Krdytkyu

let $f_1,cdots, f_m$ be a real valued functions defined on a set $S$ in $mathbb R^n$. Assume that each $f_k$ is continuous at the point $a$ of $S$. For each $x in S : f(x) = max {f_1(x),f_2(x),cdots,f_m(x) }$. Discuss continuity of $f$ at $a$.

Attempt: I am trying to prove this without using induction. Each $f_k$ is continuous $implies $

$forall epsilon>0, exists delta_1 >0$ s.t $|f_1(x)-f_1(a)|< epsilon$ whenever $|x-a|<delta_1$

$forall epsilon>0, exists delta_2 >0$ s.t $|f_2(x)-f_1(a)|< epsilon$ whenever $|x-a|<delta_2$

...

$forall epsilon>0, exists delta_m >0$ s.t $|f_m(x)-f_m(a)|< epsilon$ whenever $|x-a|<delta_m$

$forall epsilon>0, exists delta_k >0$ s.t $max |f_k(x)-f_k(a)|< epsilon$ whenever $|x-a|<delta_k$

$implies max |f_k(x)-f_k(a)|< epsilon $ whenever $|x -a|<delta = min{delta_1,cdots, delta_m}$

and $max |f_k(x)-f_k(a)| > |max f_k(x)-max f_k(a)| $

$implies |max f_k(x)-max f_k(a)| < epsilon$ whenever $|x-a|<delta = min{delta_1,cdots, delta_m}$

Am I correct?

Thank you for your help

Let $varepsilon>0$ be given, and define the set $[m]:={,j in mathbb{N} : 1 leq j leq m}$.

For each $j in [m]$, there is a positive number $delta_j$ so that $|,f_j(mathbf{x})-f_j(mathbf{a})|<frac{1}{2}varepsilon$ whenever $d(mathbf{x,a})<delta_j$.

We set $delta=min {delta_1, delta_2, ldots, delta_m}$.

Now we have that $f_j(mathbf{a})-frac{varepsilon}{2}<f_j(mathbf{x})<f_j(mathbf{a})+frac{varepsilon}{2}$ whenever $d(mathbf{x,a})<delta$, and $j in [m] .$

Since the closed ball $bar{B}(mathbf{a},delta)$ is a compact subset of $mathbb{R}^n$, we know that the set

begin{equation}
displaystyle overset{m}{underset{j=1}{bigcup}} , f_j(bar{B}(mathbf{a},delta))
end{equation}

is a compact subset of $mathbb{R}$. Thus $f(mathbf{x}):= max {, f_1(mathbf{x}), f_2(mathbf{x}), ldots, f_m(mathbf{x})}$ exists for all $mathbf{x} in bar{B}(mathbf{a},delta)$.

Furthermore, for each $mathbf{x} in B(mathbf{a},delta)$ there is $j(mathbf{x}) in [m]$ such that
begin{equation}
f(mathbf{x})-frac{varepsilon}{2} < f_{j(mathbf{x})}(mathbf{x}) leq f(mathbf{x}), text{ and so we have }
\
f(mathbf{x}) < f_{j(mathbf{x})}(mathbf{x}) + frac{varepsilon}{2} < f_{j(mathbf{x})}(mathbf{a}) + varepsilon leq f(mathbf{a})+varepsilon.
end{equation}

Moreover, there is $j in [m]$ so that $f(mathbf{a})-varepsilon < f_j(mathbf{a})-frac{varepsilon}{2} < f_j(mathbf{x}) leq f(mathbf{x})$.

We combine to conclude that $f(mathbf{a})-varepsilon < f(mathbf{x}) < f(mathbf{a}) + varepsilon$ whenever $d(mathbf{x,a})<delta$. Therefore $f(mathbf{x})=max{, f_1(mathbf{x}), f_2(mathbf{x}), ldots, f_m(mathbf{x})}$ is continuous at $mathbf{a} in S$.

Notice that while the particular family of functions is finite, a similar argument shows that $f(mathbf{x}):=sup_{alpha in A}{,f_{alpha}(mathbf{x})}$ is continuous, for any equicontinuous family (countable or uncountable) of uniformly bounded functions. To be clear, in the case of an infinite family of functions the two assumptions, "uniformly bounded and equicontinuous" are sufficient. It should be clear from the proof above how these two assumptions are essentially given in the case of a finite family of continuous functions on a compact subset of $mathbb{R}^n$.

Note: $j: B(mathbf{a},delta) longrightarrow [m]$ is a choice function that assigns to each $mathbf{x} in B(mathbf{a},delta)$ an integer $j(mathbf{x}) in [m]$ in accordance with this characterization of supremum of subset of $mathbb{R}$ which I rather mention than $max$ since $sup$ is relevant in the previously alluded to case of an infinite family (ie, $alpha: B(mathbf{a},delta) longrightarrow A$) and it is also what I had in mind when I originally wrote the preceding argument.

D. S. Bridges and L. S. Vita; "Techniques of Constructive Analysis"

So given $varepsilon>0$, we consider the set $S_{varepsilon}:=left{(mathbf{x},y) in B(mathbf{a},delta) times [m]: f(mathbf{x})-frac{1}{2}varepsilon<f_y(mathbf{x})right}$ and we know that for each $mathbf{x} in B(mathbf{a},delta)$ there exists $y in [m]$ such that $(mathbf{x},y) in S_{varepsilon}$ since $f(mathbf{x})=sup{, f_1(mathbf{x}), f_2(mathbf{x}), ldots, f_m(mathbf{x})}$ ($max$ also works in this context since $[m]$ has finite cardinality).

The context of a finite family is an important distinction since we don't technically invoke the Axiom of Choice; see: Axiom of Choice for Finite Sets.

In any case (or rather for any nonempty indexing set $A$), given an equicontinuous family of uniformly bounded functions $mathscr{F}=left{f_alpha: mathbb{R}^n longrightarrow mathbb{R}: alpha in Aright}$ we may instead show that the function $f(mathbf{x})=sup_{alpha in A}{,f_{alpha}(mathbf{x})}$ is sequentially continuous at $mathbf a$ which avoids invoking the full Axiom of Choice as we would instead invoke the Axiom of Countable Choice. In the sequential context, given a sequence ${mathbf x_n}_{n=1}^infty$ in $mathbb R^n$ converging to $mathbf a$ we would say there is a choice function $alpha: {mathbf x_n : n in mathbb N} longrightarrow A$ such that $left(mathbf x_n, alpha(mathbf x_n)right) in S_{varepsilon}$ for each $n in mathbb N$. AC cannot be derived from ACC which is good (at least in some sense, whatever one Brouwer adheres to) since AC implies the law of excluded middle.

Generalise the formula:

$$max{f,g} = frac{1}{2}(f+g+|f-g|)$$

A continues map of a continues map is continues.

to prove max, min, abs from R^n to R is continues is not complicated.

Lemma 1: consider g: (f1(), f2() ... fn()) : R^n -> R, a = (a1, a2,..an). Then if the value of g is bounded in a ball B(g(a), r): sum((fi(x) - f(ai))^2) <= r^2, then due to the continuity of fi, consider an interval Ki around ai, such that |fi(x) - f(ai)| <= r/sqrt(n), then the Cartesian product of Ki forms a cubic in R^n. Then shrink it to a smaller ball B(a, s). Then any point p in B, g(p) is bounded in B(g(a), r).

Lemma 2: since MAX2 = MAX(x1, x2) is continuous, then MAXn= MAX(MAXn-1, y) is continuous easily by induction, where MAXi is a function on R^i. The induction step is similar to the previous comment, But only needs to generate it to Cartesian product of metric spaces in a Euclidean way (that is sum of power series and then power back of the metrics to define a new metric).

by Lemma 1 and Lemma 2, MAXn * g is continuous

but this prove needs a complete space in R^n to define the g. so if discussing some subset S belonging to R^n. It should be more cautious