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Math Help - Prove for Continuity of Minimum and Maximum Functions

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by lllll View Post
    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).
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  3. #3
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    Quote Originally Posted by Laurent View Post
    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} ?
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  4. #4
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    Quote Originally Posted by lllll View Post
    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.
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