The set on which $f $ is defined, called the domain of $f $ and written $\dom{ f } $.
If the domain is not specified, then it is default to the natural domain, which is the largest subset of $\R $ on which the function is well defined real-valued function.
Let $f $ be a real-valued function whose domain is a subset of $\R $. The function $f $ is continuous at $x_0 $ in $\dom{ f } $ if
$$\forall (x_n) \in \dom{ f }, \lim (x_n) = x_0 ,\text{ we have }\lim f(x_n) = f(x_0)$$
If $f $ is continuous at each point of a set $S \subseteq \dom{ f }$, then $f $ is said to be continuous on $S $. The function $f $ is said to be continuous if it is continuous on $\dom{ f }$.
Let $f $ be a real-valued function with $\dom{ f } \subseteq \R $.
$f $ is continuous at $x_0 $ in $\dom{ f } \iff $
$$\forall \epsilon > 0, \exists \delta > 0\text{ s.t. }x \in \dom{ f }, \abs{ x - x_0 } < \delta \implies \abs{ f(x) - f(x_0)} < \epsilon$$
Let $f(x) = 2x^2 + 1$ for $x \in \R $. Prove $f $ is continuous on $\R$ by
(a) Using the definition,
(b) Using the $\epsilon - \delta $ property.
(a) Suppose $\lim x_n = x_0 $. Then we have
$\lim f(x_n) = \lim (2x_n^2 + 1) = 2 \pare{\lim x_n}^2 + 1 = 2 x_0^2 + 1 = f(x_0)$
(b) Fix $x_0 \in \R $, fix $\epsilon > 0 $. We want to show $\exists \delta $ s.t. $\abs{ x - x_0 } < \delta \implies \abs{ f(x) - f(x_0)} < \epsilon$.
$\abs{f(x) - f(x_0)} = \br \abs{ 2x^2 +1 - (2x_0^2 + 1)} = \abs{ 2x^2 - 2x_0^2 } = 2 \abs{ x -x_0 } \abs{ x + x_0 } $
We need to get a bound for $\abs{ x + x_0 }$ that does not depend on $x $. We notice that if $\abs{ x - x_0 }< 1 $, say, then $\abs{ x } < \abs{ x_0 } + 1 $ and hence $\abs{ x + x_0 } \leq \abs{ x } + \abs{ x_0 } < 2 \abs{ x_0 } + 1 $. Thus we have
$\abs{ f(x) - f(x_0)} \leq 2 \abs{ x - x_0 } (2 \abs{ x_0 } + 1) $ provided $\abs{ x - x_0 } < 1$.
To arrange for $2 \abs{ x - x_0 } (2 \abs{ x_0 } + 1) < \epsilon $, it suffices to have $\abs{ x - x_0 } < \frac{ \epsilon }{ 2 (2 \abs{ x_0 } + 1)} $ and also $\abs{ x - x_0 } < 1 $. So we put
$$\delta = \min{ \set{ 1, \frac{ \epsilon }{ 2(2 \abs{ x_0 } + 1)}}} $$
Let $f $ be a real-valued function with $\dom{ f } \subseteq \R $.
If $f $ is continuous at $x_0 $ in $\dom{ f } \implies \abs{ f } $ and $kf $, $k \in \R$, are continuous at $x_0$.
Let $f$ and $g $ be a real-valued function.
$f$ and $g$ are continuous at $x_0 \in \R$. Then
- $f+ g $ is continuous at $x_0 $.
- $fg $ is continuous at $x_0$.
- $f/g $ is continuous at $x_0$ if $g(x_0) \neq 0$.
If $f $ is continuous at $x_0 $ and $g $ is continuous at $f(x_0) $, then the composite function $g \circ f $ is continuous at $x_0 $.