# Thread: Prove for Continuity of Minimum and Maximum Functions

1. ## Prove for Continuity of Minimum and Maximum Functions

Let
a) $f_1(x) = \min\{f_2(x), \ f_3(x)\}$, show that f_1(x) is continuous.

b) $f_4(x) = \max\{f_2(x), \ f_3(x)\}$ show that f_4(x) is continuous.

where $f_2(x)$ and $f_3(x)$ are both continuous.

a) I was thinking of splitting up into 2 where I assume firstly that $f_2(x)$ is the minimum secondly where $f_3(x)$. Thus if the minimum is $f_2(x)$ then $f_1(x) = f_2(x)$ thus for it to be continuous on any point c there exist a $\delta>0$ s.t. $0<|x-c|<\delta \Rightarrow |f_1(x)-f_1(c)|= |f_2(x) - f_2(c)|<\varepsilon$.

now if I $f_3(x)$ is the minimum then just replace $f_2(x)$ by f_3(x).

b) I would think this would be the same type of procedure as in a), where we have to split them up and show that individually they are continuous.

Would this be the correct way of doing it?

2. Originally Posted by lllll
Let
a) $f_1(x) = \min\{f_2(x), \ f_3(x)\}$, show that f_1(x) is continuous.

b) $f_4(x) = \max\{f_2(x), \ f_3(x)\}$ show that f_4(x) is continuous.

where $f_2(x)$ and $f_3(x)$ are both continuous.

a) I was thinking of splitting up into 2 where I assume firstly that $f_2(x)$ is the minimum secondly where $f_3(x)$. Thus if the minimum is $f_2(x)$ then $f_1(x) = f_2(x)$ thus for it to be continuous on any point c there exist a $\delta>0$ s.t. $0<|x-c|<\delta \Rightarrow |f_1(x)-f_1(c)|= |f_2(x) - f_2(c)|<\varepsilon$.

now if I $f_3(x)$ is the minimum then just replace $f_2(x)$ by f_3(x).
If you want to show that $f_1$ is continuous at $c$, you should split according to which is smaller at $c$, first. But this minimizer may change when we pick $x$ in a neighbourhood of $c$.
Let $\varepsilon>0$. There is $\delta>0$ such that $|x-c|<\delta$ implies both $|f_2(x)-f_2(c)|<\varepsilon$ and $|f_3(x)-f_3(c)|<\varepsilon$. (By the way, the numbering of the functions is a bit strange)
Assume $f_2(c)\leq f_3(c)$, so that $f_1(c)=f_2(c)$. Then, if $|x-c|<\delta$, we have $f_1(x)-f_1(c)=f_1(x)-f_2(c)\leq f_2(x)-f_2(c)<\varepsilon$.
And we have both $f_1(x)-f_1(c)\geq f_1(x)-f_2(c)$ and $f_1(x)-f_1(c)\geq f_1(x)-f_3(c)$, and one of these (depending if $f_2(x)=f_3(x)$ or the contrary) gives the inequality $f_1(x)-f_1(c)\geq -\varepsilon$.

Another quick proof relies on the identities : $\min(a,b)=\frac{a+b-|a-b|}{2}$ and $\max(a,b)=\frac{a+b+|a-b|}{2}$, once you know that the absolute value of a continuous function is continuous (this results easily from the triangular inequality).

3. Originally Posted by Laurent
Another quick proof relies on the identities : $\min(a,b)=\frac{a+b-|a-b|}{2}$ and $\max(a,b)=\frac{a+b+|a-b|}{2}$, once you know that the absolute value of a continuous function is continuous (this results easily from the triangular inequality).
To show that absolute value function are continuous, would it look something along the lines of:

$\exists \ \delta>0 \ \mbox{and} \ \varepsilon>0$ s.t $0<|x-c|<\delta \Rightarrow |f(x) -f(c)|<\varepsilon$

$||f_2(x)-f_3(x)|-|f_2(c)-f_3(c)||\leq|(f_2(x)-f_2(c)) -(f_3(x)-f_3(c) )|\leq \varepsilon$

then for the minimum we have:

$\frac{1}{2}\bigg{[}f_2(x)+f_3(x) -(f_2(c)+f_3(c)) - ||f_2(x)+f_3(x)-|f_2(c)+f_3(c)|| \bigg{]}$

$\frac{1}{2}\bigg{[}(f_2(x)-f_2(c))+(f_3(x) -f_3(c)) - \varepsilon \bigg{]}$

$\frac{1}{2}\bigg{[}(\varepsilon)+(\varepsilon) - \varepsilon \bigg{]}=\frac{\varepsilon}{2}$ ?

4. Originally Posted by lllll
To show that absolute value function are continuous, would it look something along the lines of:

$\exists \ \delta>0 \ \mbox{and} \ \varepsilon>0$ s.t $0<|x-c|<\delta \Rightarrow |f(x) -f(c)|<\varepsilon$

$||f_2(x)-f_3(x)|-|f_2(c)-f_3(c)||\leq|(f_2(x)-f_2(c)) -(f_3(x)-f_3(c) )|$ $\leq \textcolor{red}{|f_2(x)-f_2(c)|+|f_3(x)-f_3(c)|\leq 2} \varepsilon$

then for the minimum we have:

$\frac{1}{2}\textcolor{red}{\bigg{|}}f_2(x)+f_3(x) -(f_2(c)+f_3(c)) - ||f_2(x)+f_3(x)|-|f_2(c)+f_3(c)|| \textcolor{red}{\bigg{|}}$

$\leq \frac{1}{2}\bigg{[}|f_2(x)-f_2(c)|+|f_3(x) -f_3(c)| + 2\varepsilon \bigg{]}$

$\leq \frac{1}{2}\bigg{[}(\varepsilon)+(\varepsilon) + 2 \varepsilon \bigg{]}=2\varepsilon$

But this is not exactly what I was thinking about: I meant $f_2-f_3$ is continuous, hence $|f_2-f_3|$ is continuous (cf. below), which implies that $f_1=\frac{1}{2}(f_2+f_3-|f_2-f_3|)$ is a linear combination of continuous functions, and hence is continuous.
Lemma: if $f$ is continuous, then so is $|f|$. This is because $||f(x)|-|f(c)||\leq |f(x)-f(c)|$, as you wrote.