Right-Hand Differentiable Function is Right-Continuous
Jump to navigation
Jump to search
Theorem
Let $f$ be a real function defined on an interval $I$.
Let $a$ be a point in $I$ where $f$ is right-hand differentiable.
Then $f$ is right-continuous at $a$.
Proof
By hypothesis, $\map {f'_+} a$ exists.
First we note that $a$ cannot be the right hand end point of $I$ because values in $I$ greater than $a$ need to exist for $\map {f'_+} a$ to exist.
We form the following expression:
- $\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a}$
We need to show that it is defined and to find its value.
We find:
\(\ds \) | \(\) | \(\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to a^+} \paren {\frac {\map f x - \map f a} {x - a} \paren {x - a} }\) | where the denominator is unequal to $0$ since $x > a$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to a^+} \paren {\frac {\map f x - \map f a} {x - a} } \lim_{x \mathop \to a^+} \paren {x - a}\) | Product Rule for Limits of Real Functions since (see the next step) the two limits exist | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {f'_+} a \times 0\) | Definition of Right-Hand Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
Note that this proves that $\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a}$ exists.
We continue by manipulating the result above:
\(\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a}\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a} + \map f a - \map f a\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a} + \lim_{x \mathop \to a^+} \map f a - \map f a\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a + \map f a} - \map f a\) | \(=\) | \(\ds 0\) | Sum Rule for Limits of Real Functions since the two limits in the previous expression exist | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \lim_{x \mathop \to a^+} \map f x - \map f a\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \lim_{x \mathop \to a^+} \map f x\) | \(=\) | \(\ds \map f a\) |
which means that $f$ is right-continuous at $a$.
$\blacksquare$