Is the given set open ball in metric space or not?

Question

Metric space is $(C[0,1],||.||_\infty)$ with sup norm Let $f,g:[0,1]\to R$ be continuous functions and $f(t)<g(t)\forall t\in [0,1]$.

$A=\{h\in C[0,1]|f(t)<h(t)<g(t) \forall t\in [0,1]\}$. Is $A$ is open ball in $C[0,1]$ or not? If not then what extra condition required to make it open ?

As $f,g$ are on compact set then they have max and minima attained on domain somewhere.
If $A$ is open then I have to find some radius $\epsilon $>0 such that $B(h_1,\epsilon)\subset$U for some $h_1 \in A$.As $h_1\in A$ $f(t)<h_1(t)$ hence $h_1(t)-f(t)=\delta >0$ But I think I required more smaller radius such that it lies?
Any Help will be appreciated

Answer 1

Some hints

I suppose that you equipped $V = \mathcal C([0,1], \mathbb R)$ with the $\sup$ norm, namely $\Vert f \Vert = \sup\limits_{t \in [0,1]} \vert f(t)\vert$.

Denote $m = \inf\limits_{t \in [0,1]}g(t)-f(t)$. By compacity of $[0,1]$ and continuity of $f,g$ $m$ is attained, let's say at $u \in [0,1]$

You'll be able to prove following affirmations:

1. If $A$ is an open ball $B(h,R)$, its radius $R$ has to be equal to $\dfrac{m}{2}$.
2. If $g-f$ is constant (and equal to $m$), $A$ is the open ball centered on $\dfrac{f+g}{2}$ with radius $\dfrac{m}{2}$
3. If $g-f$ is not constant, $A$ is not an open ball. To prove it, suppose that $A$ is an open ball centered on $h$ and build a continuous map that is in $A$ but not in $B(h, \dfrac{m}{2})$.