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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.

Question. 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?

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 $X$?

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If they are complemented then there are continuous projections $P:X\to V$, $Q:X\to W$. If they commute then everything is simple: Take $PQ=QP:X\to V\cap W$. Then $(PQ)^2=PQ$ and its image is the whole of $V\cap W$: $x=Px=Qx$ implies that $PQx=Px=x$. Now if they don't commute: $PQ(X)$ still contains $V\cap W$ but may not equal it. –  Yulia Kuznetsova Jan 12 '12 at 15:01
    
Why does the proof for Hilbert spaces fail in the Banach space case? As Yulia said, there are continuous projections $P_W$ and $P_V$: wouldn't the strong limit $$ lim_{n\to \infty} (P_V P_W)^n $$ work as well? –  Ruben A. Martinez-Avendano Jan 13 '12 at 3:35
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@Ruben: it's not immediately clear to me why that limit should exist ($P_VP_W$ might have norm greater than $1$, which makes convergence in the operator topology problematic; now of course we could still hope for SOT convergence, but it doesn't seem obvious to me at the moment.) –  Yemon Choi Jan 13 '12 at 4:53
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I was considering adding the "ask-Johnson" tag, but he has pre-emptied us... –  Yemon Choi Jan 13 '12 at 6:36
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1 Answer 1

up vote 13 down vote accepted

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 $\{(y,-y) | y \in Y\}$. $X\oplus 0$ 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.)

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 $\|Px\| < \|x\|$, 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 $PX\cap QX$.

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.

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I am almost literally kicking myself for not spotting the pullback/pushout argument. Very cute! –  Yemon Choi Jan 13 '12 at 6:35
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Thank you Bill! –  Pietro Majer Jan 13 '12 at 9:59
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