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Math Help - Continuous functions

  1. #1
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    Continuous functions

    Suppose f,g: D\to \mathbb{R} are both continuous on D. Define h: D\to\mathbb{R} by h(x)=\text{max}\{f,g\}. Show that h is continuous on D.

    Case 1: f(x)\geq g(x)

    Case 2: f(x)=g(x)

    Are these the two cases, and if so, I am not sure how to prove this.
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  2. #2
    Super Member girdav's Avatar
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    Re: Continuous functions

    If the case 1 is " \forall x\in D,\, f(x)\geq g(x)" and the case two is " \forall x\in D,\, f(x)\leq g(x)", we didn't cover all the cases. For example, with D=\mathbb R, f(x) =x and g(x)=-x.

    You can use the formula \max\left\{a,b\right\}=\frac{a+b+|a-b|}2 for a,b\in\mathbb R.
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    Re: Continuous functions

    Quote Originally Posted by girdav View Post
    If the case 1 is " \forall x\in D,\, f(x)\geq g(x)" and the case two is " \forall x\in D,\, f(x)\leq g(x)", we didn't cover all the cases. For example, with D=\mathbb R, f(x) =x and g(x)=-x.

    You can use the formula \max\left\{a,b\right\}=\frac{a+b+|a-b|}2 for a,b\in\mathbb R.
    Why would your case 2 be needed? After showing case 1, could we just say for f(x)\leq g(x) it follows similarly to case 1?

    Also, I don't see how \max\left\{a,b\right\}=\frac{a+b+|a-b|}2 is going to help.
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    Super Member girdav's Avatar
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    Re: Continuous functions

    Quote Originally Posted by dwsmith View Post
    Why would your case 2 be needed? After showing case 1, could we just say for f(x)\leq g(x) it follows similarly to case 1?

    Also, I don't see how \max\left\{a,b\right\}=\frac{a+b+|a-b|}2 is going to help.
    Yes, but we still don't cover all the cases (the example I gave shows that we can have neither the first case (and the similar) nor the second.

    You can write h =\frac{f+g+|f-g|}2 and you have to use the fact that the sum of two continuous functions is continuous, and if u is continuous then so is |u|.
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