I have seen the frequent use of both max and sup in for instance, :
The norm in a linear normed space of continuous functions on $C[a,b]$ $$||f||=\underset{a\le t\leq \ b}{\max}|f(t)|$$
and sup in :
$$||\tilde{f}||_X=\underset{x\in X, ||x||=1}{\sup}|\tilde{f}(x)|$$
where $\tilde{f}$ is a linear functional in Banach space $X(X,d)$, where $x\in X$.
So what is the reason for using sup in one case, and max in another?
Thanks
There is an essential reason why $\max$ is used for functions in $C[a,b]$ and $\sup$ for linear functionals on a Banach space $X.$ By a general theorem for any compact topological space $K$ and a continuous function $f:K\to \mathbb{C}$ the maximal value of $x\mapsto |f(x)|$ is attained at some point $x_0\in K$ (depending on the function $f$), i.e. $\|f\|=|f(x_0)|.$ For that reason we write $$\|f\|=\max_{x\in K}|f(x)|$$ In particular it is valid for $K=[a,b].$
Let $X=\ell^1(\mathbb{N}).$ We will denote by $e_n$ the elements of the standard basis in $\ell^1.$ Consider the linear functional on $X$ defined by $$f(x)=\sum_{n=1}^\infty {n\over n+1}x_n$$ We have $$|f(x)|\le \sum_{n=1}^\infty |x_n|=\|x\|_1$$ which implies $\|f\|\le 1.$ Moreover $$f(e_n)={n\over n+1}e_n,\quad |f(e_n)|={n\over n+1}\|e_n\|_1$$ which gives $\|f\|\ge 1.$ Thus $\|f\|=1.$ However there is no element $x\in \ell^1$ satisfying $\|x\|_1\le 1$ such that $|f(x)|=1.$ For this reason we write $$\|f\|=\sup_{\|x\|_1\le 1}|f(x)|$$
There are Banach spaces for which we may use $\max.$ Assume $X$ is a reflexive Banach space, for example $L^p$ spaces for $1<p<\infty$ are reflexive. Let ${f}$ be a bounded linear functional on $X.$ Then by a corollary of the Hahn-Banach theorem there is an element $x_0^{**}\in X^{**}$ such that $\|x_0^{**}\|=1$ and $x_0^{**}(f)=\|f\|.$ By reflexivity $x_0^{**}$ corresponds to an element $x_0\in X$ such that $x_0^{**}(f)=f(x_0)$ and $\|x_0\|=\|x_0^{**}\|=1.$ Therefore we may write $$\|f\|=\max_{\|x\|=1}|f(x)|$$