bio  website  math.tamu.edu/~johnson 

location  Texas A&M  
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Distinguished Professor of Mathematics at Texas A&M University
1d

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 strictlysingular operators and complemented subspaces
Kevin, in my email aliases I have "askSpiros". 
Jul 27 
comment 
Non strictlysingular 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 strictlysingular 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 strictlysingular operators and complemented subspaces
added 314 characters in body 
Jul 26 
answered  Non strictlysingular operators and complemented subspaces 
Jul 26 
comment 
Non strictlysingular 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 $11$, 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 
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Noncommutative analogs of classical Banach geometric properties
No, I have not seen anything about a non commutative Dunford Pettis property. 
Jul 18 
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Noncommutative analogs of classical Banach geometric properties
You can start with Pisier, Gilles; Xu, Quanhua Noncommutative Lpspaces. Handbook of the geometry of Banach spaces, Vol. 2, 1459–1517, NorthHolland, Amsterdam, 2003. 
Jul 16 
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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 
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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 
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$L_\infty(\mu)$ spaces nonisomorphic 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. 
Jul 13 
revised 
Is a specific sequentially closed subset of $M([0,1])$ closed?
edited tags 
Jul 9 
comment 
Is a specific sequentially closed subset of $M([0,1])$ closed?
To show that a convex subset of a dual space is weak* closed it is sufficient to show that its intersection with every closed ball around 0 is weak* closed. This is sometimes called the KreinSmulian theorem. For a proof see math.ksu.edu/~nagy/funcan20072008/bs5.pdf 
Jul 3 
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Is a specific sequentially closed subset of $M([0,1])$ closed?
It's true if $f(x,y)$ is of the form $g(x)h(y)$. Linear combinations of such functions are dense in $C[a,b]^2$ by StoneWeierstrass, so it is true in general. 
Jul 2 
awarded  Nice Answer 