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Distinguished Professor of Mathematics at Texas A&M University


2d
comment When do block sequences yield disjoint subspaces?
An $M$ basis is a biorthogonal system s.t. the vectors have dense linear span and the functionals separate points. A minimal sequence is a sequence of vectors s.t. there exist biorthogonal functionals for the sequence (but the biorthogonal functionals need not separate points of the closed linear span of the sequence of vectors).
Aug
25
comment When do block sequences yield disjoint subspaces?
Let $(y^∗_n)$ be the functionals biorthogonal to $(y_n)$. Extend these to $Y' + Z'$ to be zero on $Z'$. I guess the most natural sufficient condition is that for each $n$ these extensions are all continuous (with a similar condition on the functionals biorthogonal to $(z_n)$). This is the same as saying that $(x_n)\cup (y_n)$ is a minimal sequence, which is weaker than saying that $(x_n)\cup (y_n)$ is an $M$-basis for its closed linear span. – Bill Johnson
Aug
16
comment Well-complemented copies of $\ell_p^n$
You can do an ultraproduct argument to reduce the problem to infinite dimensional subspaces of super reflexive lattices, but that looks like a silly way of proving it and gives no estimates.
Aug
14
comment Well-complemented copies of $\ell_p^n$
The OP wants a well complemented $\ell_2^k$ in an ARBITRARY high dimensional subspace of $\ell_p$.
Aug
3
answered On the Lorentz sequence space $d(w,1)$
Jul
29
comment Is there a quotient of $c_0$ without the approximation property?
Yes. Probably it is mentioned in A. Szankowski's most recent single author paper.
Jul
29
comment Non strictly-singular operators and complemented subspaces
Kevin, in my email aliases I have "askSpiros".
Jul
27
comment Non strictly-singular operators and complemented subspaces
The answer to your last question is yes. Consider $X$ to be a subspace of $L_\infty$. Use a back and forth argument to extend the operator on $X$ to an operator on some separable sublattice $Y$ of $L_\infty$ that contains $X$ and the constant functions. $Y$ is isometric to a separable $C(K)$ space and thus is norm one complemented in $C[0,1]$.
Jul
27
comment Non strictly-singular operators and complemented subspaces
For general spaces there is no difference between the OP's question for operators on a space and for operators between two spaces. because a counterexample for $L(X,Y)$ gives a counterexample for $(L(X\oplus Y)$.
Jul
26
revised Non strictly-singular operators and complemented subspaces
added 314 characters in body
Jul
26
answered Non strictly-singular operators and complemented subspaces
Jul
26
comment Non strictly-singular operators and complemented subspaces
If you send me an email I'll write up an answer to your question.
Jul
23
comment Continuous maps on compact topological spaces which induce compact (Fredholm) operators
The only way for $T_f$ to be Fredholm when $X=[0,1]$ is for it to be a surjective homeomorphism. If $f$ is not surjective, then $T_f(\phi)=0$ when every $\phi$ is supported off of the interval $f[0,1]$. If $f$ is not $1-1$, then there are infinitely many pairs $(a,b)$ of distinct points s.t. $f(a)=f(b)$, which implies that $T_f$ cannot have finite codimensional range.
Jul
23
comment Continuous maps on compact topological spaces which induce compact (Fredholm) operators
It is easy to check that $T_f$ is compact iff it has finite rank iff $f[X}$ is a finite set. For the not completely obvious implication, compose $T_f$ with the restriction mapping $R$ from $C(X)$ to $C(f[X])$. The composition $RT_f$ is a quotient map by Tietze, and $C(f[X])$ is infinite dimensional if $f[X}$ is infinite. In particular, if $X$ is connected then $T_f$ is not compact unless $f$ is constant. As for your second question, let $X$ be the range of a sequence of distinct points together with its limit and let $f$ act as a shift.
Jul
20
comment Noncommutative analogs of classical Banach geometric properties
No, I have not seen anything about a non commutative Dunford Pettis property.
Jul
18
comment Noncommutative analogs of classical Banach geometric properties
You can start with Pisier, Gilles; Xu, Quanhua Non-commutative Lp-spaces. Handbook of the geometry of Banach spaces, Vol. 2, 1459–1517, North-Holland, Amsterdam, 2003.
Jul
16
comment Existence of topology on the space of continuous functions
If you want a Hausdorff locally convex topology, the answer is no, because it would make $C[0,1]$ into a dual Banach space, and $C[0,1]$ is not even isomorphic to a dual Banach space.
Jul
16
comment Existence of topology on the space of continuous functions
Sure. Take the weak topology generated by the three functionals ptwise evaluation at 0, ptwise evaluation at 1, and integration.
Jul
14
comment $L_\infty(\mu)$ spaces non-isomorphic to a dual space
If $L_\infty(\mu)$ is isomorphic to a dual space, must $L_1(\mu)$ be a predual?
Jul
13
comment Is a specific sequentially closed subset of $M([0,1])$ closed?
Oh, of course, Nate. So I don't know the answer to the OP's question.