Let $C[0,1]$ be the space of all continuous real valued functions defined on $[0,1]$ and let $Y$ be the closed subspace of $C[0,1]$ consisting of constant functions. Then I am asked to explicitly compute the quotient norm for any function(coset representative) of the quotient space $C[0,1]/Y$ .
Quotient norm:- If $q$ is the quotient map then $||q(x)||=\inf_{y\in Y} ||x-y||= d(x,Y)$
My attempt:-
What we want is $\displaystyle d(f,Y)=\inf_{c\in Y}\sup_{x\in [0,1]}|f(x)-c|$ . Now as $f$ attains it's maximum and minimum for some points $x_{min}$ and $x_{max}$ in $[0,1]$ . After drawing a few graphs I was able to see that
$$\sup|f(x)-c| = \begin{cases} |c-f(x_{min})|\,, c\geq \frac{f(x_{max})+f(x_{min})}{2}\\ |c-f(x_{max})|\,,c\leq\frac{f(x_{min})+f(x_{max})}{2}\end{cases} $$
Therefore I conclude that $d(f,Y)=\frac{|f(x_{max})-f(x_{min})|}{2}$ . It matches with the fact that if $d(f,Y)=0$ then $f(x_{max})=f(x_{min})\implies$ f is a constant.
I am however not hundred percent sure of it's rigor or if I am missing some big detail. Can anyone please comment on my attempt and guide me to the correct answer?
The answer is correct, and your approach is great. However, after finding the solution you should find a rigorous explanation and this is what is missing.
For any $c\in\Bbb R$ we have $$ \|f-c\|_\infty\geq |f(x_{\mathrm{max}})-c| \text{ and } \|f-c\|_\infty\geq |f(x_{\mathrm{min}})-c| $$ so $$ 2\|f-c\|_\infty\geq |f(x_{\mathrm{max}})-c|+|f(x_{\mathrm{min}})-c| \geq |f(x_{\mathrm{min}})-f(x_{\mathrm{min}})|. $$
This shows that $\|f-c\|_\infty\geq \frac 12|f(x_{\mathrm{min}})-f(x_{\mathrm{min}})| =:A$ and therefore $d(f,Y)\geq A$.
On the other hand, taking $c=\frac 12\left(f(x_{\mathrm{min}})+f(x_{\mathrm{min}}\right)$ we get $f(x)\in [f(x_{\mathrm{min}}),c]\cup[c,f(x_{\mathrm{max}})]$ (two segments of tte length $A$) so $|f(x)-c|\leq A$. This implies that $\|f-a\|_\infty\leq A$ and therefore $d(f,Y)\leq A$.