Skip to main content
added 192 characters in body
Source Link
Bill Johnson
  • 31.5k
  • 5
  • 89
  • 138

After Phil's answer and the ensuing discussion, the remaining question can be formulated as:

Estimate $\lambda$ s.t. if $F\subset E$ and $E/F$ is finite dimensional, then every norm one linear operator $T$ from $F$ into a $C(K)$ space can be extended to an operator from $E$ into the $C(K)$ space which has norm at most $\lambda$.

I want to point out that $\lambda$ cannot be less than two. I think that it is known that $\lambda$ can be anything larger than two, but I did not find a reference after a quick search. I called Morry Zippin's attention to the problem; maybe he will comment. Or maybe this was proved after he worked on almost complementation; perhaps by Jesus Castillo and colleagues. I more or less thought through that $\lambda$ can be anything larger than three.

Before saying why $\lambda$ cannot be less than two, let's further reformulate the problem. The second dual of every $C(K)$ is $1$-injective, so the problem reduces to the case where $F$ is a $C(K)$ space and $T$ is the identity operator on $F$. That is, we need to estimate how well complemented a $C(K)$ must be in a superspace in which the $C(K)$ space has finite codimension. In fact, the superspace of $C(K)$ can be assumed to be in $C(K)^{**}$, but that is not particularly useful in getting the lower bound on $\lambda$.

Consider the $C(K)$ space $c$, the space of convergent sequences under the sup norm. (I'll treat the real case, but the complex case is only slightly more complicated.) The superspace $E$ is the span in $\ell_\infty$ of $c$ and the sequence $x$ defined by $x(n)= (-1)^n$. Let $P_n$ be the projection onto the span of the first $n$ unit vectors in $\ell_\infty$ and let $P$ be any projection from $E$ onto $c$. Write $Px = a\cdot 1 + y$ with $y$ in $c_0$ and where $1$ is the constant one function. Now for every $n$, the norm of $x-2P_n x$ is one, and $P(x- 2P_n x)=a 1 + y - 2P_n x$. Since $y$ is in $c_0$, letting $n\to \infty$ gives $\|P\| \ge |a|+2$. The choice $a=0$, $y=0$ of course gives a projection of norm two.

EDIT Nov. 29, 2011: Zippin's article in Handbook of the Geometry of Banach Spaces, vol. 2, contains information about almost complementation (differently named) including many open problems.

After Phil's answer and the ensuing discussion, the remaining question can be formulated as:

Estimate $\lambda$ s.t. if $F\subset E$ and $E/F$ is finite dimensional, then every norm one linear operator $T$ from $F$ into a $C(K)$ space can be extended to an operator from $E$ into the $C(K)$ space which has norm at most $\lambda$.

I want to point out that $\lambda$ cannot be less than two. I think that it is known that $\lambda$ can be anything larger than two, but I did not find a reference after a quick search. I called Morry Zippin's attention to the problem; maybe he will comment. Or maybe this was proved after he worked on almost complementation; perhaps by Jesus Castillo and colleagues. I more or less thought through that $\lambda$ can be anything larger than three.

Before saying why $\lambda$ cannot be less than two, let's further reformulate the problem. The second dual of every $C(K)$ is $1$-injective, so the problem reduces to the case where $F$ is a $C(K)$ space and $T$ is the identity operator on $F$. That is, we need to estimate how well complemented a $C(K)$ must be in a superspace in which the $C(K)$ space has finite codimension. In fact, the superspace of $C(K)$ can be assumed to be in $C(K)^{**}$, but that is not particularly useful in getting the lower bound on $\lambda$.

Consider the $C(K)$ space $c$, the space of convergent sequences under the sup norm. (I'll treat the real case, but the complex case is only slightly more complicated.) The superspace $E$ is the span in $\ell_\infty$ of $c$ and the sequence $x$ defined by $x(n)= (-1)^n$. Let $P_n$ be the projection onto the span of the first $n$ unit vectors in $\ell_\infty$ and let $P$ be any projection from $E$ onto $c$. Write $Px = a\cdot 1 + y$ with $y$ in $c_0$ and where $1$ is the constant one function. Now for every $n$, the norm of $x-2P_n x$ is one, and $P(x- 2P_n x)=a 1 + y - 2P_n x$. Since $y$ is in $c_0$, letting $n\to \infty$ gives $\|P\| \ge |a|+2$. The choice $a=0$, $y=0$ of course gives a projection of norm two.

After Phil's answer and the ensuing discussion, the remaining question can be formulated as:

Estimate $\lambda$ s.t. if $F\subset E$ and $E/F$ is finite dimensional, then every norm one linear operator $T$ from $F$ into a $C(K)$ space can be extended to an operator from $E$ into the $C(K)$ space which has norm at most $\lambda$.

I want to point out that $\lambda$ cannot be less than two. I think that it is known that $\lambda$ can be anything larger than two, but I did not find a reference after a quick search. I called Morry Zippin's attention to the problem; maybe he will comment. Or maybe this was proved after he worked on almost complementation; perhaps by Jesus Castillo and colleagues. I more or less thought through that $\lambda$ can be anything larger than three.

Before saying why $\lambda$ cannot be less than two, let's further reformulate the problem. The second dual of every $C(K)$ is $1$-injective, so the problem reduces to the case where $F$ is a $C(K)$ space and $T$ is the identity operator on $F$. That is, we need to estimate how well complemented a $C(K)$ must be in a superspace in which the $C(K)$ space has finite codimension. In fact, the superspace of $C(K)$ can be assumed to be in $C(K)^{**}$, but that is not particularly useful in getting the lower bound on $\lambda$.

Consider the $C(K)$ space $c$, the space of convergent sequences under the sup norm. (I'll treat the real case, but the complex case is only slightly more complicated.) The superspace $E$ is the span in $\ell_\infty$ of $c$ and the sequence $x$ defined by $x(n)= (-1)^n$. Let $P_n$ be the projection onto the span of the first $n$ unit vectors in $\ell_\infty$ and let $P$ be any projection from $E$ onto $c$. Write $Px = a\cdot 1 + y$ with $y$ in $c_0$ and where $1$ is the constant one function. Now for every $n$, the norm of $x-2P_n x$ is one, and $P(x- 2P_n x)=a 1 + y - 2P_n x$. Since $y$ is in $c_0$, letting $n\to \infty$ gives $\|P\| \ge |a|+2$. The choice $a=0$, $y=0$ of course gives a projection of norm two.

EDIT Nov. 29, 2011: Zippin's article in Handbook of the Geometry of Banach Spaces, vol. 2, contains information about almost complementation (differently named) including many open problems.

Source Link
Bill Johnson
  • 31.5k
  • 5
  • 89
  • 138

After Phil's answer and the ensuing discussion, the remaining question can be formulated as:

Estimate $\lambda$ s.t. if $F\subset E$ and $E/F$ is finite dimensional, then every norm one linear operator $T$ from $F$ into a $C(K)$ space can be extended to an operator from $E$ into the $C(K)$ space which has norm at most $\lambda$.

I want to point out that $\lambda$ cannot be less than two. I think that it is known that $\lambda$ can be anything larger than two, but I did not find a reference after a quick search. I called Morry Zippin's attention to the problem; maybe he will comment. Or maybe this was proved after he worked on almost complementation; perhaps by Jesus Castillo and colleagues. I more or less thought through that $\lambda$ can be anything larger than three.

Before saying why $\lambda$ cannot be less than two, let's further reformulate the problem. The second dual of every $C(K)$ is $1$-injective, so the problem reduces to the case where $F$ is a $C(K)$ space and $T$ is the identity operator on $F$. That is, we need to estimate how well complemented a $C(K)$ must be in a superspace in which the $C(K)$ space has finite codimension. In fact, the superspace of $C(K)$ can be assumed to be in $C(K)^{**}$, but that is not particularly useful in getting the lower bound on $\lambda$.

Consider the $C(K)$ space $c$, the space of convergent sequences under the sup norm. (I'll treat the real case, but the complex case is only slightly more complicated.) The superspace $E$ is the span in $\ell_\infty$ of $c$ and the sequence $x$ defined by $x(n)= (-1)^n$. Let $P_n$ be the projection onto the span of the first $n$ unit vectors in $\ell_\infty$ and let $P$ be any projection from $E$ onto $c$. Write $Px = a\cdot 1 + y$ with $y$ in $c_0$ and where $1$ is the constant one function. Now for every $n$, the norm of $x-2P_n x$ is one, and $P(x- 2P_n x)=a 1 + y - 2P_n x$. Since $y$ is in $c_0$, letting $n\to \infty$ gives $\|P\| \ge |a|+2$. The choice $a=0$, $y=0$ of course gives a projection of norm two.