3
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ I think you mean $X = \ker(f)\oplus \mathbb{C}x$. $\endgroup$ – Michael Lee Nov 15 '17 at 21:38
  • $\begingroup$ Indeed. Thanks! $\endgroup$ – Martin Argerami Nov 15 '17 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.