and $X=N+R$). If $N$ is finite dimensional, then does $R$ have to be closed?

There is a famous "weaker" resut as follows:

Let $X,Y$ be Banach spaces and $T\in L(X,Y)$. If there exists some finite dimensional subspace $N$ such that $Y=T(X)\oplus N$, then $T(X)$ is closed.

This result is very useful in the theory of Fredholm operators. So I guess the previous result is NOT true. Otherwise, the textbook will prove the first one instead of the second one. I'm trying to construct a counterexample but can't find it.


The canonical example would be to take an unbounded functional $f:X\to\mathbb C$. Then for any $x\in X$ with $f(x)\ne0$, you have $$ X=\ker f\oplus \mathbb C x.$$ Because $f$ is unbounded, $\ker f$ is not closed (it's actually dense).

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    $\begingroup$ I think you mean $X = \ker(f)\oplus \mathbb{C}x$. $\endgroup$ – Michael L. Nov 15 '17 at 21:38
  • $\begingroup$ Indeed. Thanks! $\endgroup$ – Martin Argerami Nov 15 '17 at 21:50

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