Extendability Theorem for Derivatives Continuous on Open Intervals

Theorem
Let $f$ be a continuous real function defined on an interval $\closedint a b$ where $a < b$.

Then $f$ is continuously differentiable on $\closedint a b$ :
 * $f$ is continuously differentiable on $\openint a b$

and:


 * $\ds \lim_{x \mathop \to a^+} \map {f'} x$ and $\ds \lim_{x \mathop \to b^-} \map {f'} x$ exist.

Necessary Condition
Suppose that $f$ is continuously differentiable on $\closedint a b$.

We need to show that:


 * $f$ is continuously differentiable on $\openint a b$

and:


 * $\ds \lim_{x \mathop \to a^+} \map {f'} x$ and $\ds \lim_{x \mathop \to b^-} \map {f'} x$ exist.

$f$ is continuously differentiable on $\openint a b$ because:
 * $f$ is continuously differentiable on $\closedint a b$
 * $\openint a b$ is a subset of $\closedint a b$.

What remains to show is that $\ds \lim_{x \mathop \to a^+} \map {f'} x$ and $\ds \lim_{x \mathop \to b^-} \map {f'} x$ exist.

We shall only make a proof of the existence of $\ds \lim_{x \mathop \to a^+} \map {f'} x$ as the proof of the existence of $\ds \lim_{x \mathop \to b^-} \map {f'} x$ is similar.

Since $f$ is differentiable at $a$, $\map {f'} a$ exists.

Since $f'$ is continuous at $a$, the following equation is true:


 * $\ds \map {f'} a = \lim_{x \mathop \to a^+} \map {f'} x$

Therefore the of this equation exists.

Sufficient Condition
Suppose that:


 * $f$ is continuously differentiable on $\openint a b$

and:


 * $\ds \lim_{x \mathop \to a^+} \map {f'} x$ and $\ds \lim_{x \mathop \to b^-} \map {f'} x$ exist.

We need to show that $f$ is continuously differentiable on $\closedint a b$.

We know that $f$ is continuously differentiable on $\openint a b$ so what remains to show is the following:


 * $\map {f'} a$ exists


 * $f'$ is continuous at $a$


 * $\map {f'} b$ exists


 * $f'$ is continuous at $b$.

We shall limit ourselves to constructing a proof of the first two of these statements as the proof of the last two statements is similar.

Since $a$ is the left-most end point of the domain of $f$, $\map {f'} a$ is the right-hand derivative of $f$ at $a$, denoted by $\map {f'_+} a$.

We have per definition:


 * $\ds \map {f'_+} a = \lim_{\delta \mathop \to 0^+} \frac {\map f {a + \delta} - \map f a} \delta$

Since $f$ is continuous on $\closedint a b$ and differentiable on $\openint a b$, the Mean Value Theorem gives that a point $c$ in $\openint a {a + \delta}$ exists such that


 * $\map f {a + \delta} - \map f a = \map {f'} c \delta$

We use this equation in the expression for $\map {f'_+} a$:

Since $c$ is in $\openint a {a + \delta}$, we have $\size {c - a} < \delta$.

Therefore, $c$ approaches $a$ when $\delta$ approaches 0.

Moreover, since $c > a$, $c$ approaches $a$ from above.

Since $\ds \lim_{x \mathop \to a^+} \map {f'} x$ exists, so does $\ds \lim_{\delta \mathop \to 0^+} \map {f'} c$ because $c \to a^+$ when $\delta \to 0^+$.

Furthermore, they are equal.

Hence, since we proved above that $\ds \lim_{\delta \mathop \to 0^+} \map {f'} c$ equals $\map {f'_+} a$


 * $\ds \map {f'_+} a = \lim_{x \mathop \to a^+} \map {f'} x$

Since $\ds \lim_{x \mathop \to a^+} \map {f'} x$ exists per presupposition, we conclude that $\map {f'_+} a$ exists, and that their values are equal.

Hence $\map {f'} a$ exists and equals $\ds \lim_{x \mathop \to a^+} \map {f'} x$.

Thus, the first of the two statements we needed to prove is proven.

Now we shall prove that $f'$ is continuous at $a$.

Since $a$ is the left-most end point of the domain of $f$, $f'$ being continuous at $a$ means that $\ds \map {f'} a = \lim_{x \mathop \to a^+} \map {f'} x$.

Since we just proved this, we conclude that $f'$ is continuous at $a$.

Also see

 * Right-Hand Derivative not Limit of Derivative from Right
 * Extendability Theorem for Function Continuous on Open Interval