Intersection of complemented subspaces of a Banach space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:44:52Z http://mathoverflow.net/feeds/question/85492 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85492/intersection-of-complemented-subspaces-of-a-banach-space Intersection of complemented subspaces of a Banach space Pietro Majer 2012-01-12T14:29:49Z 2012-01-13T13:17:21Z <p>The following seems a very basic question in the theory of complemented subspaces of Banach spaces, but I was not able to find a reference, so I wish to ask it here.</p> <blockquote> <p><strong>Question.</strong> Let $X$ be a Banach space, and let $V$ and $W$ be complemented subspaces of $X$. Is it true that $V \cap W$ is a complemented subspace? If not, is it true under (nontrivial) additional assumptions?</p> </blockquote> <p>In the case of a Hilbert space $X$, where the answer is of course yes, the orthogonal projector onto $V \cap W$ may be found as a strong limit of operators $P_{V\cap W}=\lim_{n\to \infty}(P_V P_W)^n$ . Is there a similar procedure to obtain a linear projector onto $V\cap W$ in the general case of a Banach space $X$?</p> http://mathoverflow.net/questions/85492/intersection-of-complemented-subspaces-of-a-banach-space/85557#85557 Answer by Bill Johnson for Intersection of complemented subspaces of a Banach space Bill Johnson 2012-01-13T06:12:54Z 2012-01-13T12:27:44Z <p>The answer to the first question is "no". You can see this with specific examples, but here is a more conceptual approach: Take $Y$ an uncomplemented subspace of $X$ and in $Z:= X\oplus_1 X$ identify $Y\oplus 0$ with $0 \oplus Y$ in the obvious way; that is, mod out from $Z$ the subspace <code>$\{(y,-y) | y \in Y\}$</code>. $X\oplus 0$ and $0 \oplus X$ are norm one complemented in the resulting quotient space of $Z$ but their intersection $Y \oplus 0 = 0 \oplus Y$ is not complemented. (This is just a categorical push out construction specialized to the appropriate category of Banach spaces.) </p> <p>The answer is yes if the subspaces are norm one complemented and the space $X$ is uniformly convex. This is intuitive, because if $P$ is a norm one projection on a uniformly convex space and $x$ is not in the range of $P$, then $\|Px\| &lt; \|x\|$, since otherwise all vectors on the line segment from $x$ to $Px$ would have norm $\|x\|$. Hence one guesses that playing ping pong with two norm one projections $P$ and $Q$ will produce a norm one projection onto $PX\cap QX$. To see that this works without doing any computations or calculating rates of convergence (at the risk of making experts cringe), set $P_1=P$, $P_{2n} = QP_{2n-1}$, $P_{2n+1}=PP_{2n}$. Let $x\in X$ and let $a=a(x)$ be the limit of the nonincreasing sequence $\|P_n x\|$. I claim that $\|P_{n+1}x - P_{n}x\| \to 0$. Indeed, (1/2)$\|P_{n+1}x + P_{n}x\|$ also converges to $a$, so the claim follows from the uniform convexity of $X$. Let $V$ be a limit in the weak operator topology of some subnet of $P_{2n}$. By the claim, the corresponding subnet of $P_{2n+1}$ also converges to $V$ in the weak operator topology. From this it is evident that $V$ is a norm one projection onto $PX\cap QX$.</p> <p>ADDED 13 Jan. 2012: Notice that in the first construction $X$ can be uniformly convex, in which case $Z$ (and therefore also every quotient of $Z$) is isomorphic to a uniformly convex space. </p>