# Boundedness of nonlinear continuous functionals

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.

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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 $i$).

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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
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
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.

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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 $C[0,1]$.

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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 $F(v)\geq M$.

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@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