User anonymous - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T23:13:52Z http://mathoverflow.net/feeds/user/2858 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9773/boundedness-of-nonlinear-continuous-functionals/9959#9959 Answer by Anonymous for Boundedness of nonlinear continuous functionals Anonymous 2009-12-28T16:06:38Z 2009-12-28T16:53:31Z <p>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</p> <p>U_n={g in K: min{|g(t)|:t in C_n} > \|g\| - 1/4}</p> <p>where \|g\|=sup{|g(t)|:t in [0,1]}.</p> <p>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 supp(F_m)={g in K: F_m(g)>0} is contained in U_n.</p> <p>Now define F:K\to \mathbb{R} by</p> <p>F(g)= \sum_{m=1}^{\infty} n_m\cdot F_m(g)</p> <p>Notice that F is well-defined and continuous.</p> <p>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 &lt; 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).</p> http://mathoverflow.net/questions/9773/boundedness-of-nonlinear-continuous-functionals/9959#9959 Comment by Anonymous Anonymous 2009-12-29T16:32:24Z 2009-12-29T16:32:24Z For every n &lt; 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) &lt; 1/5 for all n &lt; 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 &lt; 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 &lt; 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 &lt; i.