Thread: Analysis problem

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

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

Anyone have any ideas on solving this?

Hello,

Originally Posted by Cato

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

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

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

Originally Posted by Cato

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

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