# Prove for Continuity of Minimum and Maximum Functions

• Nov 11th 2008, 11:43 AM
lllll
Prove for Continuity of Minimum and Maximum Functions
Let
a) $\displaystyle f_1(x) = \min\{f_2(x), \ f_3(x)\}$, show that f_1(x) is continuous.

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

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

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

now if I $\displaystyle f_3(x)$ is the minimum then just replace $\displaystyle f_2(x)$ 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?
• Nov 11th 2008, 12:17 PM
Laurent
Quote:

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

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

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

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

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

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

Another quick proof relies on the identities : $\displaystyle \min(a,b)=\frac{a+b-|a-b|}{2}$ and $\displaystyle \max(a,b)=\frac{a+b+|a-b|}{2}$, once you know that the absolute value of a continuous function is continuous (this results easily from the triangular inequality).
• Nov 11th 2008, 09:31 PM
lllll
Quote:

Originally Posted by Laurent
Another quick proof relies on the identities : $\displaystyle \min(a,b)=\frac{a+b-|a-b|}{2}$ and $\displaystyle \max(a,b)=\frac{a+b+|a-b|}{2}$, 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:

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

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

then for the minimum we have:

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

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

$\displaystyle \frac{1}{2}\bigg{[}(\varepsilon)+(\varepsilon) - \varepsilon \bigg{]}=\frac{\varepsilon}{2}$ ?
• Nov 11th 2008, 10:12 PM
Laurent
Quote:

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

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

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

then for the minimum we have:

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

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

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

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