3
$\begingroup$

I'll begin by writing down the definitions I'm using, to avoid confusion.

Let $X$ be a Banach space and let $Y$ be a subspace of $X$. We say that $Y$ is complemented in $X$ if there exists a linear continuous operator $P : X \to Y$ such that Im$(P) = Y$ and $P \circ P =P$. Equivalently, $Y$ is complemented if $Y$ is closed and there is a closed subspace $W$ of $X$ such that $X = Y \oplus W$.

My question is this: Let $X$ be a Banach space, $Y$ a complemented subspace of $X$ and $Z$ a (closed) subspace of $X$ isomorphic to $Y$. Must $Z$ be complemented in $X$ as well?

The natural idea would be to take $P : X \to Y$ a projection and $T : Y \to Z$ an isomorphism, and define $Q = T \circ P : X \to Z$. The problem I'm having is showing that $Q(z)=z$, for all $z \in Z$, or $Q^2(x)=Q(x)$, or all $x \in X$. It's enough that $P(z)=T^{-1}(z)$, but I don't know if I can guarantee this - and this might be the clue to finding a counterexample.

Thank you in advance.

$\endgroup$
3
$\begingroup$

A simple counterexample is $X=\ell_\infty\oplus c_0$. The second summand, $c_0$, is complemented in $X$. But the copy of $c_0$ sitting in $\ell_\infty$ is not complemented in $\ell_\infty$, hence not complemented in $X$.

$\endgroup$
  • $\begingroup$ I don't understand your argument. So, if $S \subset \ell_\infty \oplus c_0$ was a complement for $\{0\} \oplus c_0 \subset c_0 \oplus \ell_\infty$, are you suggesting that $\pi_2(S)$ would be a complement for $c_0 \subset \ell_\infty$ where $\pi_2$ is the factor projection $c_0 \oplus \ell_\infty$? I don't even see why $\pi_2(S)$ would be closed.... $\endgroup$ – Mike F Nov 18 '14 at 1:30
  • $\begingroup$ Indeed, if $T : X \to Y$ is a closed operator, then graph $\Gamma \subset X \times Y$ of $T$ would have $\pi_2(\Gamma) = ran(T)$, which may or may not be closed in $Y$. $\endgroup$ – Mike F Nov 18 '14 at 1:33
  • 1
    $\begingroup$ @MikeF No. It's easiest to see using the projection, I think. If there were a continuous projection $P \colon X \to X$ with image $c_0 \oplus \{0\}$, the restriction to $\ell_{\infty} \oplus \{0\}$ would induce a continuous projection $Q \colon \ell_{\infty} \to \ell_{\infty}$ with image $c_0$. Thus $c_0$ would be complemented in $\ell_{\infty}$ (which it isn't, so $P$ doesn't exist). $\endgroup$ – Daniel Fischer Aug 12 '18 at 12:47

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.