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

  • $\begingroup$ i do not get what you did there. you need to show that $f$ is continues at $a$ not $max f_k$ $\endgroup$
    – sha
    Oct 6, 2014 at 22:33
  • 1
    $\begingroup$ How do you justify $\max |f_k(x)-f_k(a)| > |\max f_k(x)-\max f_k(a)|?$ $\endgroup$
    – Lemon
    Oct 7, 2014 at 0:30
  • 1
    $\begingroup$ @Hawk .It's elementary, and perhaps obvious, although it should be $\geq ,$ not $>$. But perhaps a justification of it should be included in the proof. $\endgroup$ Aug 12, 2018 at 6:42

3 Answers 3


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 $d(\mathbf{x,a})<\delta_j$ (Principle of Finite Choice).

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.

$(*)$ We could also consider the set $\left\{\{f_1(\mathbf x),f_2(\mathbf x),\ldots, f_m(\mathbf x)\}:\mathbf x \in B(\mathbf a, \delta) \right\}$ to be the domain of $j$ and the set $\bigcup_{\mathbf x \in B(\mathbf a, \delta)}\{f_1(\mathbf x),f_2(\mathbf x),\ldots, f_m(\mathbf x)\}$ to be the codomain of $j$.

enter image description here 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. Moreover, the proof from the ordering principle of the axiom of choice for finite sets suggests why we should have an explicit definition such as $\max\{f( \mathbf x),g(\mathbf x)\}:=\frac12\left[f(\mathbf x)+g(\mathbf x)+|f(\mathbf x)-g(\mathbf x)|\right]$ since $\bigcup_{\mathbf x \in \mathbb R^n} \{f(\mathbf x),g(\mathbf x)\} \subset \mathbb R$ is already a totally ordered set without the ordering principle.

  • $\begingroup$ Showing that $F(x)=\max\{f(x),g(x)\}$ is continuous ($f$ and $g$ both assumed to be continuous) was one of my favorite problems in undergrad analysis: (1) imgur.com/dxWQuH7; (2) imgur.com/ih1edun. (don't judge this was from our first hw in the course, and predates my answer here by about year) $\endgroup$
    – M A Pelto
    Dec 7, 2018 at 4:16

Generalise the formula:

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

  • 3
    $\begingroup$ This formula generalizes to any finite family of continuous functions without induction. Likewise with the fact that such a sum will be continuous. So for a finite family of continuous functions this is definitely the method to use. $\endgroup$
    – M A Pelto
    Dec 7, 2018 at 6:28

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

  • $\begingroup$ This is more a comment than an answer. $\endgroup$ Aug 13, 2018 at 17:10

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