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I was wondering if someone could explain what is the difference between the max of a finite number of functions and the max of a finite number of function values.

For instance, as shown here, given real-valued continuous functions $f,g$, the following inequality holds.

$\max(f+g)(z)\leq \max f(z)+\max g(z), \forall z \in Z$.

However, if I define the functions $ g= \max_{i \in I}\{g_i\}$, $ h= \max_{j \in J}\{h_j\}$, and let $f= g+h$, then $f$ can be represented as

$ f = \max_{i \in I,j \in J}\{g_i+h_j\}$

rather than writing as

$f = \max_{i \in I}\{g_i\} + \max_{j \in J}\{h_j\} \leq \max_{i \in I,j \in J}\{g_i+h_j\}$.

My question is why equality holds when we work with indices. What exactly am I misinterpreting?

ball_jan
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  • It has to do with when equality holds. It need not be for the same $z$ value, or corresponding $i$ index. For example, let $f(z) = -x^2, g(z) = -(x-1)^2$, clearly their max is 0 and 0 (at 0 and 1 resp), but what can you say about max $ (f+g)(z)?$ Now, discretize it by letting $ f_n = f(n)$. – Calvin Lin Jul 03 '21 at 17:52
  • It is clear to me why the inequality holds for each input vector z. The issue is when we work with indices, for some reason, we directly go for equality. I'm trying to understand the logical difference between those two. – ball_jan Jul 03 '21 at 18:25
  • When you go for indicies, it's likely that the same holds. You can't use distinct indices (despite what was written). Having said that, you should share the context where where you see that definition. Maybe there's something else going on, like that the functions aren't indexed with the same "base set". – Calvin Lin Jul 03 '21 at 22:08
  • I don't have springer access. – Calvin Lin Jul 03 '21 at 22:24

1 Answers1

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The inequality stated in the referred question is given as \begin{align*} \color{blue}{\max\{f+g\}\leq \max\{f\}+\max\{g\}\qquad\qquad x\in[a,b]}\tag{1} \end{align*} where we assume $f,g$ being real-valued continuous functions with domain $[a,b]$ to assure that a maximum exists.

Note that (1) is not an inequality of functions but an inequality of real numbers. We have \begin{align*} \max\{f+g\}&=\max_{x\in[a,b]}\{(f+g)(x)\}=\max_{x\in[a,b]}\{f(x)+g(x)\}\in\mathbb{R}\\ \max\{f\}&=\max_{x\in[a,b]}\{f(x)\}\in\mathbb{R}\\ \max\{g\}&=\max_{x\in[a,b]}\{g(x)\}\in\mathbb{R}\\ \end{align*} Each of the three maxima above specifies a maximum of the values of a single function with domain $[a,b]$, namely $f+g,f$ and $g$.

We have a different situation when we consider a (finite) family $\left(g_i\right)_{i\in I}$ of functions $g_i$ with $I$ finite. In this case the symbol \begin{align*} \max_{i\in I}\{g_i\}\tag{2} \end{align*} denotes often not a maximum of real values, but rather a maximum of functions $g_i, i\in I$ instead. We also need an order relation on the function-space $\mathcal{C}_{\mathbb{R}}\left([a,b]\right)$ of real-valued continuous functions on $[a,b]$ in order to be able to talk about maxima.

But here we want to use $\max_{i\in I}\{g_i\}$ somewhat differently, namely as \begin{align*} \max_{i\in I}\{g_i\}:= \max_{{i\in I}\atop {x\in[a,b]}}\{g_i(x)\}\in\mathbb{R}\tag{3} \end{align*} In (3) we have as in (1) the maximum being a real number of just a set of real numbers \begin{align*} \{g_i(x):i\in I, x\in[a,b]\}. \end{align*} Next we consider two (finite) families $(g_i)_{i\in I}$ and $(h_j)_{j\in J}$. We have \begin{align*} \max_{i\in I,j\in J}\{g_i+h_j\}&=\max_{{i\in I,j\in J}\atop {x\in [a,b]}}\{(g_i+h_j)(x)\}\in\mathbb{R}\\ \max_{i\in I}\{g_i\}&=\max_{{i\in I}\atop {x\in [a,b]}}\{g_i(x)\}\in\mathbb{R}\\ \max_{j\in J}\{h_i\}&=\max_{{j\in J}\atop {x\in [a,b]}}\{h_j(x)\}\in\mathbb{R}\\ \end{align*} and derive consequently in the same way as in (1) \begin{align*} \color{blue}{\max_{i\in I,j\in J}\{g_i+h_j\}\leq \max_{i\in I}\{g_i\}+\max_{j\in J}\{h_j\}\qquad\qquad x\in[a,b]} \end{align*}

Markus Scheuer
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  • If your last inequality holds, then how can one claim that $f= \max_{i,j \in I,J} {(g_i+h_j)}$? When $f= g+h$, then it implies that $f = \max_{i, \in I} {g_i} + \max_{j \in J}(h_j)$. Then, it is derived that $f =\max_{i,j \in I,J} {(g_i+h_j)} $. So my question is what exactly happens when we work with a finite number of function rather than their values? – ball_jan Jul 04 '21 at 22:19
  • @ball_jan: If you let $f=g+h$ then the conclusion $f=\max_{i\in I,j\in J}{g_i+h_j}$ is not sound. This can be checked by considering the special case $|I|=|J|=1$ which gives (1). – Markus Scheuer Jul 04 '21 at 22:34