Continuous functions

**dwsmith** · July 12th 2011, 10:19 AM

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.

**girdav** · July 12th 2011, 10:37 AM

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 $a,b\in\mathbb R$ .

**dwsmith** · July 12th 2011, 11:13 AM

Originally Posted by girdav

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.

**girdav** · July 12th 2011, 11:23 AM

Originally Posted by dwsmith

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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