7

Let $X$ be a complex Banach space of infinite dimension. Does there exist a finite dimensional subspace of $X$ of arbitrary (finite) dimension which is complemented by a projection of norm 1?

Rudy the Reindeer
  • 46,359
krosprodakt
  • 71
  • 1

1 Answers1

-2

I am not entirely sure what you mean by "which is complemented by a projection of norm 1", but I suppose you mean: If $Y$ is a subspace of $X$, is there a non-trivial projection $P$ such that $Y\oplus P(X) = X$. If this is not what you mean, then I am sorryt for the misunderstanding.

There exists such a projection if , and only if, the is a projection $\hat P$ such that $\hat P (X) = Y$, because then $P$ can be chosen as $1-\hat P$.

To find out for which subspaces $Y$ this is true, take the map $x \mapsto y$, where $y$ is the unique vector out of $Y$ such that $x$ can be written as $x=y+u$, for some $u$ in a complementary subspace. This map is a projection if and only if it is continuous, but now it is easy to see that this is true if and only if $Y$ is a closed subspace.

Hence for all finite dimensional subspaces and for all infinite dim. subspaces which are closed the answer to your question is yes, otherwise no.

fk44
  • 107
  • 3
  • 2
    A subspace $Y$ of $X$ is said to be complemented if it is the image of a continuous projection $P: X \to X$. By Hahn-Banach every finite-dimensional subspace is complemented, but if you estimate the norm you get for a $P$ obtained this way, you can't do better than $|P| \leq \dim{Y}$. The question is: Is there for any $n$ an $n$-dimensional subspace $Y$ of $X$ admitting a projection $P$ with $|P| = 1$? – t.b. May 04 '12 at 12:36
  • This is more or less contained in t.b.'s comment, but just to state it in plain words: the last sentence is demonstrably false, as there are examples of infinite dimensional closed subspaces that are not complemented (for example, $c_0$ is not complemented in $\ell^\infty$). – Martin Wanvik May 04 '12 at 13:22
  • In case anyone else was wondering, $c_0$ is the subspace of sequences with limit $0$. – joriki May 04 '12 at 15:31
  • @joriki: in case you're interested two proofs of this fact (called Phillips's lemma) are here. – t.b. May 04 '12 at 17:42