Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $K$ be the closed unit ball of $C[0,1]$, and let $f$ in $C(K,\mathbb{\, R})$. Is it true that there exists an infinite dimensional reflexive subspace $E$ of $C[0,1]$ s.t. $f(K\cap E)$ is bounded ?

If the answer is affirmative, this would be a very weak kind of Weierstrass-type theorem [and also a very general one, due to the "universality" of $C[0,1]$ (i.e., the Banach-Mazur Embedding Theorem)].


One may also replace $C[0,1]$ by $B[0,1]$, the space of all bounded functions on $[0,1]$, endowed with the sup-norm.

share|improve this question

4 Answers 4

up vote 4 down vote accepted

Ady, I think there is a counterexample to your question. To describe it, let $(V_n)$ be a basis of $[0,1]$ consisting of non-empy open sets; $K$ stands for the closed unit ball of $C[0,1]$. For every $n$ let $C_n$ be the closure of $V_n$ and define

$U_n={g \in K: \min{|g(t)|:t \in C_n} > \|g\| - 1/4}$

where $\|g\|=\sup{|g(t)|:t \in [0,1]}$.

The family $(U_n)$ is an open cover of $K$. Let $(F_m)$ be a partition of unity subordinate to $(U_n)$. For every m let $n_m$ be the least integer $n$ such that $\sup(F_m)={g \in K: F_m(g)>0}$ is contained in $U_n$.

Now define $F:K\to \mathbb{R}$ by

$F(g)= \sum_{m=1}^{\infty} n_m\cdot F_m(g)$

Notice that F is well-defined and continuous.

Finally notice that $F(K\cap E)$ is unbounded for every infinite-dimensional subspace E of $C[0,1]$. This follows from the following fact: for every integer i and every infinte-dimensional subspace $E$ of $C[0,1]$ there is a norm-one vector $e \in E$ such $e$ is NOT in $U_n$ for every $n < i$ (and therefore, if $m$ is such that $F_m(e)>0$, then necessarily $n_m$ is greater or equal to $i$ which gives that $F(e)$ is also greater or equal to $i$).

share|improve this answer
    
Why "for every integer i and every infinte-dimensional subspace E of C[0,1] there is a norm-one vector e in E such e is NOT in U_n for every n < i" ? –  Ady Dec 29 '09 at 13:19
1  
For every n < i fix some t_n in V_n. Since E is infinite-dimensional you can find a vector e in E such that e has norm one and satisfies e(t_n) < 1/5 for all n < i [indeed, you may select a basic sequence (e_k) in E with basis constant, say, 2 and such that lim_k e_k(t_n) exists for all n < i; so, if k is large enough you have that e_k(t_n)-e_{k+1}(t_n) is almost zero for all n < i; so, for a sufficiently large k, the vector e=(e_k-e_{k+1})/\|e_k-e_{k+1}\| is as desired]. By definition, the vector e is not in U_n for all n < i. –  Anonymous Dec 29 '09 at 16:32
1  
Nice construction, Anon. A simpler answer to Ady's question about getting e in E s.t. ... is that the functions in C[0,1] that vanish at t_n for all n<i is a finite codimensional subspace of C[0,1]. –  Bill Johnson Dec 29 '09 at 17:49

Ady, I don't have an answer to the new version of your question but let me make some remarks which might be useful.

The new version is about non-linear real-valued continuous functions on $\ell_\infty(\Gamma)$ where $\Gamma$ has the cardinality of the continuum. This can be slightly generalized as follows:

Let $\kappa$ be an infinite cardinal and set $K$ to be the closed unit ball of $\ell_\infty(\kappa)$. Let $f:K\to\mathbb{R}$ be a continuous map. Does there exist an infinite-dimensional subspace $E$ of $\ell_\infty(\kappa)$ such that $f(K\cap E)$ is bounded?

If $\kappa=\aleph_0$, then a counterexample can be constructed.

On the other hand, if $\kappa$ is a measurable cardinal, then there exists a subspace $E$ of $\ell_\infty(\kappa)$ which is isomorphic to $c_0(\kappa)$ and such that $f(K\cap E)$ is bounded. The argument goes back to Ketonen. Let $FIN(\kappa)$ be the set of all non-empty finite subsets of $kappa$ and define a coloring $c:FIN(\kappa)\to\mathbb{N}$ as follows. Let $c(F)$ be $n$ if $n$ is the least integer $m$ such that

$ \max\{ |f(x)|: x\in span\{e_t: t\in F\} and x\in K \} \leq m $

where $e_t$ is the dirac function at $t$. Notice that $c$ is well-defined. There exist $n_0\in\mathbb{N}$ and a subset $A$ of $\kappa$ with $|A|=\kappa$ and such that $c$ is constant on $FIN(A)$ and equal to $n_0$. If we set $E$ to be the closed linear span of $\{e_t: t\in A\}$, then $E$ is isomorphic to $c_0(\kappa)$ and $F(K\cap E)$ is in the interval $[-n_0, n_0]$.

Concerning the continuum: it might be that there are set-theoretic issues. Firstly, let me recall that it is consistent that the the continuum is real-valued measurable (R. M. Solovay). On the other hand, if CH holds, then there is heavy (and quite advanced) machinery for ``killing" various Ramsey properties on $\omega_1$ (largely due to S. Todorcevic).


A quick remark: there exists a non-linear continuous map $f:K\to\mathbb{R}$, where $K$ is the closed unit ball of $c_0(\kappa)$ and $\kappa$ is the continuum, such that for every infinite-dimensional subspace $E$ of $c_0(\kappa)$ the set $f(K\cap E)$ is unbounded.

share|improve this answer

There is a simpler counterexample for the $C[0,1]$ case. Namely, $f(x):=$ $\log\left(1-\left\Vert x\right\Vert _{\infty}+\left\Vert Vx\right\Vert _{\infty}\right)$

,where $V$ is the classical Volterra operator acting on $C[0,1]$.

share|improve this answer
1  
All you need is a one to one strictly singular operator from $X$ into some space. There isn't one when $X$ is $c_0(\kappa)$ with $\kappa$ uncountable, but Pandelis took care of those spaces. It certainly looks like the answer to your question is negative for every infinite dimensional space. –  Bill Johnson Feb 13 '10 at 21:15

Firstly, let me give the details for $\ell_\infty(\aleph_0)$; $K$ stands for the closed unit ball of $\ell_\infty(\aleph_0)$. For every $n$ let $U_n=\{ x\in K: |x(n)| > 1/4 - \|x\| \}$. The family $(U_n)$ is an open cover of $K$. Let $(F_m)$ be a partition of unity subordinate to $(U_n)$. For every $m$ let $n_m$ be the least integer $m$ such that $supp(F_m)$ is contained in $U_n$ and define $$F(x)=\sum_m n_m \cdot F_m(x)$$. Then using the arguments outlined above, one can show that $F(K\cap E)$ is unbounded for every infinite-dimensional subspace $E$ of $\ell_\infty(\aleph_0)$.

Secondly, let me remark that my argument for $\ell_\infty(\kappa)$ with $\kappa$ measurable is not correct; I apologize for that (I have a remark at the end). What I can show is that for every $\kappa$ (even measurable) there exists a continuous function $F:K_0\to\mathbb{R}$, where $K_0$ is the closed unit ball of $c_0(\kappa)$, such that $F(K_0\cap E)$ is unbounded for every infinite-dimensional subspace $E$ of $c_0(\kappa)$. The argument is a variation of the previous one. For every pair of rationals $0 < a < b < 1/4$ let $U_{a,b}$ be the set of all $x\in c_0(\kappa)$ such that for every $t\in\kappa$ either $|x(t)| < a$ or $|x(t)| > b$. Notice that $U_{a,b}$ is open in $K_0$ and for every $x\in K_0$ there exists such a pair $(a,b)$ such that $x\in U_{a,b}$. Now for every $n$ (including zero) and every pair $0 < a < b < 1/4$ let $U_{a,b,n}$ be the set of all $x\in U_{a,b}$ for which the cardinality of the set $\{t: |x(t)| > b\}$ is $n$. The family $(U_{a,b,n})$ is an open cover of $K_0$. Let $(F_i) (i\in I)$ be a partition of unity subordinate to a locally finite refinement of $(U_a,b,n)$. For every $i\in I$ set $L_i=\{n: there exist 0 < a < b < 1/4 s.t. supp(F_i) is contained in U_{a,b,n}\}$ and let $n_i$ be the least element of $L_i$. Now define $F:K_0\to\mathbb{R}$ by $$F(x)=\sum_i n_i \cdot F_i(x)$$. It is continuous.

Now we check that $F(K_0\cap E)$ is unbounded for every infinite-dimensional subspace $E$ of $c_0(\kappa)$. So let $E$ be one. Since $c_0(\kappa)$ is hereditarily $c_0$, by James, we can find a normalized sequence $(e_n)$ in $E$ which a $2$-equivalent to the standard unit vector basis of $c_0$ (in particular, $(e_n)$ is weakly null). Fix some integer $M$. We may recursively select a sequence $(n_k)$ in $\mathbb{N}$ such that for all $k < m$ the sets $\{t: |e_{n_k}(t)| > 1/4M\}$ and $\{t: |e_{n_m}(t)| > 1/4M\}$ are disjoint. Consider that vector $e= \sum_{k=1}^M e_{n_k}$. Observe, first, that $1/2\leq \|e\| \leq 2$. Also notice that the set $\{t: |e(t)|\geq 3/4\}$ has cardinality at least $M$. Let us normalize $e$ and denote the normalized vector by $v$. The set $\{ t: |v(t)| \geq 3/8 \}$ has cardinality at least $M$. Let $i\in I$ be such that $F_i(v)>0$. Let $0 < a < b < 1/4$ and $n$ be arbitrary such that $supp(F_i)$ is contained in $U_{a,b,n}$. Notice that the set $\{t: |v(t)| \geq 3/8\}$ is contained in the set $\{t: |v(t)|> b\}$, and so, the cardinality of the set $\{t: |v(t)| > b\}$ is at least $M$. It follows that $n_i\geq M$ yielding that $F(v)\geq M$.

share|improve this answer
    
@Pandelis Why it [i.e., the "colouring" argument] is not correct ? At least, do yo have an affirmative result for some "big" spaces ? –  Ady Feb 7 '10 at 15:07
    
Ady, if you do the coloring argument then you will "canonize" the function $F$, i.e. $F(x)$ will only depend of the size of the support of $x$. I thought that this was enough to "freeze" the values of $F$ on a large set and this was wrong. What the coloring argument yielded was a hint in order to get the counterexample above. For "bigger" spaces I honestly don't know. –  Pandelis Dodos Feb 7 '10 at 19:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.