Left-Hand Differentiable Function is Left-Continuous

Theorem

Let $f$ be a real function defined on an interval $I$.

Let $a$ be a point in $I$ where $f$ is left-hand differentiable.

Then $f$ is left-continuous at $a$.

Proof

By hypothesis, $\map {f'_-} a$ exists.

First we note that $a$ cannot be the left-hand end point of $I$ because values in $I$ less 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 Left-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 left-continuous at $a$.

$\blacksquare$