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Thread: Continuous functions

  1. #1
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    Continuous functions

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

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

    Case 2: $\displaystyle f(x)=g(x)$

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

    You can use the formula $\displaystyle \max\left\{a,b\right\}=\frac{a+b+|a-b|}2$ for $\displaystyle a,b\in\mathbb R$.
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  3. #3
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    Re: Continuous functions

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

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

    Also, I don't see how $\displaystyle \max\left\{a,b\right\}=\frac{a+b+|a-b|}2$ 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 $\displaystyle f(x)\leq g(x)$ it follows similarly to case 1?

    Also, I don't see how $\displaystyle \max\left\{a,b\right\}=\frac{a+b+|a-b|}2$ 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 $\displaystyle h =\frac{f+g+|f-g|}2$ and you have to use the fact that the sum of two continuous functions is continuous, and if $\displaystyle u$ is continuous then so is $\displaystyle |u|$.
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