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)].




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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.



Boundedness of nonlinear continuous functionals
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 $i$).


