Preliminary:
A={f $\in C(X); f(a)=0$} is a banach space with norm the following:
$\Vert f\Vert=sup\vert f(x)-f(y)\vert; x,y \in X$
( X is Hausdorff and compact space. 'a' is a point in X)
$\tilde f (x,y)= f(x)-f(y)$
$\tilde A=\{\tilde f; f\in A\}$ is banach space with sup norm ( $\Vert \tilde f\Vert=sup \vert \tilde f(x,y)\vert ; x,y \in X$)
notice that $\tilde A $ is not a algebra
for any x,y in X where $x\neq y, \tilde A$ satisfy the following property:
for any $ \varepsilon > 0 $ andopen neighborhoods U, V of x, y There exists a $\tilde f$ in the unit sphere of $\tilde A$ such that $\tilde f(x,y)=1=\Vert \tilde f\Vert$ and $\vert\tilde f(z,t)\vert\leq\epsilon$ ;
z,t $\in U \cup V $
Problem:
Now, let $x_1,y_1$ are in X and $\tilde f $is in the unit sphere of$\tilde A $ such that $0\leq\tilde f(x_1,y_1)<1$ then there exists $\tilde h$ in the unit sphere of $\tilde A$ such that $\tilde h (x_1,y_1)=1$ and $\Vert\tilde h- \tilde f\Vert\leq 1-\tilde f(x_1,y_1)$
My proof:
Using Oryshon lemma there exists a function of g in C(X) such that $0\leq g\leq 1$ and $g(x_1)=1, g(y_1)=0,g(a)=0$ so $\tilde g $ is in the unit sphere of $\tilde A $ and $\tilde g(x_1,y_1)=1$
Let's consider the function of h like the following :
$h=f+(1-\tilde f(x_1,y_1))g$, clearly $\tilde h$ is in $\tilde A$ and $\tilde h(x_1,y_1)=1$ and $\Vert \tilde h-\tilde f\Vert <1-\tilde f(x_1,y_1)$ but $\tilde h$ is not in the unit sphere of $\tilde A$
Could someone help me?