Let $f, g$ be functions
- $f$ is continuous at a,
- $\operatorname{f}(a) = b \in \operatorname{Dom}(f)$,
- $g$ is continuous at b.
Then $g\circ f$ is continuous at $a$.
Proof: Let $\varepsilon>0$,
$$\exists\delta'>0, \text{ s.t. } \lvert g(y)-g(b)\rvert<\varepsilon\qquad\forall y\in\operatorname{Dom}(g) \text{ s.t. } \lvert\ y-b\rvert < \delta'$$
$$\exists\delta>0, \text{ s.t. } \lvert f(x)-f(a)\rvert<\delta'\qquad\forall x\in\operatorname{Dom}(f) \text{ s.t. } \lvert\ x-a\rvert < \delta$$
$$\implies \forall x\in\operatorname{Dom}(g\circ f) \text{ s.t. } \lvert x-a \rvert < \delta, \text{ we have:}$$
$$\lvert(g\circ f)(x) – (g\circ f)(x)\rvert = \lvert g(f(x)) – g(f(a)) \rvert < \varepsilon, \qquad \text{ since } \lvert f(x)-f(a)\rvert<\delta'$$
Why does $\lvert f(x)-f(a)\rvert<\delta'\implies \lvert g(f(x)) – g(f(a)) \rvert < \varepsilon$?
Best Answer
Because $\delta'$ was chosen so that whenever an element $y$ of the domain of $g$ is $\delta'$-close to $b$, i.e.
$$|y-b|<\delta' $$ then the corresponding images under $g$ are automatically $\epsilon$-close, i.e.
$$|g(y)-g(b)|<\epsilon $$ because of the continuity of $g$.
Here, the two domain elements of $g$ are $\underbrace{f(x)}_{y}$ and $\underbrace{f(a)}_{b}$, so the guaranteed result is that
$$ |g(\;\underbrace{f(x)}_{y}\;)-g(\;\underbrace{f(a)}_{b}\;)|<\epsilon $$