Minimum of Finitely Many Continuous Real Functions is Continuous
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Theorem
Let $n \ge 2$ be a natural number.
Let $X \subseteq \R$.
Let $f_1, f_2, \ldots, f_n$ be functions $X \to \R$.
Define the function $m : X \to \R$ by:
- $\ds \map m x = \min_i \map {f_i} x$
for all $x \in X$.
Then $m$ is continuous.
Proof
We proceed by induction.
For all natural numbers $n \ge 2$, let $\map P n$ be the proposition:
- for every collection of $n$ functions $f_1, f_2, \ldots, f_n : X \to \R$, $m$ is continuous.
Basis for the Induction
Take $n = 2$.
We have, for each $x \in X$:
\(\ds \map m x\) | \(=\) | \(\ds \min \set {\map {f_1} x, \map {f_2} x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {\paren {\map {f_1} x + \map {f_2} x} - \size {\map {f_1} x - \map {f_2} x} }\) | Minimum Function in terms of Absolute Value |
From Combined Sum Rule for Continuous Real Functions:
- $f_1 + f_2$ and $f_1 - f_2$ are continuous.
From Absolute Value of Continuous Real Function is Continuous:
- $\size {f_1 - f_2}$ is continuous.
Applying Combined Sum Rule for Continuous Real Functions again, we obtain:
- $m$ is continuous.
So $\map P 2$ is true.
This is our base case.
Induction Hypothesis
Suppose that:
- for every collection of $N$ functions $f_1, f_2, \ldots, f_N : X \to \R$, $m$ is continuous.
That is:
- $\map P N$ is true for some $N$.
We aim to show that:
- $\map P {N + 1}$ is true.
Induction Step
For each $x \in X$, we have:
\(\ds \map m x\) | \(=\) | \(\ds \min_i \map {f_i} x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \min \set {\min_{i \mathop \le N} \map {f_i} x, \map {f_{N + 1} } x}\) |
By the induction hypothesis:
- $\ds \min_{i \mathop \le N} f_i$
is continuous.
Since $m$ is the minimum of two continuous functions, it is continuous by the base case.
So $\map P N \implies \map P {N + 1}$ and the result follows by the Principle of Mathematical Induction.
$\blacksquare$