Vrftsjtryk

Let $X$ be a Banach space, $S_X$ its unit sphere and $B_X$ its unit ball.

The space $X$ is said to has $beta$ property if there exists a system ${(x_i,f_i):i in I} subset S_X times S_{X^*}$ and $0 leq rho <1$ such that

(i) $f_i(x_i)=1$, for all $i in I$

(ii) $|f_i(x_j)| leq rho$, for all $i neq j$

(iii) $||x|| = sup{|f_i(x)|:i in I}$, for all $x in X$

A point $x in S_X$ is called locally uniformly convex (LUR) if for every $epsilon>0$ there exists $delta(epsilon)>0$ such that

$y in S_X, cfrac{||x+y||}{2}>1-delta(epsilon) Rightarrow ||x-y||<epsilon $

I'm trying to prove that if $X$ has $beta$ property and $dim X >1$, such points $(x_i)_{i}$ cannot be LUR points.

My first attempt at a solution was try to define a sequence $(x_n)$ in $S_X$ such that $||x_n+x_i||$ approaches to $2$ and $||x_n-x_i||$ does not approach to $0$. It is easy to see that if $x in S_X cap ker{f_i}$, we have that $||x-x_i|| geq 1$, then maybe it is a good idea to define such sequence in $S_X cap ker{f_i}$.

Is it a good approach? Any other hints?

Assume $X$ has the $beta$ property and $dim X>1.$ Fix $iin I.$ The point $x_i$ fails the LUR property with $epsilon=1-rho.$

For any $kinker f_i,$
$$|k|leq 1-rhoimplies|x_i+k|=1tag{*}$$
because $f_i(x_i+k)=f_i(x_i)=1$ and $|f_j(x_i+k)|leq|f_j(x)|+|f_j(k)|leq rho+|k|leq1.$ (The inequality $|f_j(k)|leq |k|$ comes from property (iii).)

Since $dim X>1,$ there exists $kin ker f_i$ with $|k|=1-rho.$ We get a counterexample to the LUR property with $epsilon=1-rho$ and $y=x+k,$ because $|(x+y)/2|=1$ and $|y|=1$ by (*), but $|x-y|=|k|=epsilon.$

Assume for contradiction that $X$ has the $beta$ property and that the points $x_i$ are LUR,
define
$$ varepsilon = frac12 , sup_{x in S_X} left( inf_{i in I} | pm x - x_i| right), $$
and choose $x_{varepsilon}$ such that
$$ inf_{i in I} |pm x_{varepsilon} - x_i| geq varepsilon, quad $$
If $varepsilon = 0$, then there exists a subsequence of ${x_i}$, say $x_{k_i}$,
such that $x_{k_i} to pm x_{varepsilon}$
(either to $+ x_{varepsilon}$, or to $- x_{varepsilon}$) in $X$.
But then, using the $beta$ property, (i),
begin{align}
|f_{k_i}(x_{k_{i-1}})| &= |f_{k_i}(x_{k_i}) + f_{k_i}(x_{k_{i-1}}-x_{k_i})| \
&geq |f_{k_i}(x_{k_i})| - |f_{k_i}| , |x_{k_i} - x_{k_{i-1}}| \
&= 1 - |x_{k_i} - x_{k_{i-1}}| to 1 quad text{as} quad i to infty,
end{align}
contradicting the $beta$ property, (ii); therefore, $varepsilon > 0$.
Using the $beta$ property, (iii),
for any $alpha > 0$ there exists $i_{alpha} in I$ such that
begin{align}
|f_{i_{alpha}}(x_{varepsilon})| geq 1 - 2alpha = |x_{varepsilon}| - 2alpha.
end{align}
Assume $f_{i_{alpha}}(x_{varepsilon}) geq 0$ (renaming $x_{varepsilon} rightarrow -x_{varepsilon}$ if necessary).
Then we have
begin{align}
frac{1}{2}| x_{varepsilon} + x_{i_{alpha}} | &geq frac{1}{2} f_{i_{alpha}} (x_{varepsilon} + x_{i_{alpha}}) \
&geq 1 - alpha.
end{align}
To obtain a contradiction using this reasoning,
there must exist $delta(varepsilon)$ independent of $i$ such that the convexity condition is satisfied.
If $X$ is finite dimensional,
this is possible,
since there can be only finitely many $x_i$.
If $X$ is infinite-dimensional, I suppose this argument isn't enough for non-uniformly convex norms...