MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Recall that a closed subspace $Y$ of a Banach space $X$ is weakly complemented if the set $$Y^{\bot}:= \{ f\in X^\*| f(y) = 0 \forall y\in Y\}$$ is a complemented subspace of $ X^*$. For example, $c_0$ is a weakly complemented subspace of $l_{\infty}$.

Question: Is there a Banach space $X$ such that there is a weak${}^*$-closed subspace $Y$ which is weakly complemented but not complemented in $X$.

share|cite|improve this question
up vote 7 down vote accepted

No. You get $Y^{**}=Y^{\perp\perp}$ complemented in $X^{**}$ and $Y$, being a dual space, is norm one complemented in $Y^{**}$.

share|cite|improve this answer
Dear Professor Johnson, that is a nice observation. Thank you. – Denis Poulin Mar 7 '12 at 2:28
As an aside, this would seem to show that one can relax the hypotheses that $X$ is a dual space and $Y$ a weakly complemented, weak-star closed subspace to: $X$ is complemented in its second dual, $Y\subseteq X$ is complemented in its second dual, and $Y^{**}$ is complemented in $X^{**}$. (This might be of interest when $X$ is, for instance, the predual of a von Neumann algebra.) – Yemon Choi Mar 7 '12 at 2:57
In fact, we don't even need the assumption that $X$ is complemented in its second dual. – Yemon Choi Mar 7 '12 at 2:58
Right, Yemon: If X is complemented in any dual space, then it is complemented in its bidual. – Bill Johnson Mar 7 '12 at 21:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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