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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

Davide Giraudo
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Luthier415Hz
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    The max is the greatest element which is actually in the given set. So, the max need not exist. The sup is an element which bounds the set, so it exists (assuming the set is bounded at all) but it need not be in the set. – lulu Dec 23 '22 at 12:54
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    See Difference between supremum and maximum – Mauro ALLEGRANZA Dec 23 '22 at 12:54
  • Thanks for this – Luthier415Hz Dec 23 '22 at 12:59
  • Recall if $A \subset \mathbb{R}$ is nonempty and bounded above, then $\sup(A)$ is defined to be the least upper bound of $A.$ Note there is no requirement that $\sup(A)$ is an element of $A.$ For example, let $A=(0,1).$ We have $\sup(A)=1 \notin A.$ On the other hand, if $\sup(A) \in A,$ then $\sup(A)$ coincides with the maximum of $A.$ For example, $A=(0,1].$ So the main difference is whether the supremum is in the set itself or not. If it is, then the maximum and supremum coincide. Otherwise, the supremum still exists, but the maximum doesn't. – Vivek Kaushik Dec 23 '22 at 20:53
  • By the way, why is that first norm involving max on $C([a,b])$ well-defined ? In other words, if $f \in C([a,b]),$ why does the number $\max_{a \le t \le b} |f(t)|$ exist ? – Vivek Kaushik Dec 23 '22 at 20:55
  • @lulu couldn't be clearer! – Luthier415Hz Dec 24 '22 at 09:45
  • @VivekKaushik https://en.m.wikipedia.org/wiki/Extreme_value_theorem – Ryszard Szwarc Dec 24 '22 at 11:37
  • @RyszardSzwarc I was intending that question to be for the OP, but yes the answer is the Extreme Value Theorem. – Vivek Kaushik Dec 27 '22 at 05:38
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    @VivekKaushik Sorry, I have spoiled you didactic effort. – Ryszard Szwarc Dec 27 '22 at 09:43

1 Answers1

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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 $$\|f\|=\max_{\|x\|=1}|f(x)|$$

Ryszard Szwarc
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