Ex. 1.2.16. - Topology of metric spaces by Kumaresan 
$\textbf{Ex. 1.2.16.}$ Let $f,g: [0,1] \longrightarrow \mathbb{R}$ be continuous and $f(t) < g(t)$ for all $t \in [0,1]$. Consider the set
$U := \{ h \in \mathcal{C}[0,1] \ ; \ f(t) < h(t) < g(t), t \in [0,1] \}$
in the space $X = \left( \mathcal{C}[0,1], || \ ||_{\infty} \right)$. Is $U$ a ball in $X$? If not, can you think of a condition of  $f$ and $g$ will ensure that the set $U$ is an open ball?

My attempt:
I thought if even I define $\epsilon := \inf_{h \in \mathcal{C}[0,1]} \ \{ \sup_{t \in [0,1]} \{ |(g-h)(t)| \}, \sup_{t \in [0,1]} \{ |(f - h)(t)| \} \}$, I won't conclude that have that $U$ is an open ball, but I don't know how to show this.
Anyone can help me? Thanks in advance!
 A: $U$ is a ball if and only if $f-g=\text{constant function}$!
Pr00f: By translating $U$ toward the direction $-\frac{g+f}{2}$, we may assume $g >0$ and  $f= -g.$
Therefore in this case $U = \{ h \in C[0,1] ~|\quad  -g < h < g \}. $ Since $U$ is symmetric about origin, then $U$ is a ball if and only if $U$ is a ball centered origin. This means there is a positive real number, say $r >0$ such that
$$U = \{ h \in C[0,1] ~|\quad  -r < h < r \}$$
A simple comparing between later set and $U$, we get $g(x) =r$ for all $x \in [0,1],$ (note that in this argument the compactness of $[0.1]$ is the key ingredient)  
A: Let $B(j,r)=\{h\in C[0,1]: \|h-j\|<r\}$ for  $r>0$ and  $j\in C[0,1].$
Suppose there exist $x,x'$ with $A=f(x)-g(x)\ne f(x')-g(x')=B.$ 
(I).  If $r> B/2$ and  $j(x')\geq (f(x')+g(x'))/2$ there exists $h\in B(j,r)$ with  $h(x')=j(x')+B/2\geq f(x')$. So $h\not\in U.$
(II).  If $r>B/2$ and $j(x')\leq (f(x')+g(x'))/2$ there exists $h\in B(j,r)$ with $h(x')=j(x')-B/2\leq g(x')$. So $h\not \in U.$
(III).  If $r\leq B/2$ and $j(x)\geq (f(x)+g(x))/2$ there exists $h\in U$ with $h(x)=j(x)-B/2$. So $h\not \in B(j,r).$
(IV).   If $r\leq B/2$ and $j(x)\leq (f(x)+g(x))/2$ there exists $h\in U$ with $h(x)=j(x)+B/2$. So $h\not \in B(j,r).$
.......................In all cases we have $U\ne B(j,r).$......................
We can also see it this way: For all $y\in [0,1]$ we have $\sup \{h_1(y)-h_2(y): h_1,h_2\in  B(j,r)\}=2r.$ Then:
(I'):  If  $r>B/2$ there exist $h_1,h_2 \in B(j,r)$ with $h_1(x')-h_2(x')>B,$ so $h_1,h_2$ can't both belong to $U.$ 
(II'): If $r\leq B/2$ there exist $h_1,h_2\in U$ with $h_1(x)-h_2(x)>B\geq 2r$ so $h_1,h_2$ can't both belong to $B(j,r).$
Remark. The set $U$ is open: For $h\in U,$  let $a=\min_{x\in [0,1]}h(x)-g(x)$ and $b=\min_{x\in [0,1]}f(x)-g(x).$ Then $c=\min (a,b)$ is positive, and $B(h,c/2)\subset U.$
