Analysis problem

**Cato** · Oct 29th 2008, 10:30 AM

Let g: [a,b] --> [c,d] be a continuous function in $x_0\in$ [a,b], and f: [c,d] --> R a continuous function in g( $x_0$ ) $\in$ [c,d]

Show that f o g: [a,b] --> R defined by (f o g)(x) = f(g(x)) $\forall$ x $\in$ [a,b] is continuous in $x_0 \in$ [a,b]

**Cato** · Oct 29th 2008, 04:11 PM

Anyone have any ideas on solving this?

**Moo** · Oct 30th 2008, 12:06 PM

Hello,

Originally Posted by Cato

Let g: [a,b] --> [c,d] be a continuous function in $x_0\in$ [a,b], and f: [c,d] --> R a continuous function in g( $x_0$ ) $\in$ [c,d]

Show that f o g: [a,b] --> R defined by (f o g)(x) = f(g(x)) $\forall$ x $\in$ [a,b] is continuous in $x_0 \in$ [a,b]

Use the definition of continuity for f in $x_0$ :
$\forall \epsilon > 0,~ \exists \delta > 0,~ |x-x_0|< \delta \Rightarrow |f(x)-f(x_0)|< \epsilon$

**ThePerfectHacker** · Oct 30th 2008, 01:02 PM

Originally Posted by Cato

Show that f o g: [a,b] --> R defined by (f o g)(x) = f(g(x)) $\forall$ x $\in$ [a,b] is continuous in $x_0 \in$ [a,b]

Let me rephrase the problem a little. Let $g: D_1 \to D_2$ and $f: D_2 \to \mathbb{R}$ . Let $g$ be continous at $x_0 \in D_1$ and $f$ be continous at $g(x_0) \in D_2$ . Let $\epsilon > 0$ . There is $\delta > 0$ so that if $y\in D_2 \text{ and }|y-g(x_0)| < \delta \implies |f(y) - f(g(x_0))| < \epsilon$ . There is $\delta_1 > 0$ so that $x\in D_1 \text{ and }|x-x_0| < \delta_1 \implies |g(x) - g(x_0)| < \delta \implies |f(g(x)) - f(g(x_0))| < \epsilon$ . Thus, we see that $f\circ g$ is continous at $x_0$ .

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