Extendability Theorem for Derivatives Continuous on Open Intervals

Theorem
Let $f$ be a continuous real function defined on an interval $\left[{a \,.\,.\, b}\right]$, $a<b$.

Then $f$ is continuously differentiable on $\left[{a \,.\,.\, b}\right]$ iff:
 * $f$ is continuously differentiable on $\left({a \,.\,.\, b}\right)$

and:
 * $\displaystyle \lim_{x \to a^+} f'\left({x}\right)$ and $\displaystyle \lim_{x \to b^-} f'\left({x}\right)$ exist.

Proof
We start with the "if" part.

We need to show that $f$ is continuously differentiable on $\left[{a \,.\,.\, b}\right]$.

We know that $f$ is continuously differentiable on $\left({a \,.\,.\, b}\right)$ so what remains to show is the following:


 * $f'\left({a}\right)$ exists.


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


 * $f'\left({b}\right)$ 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 $f$, $f'\left({a}\right)$ is the right-hand derivative of $f$ at $a$, denoted by $f'_+\left({a}\right)$.

We have per definition:


 * $\displaystyle f'_+\left({a}\right) = \lim_{\delta \to 0^+} \frac {f\left({a + \delta}\right) - f\left({a}\right)} {\delta}$

Since $f$ is continuous on $\left[{a \,.\,.\, b}\right]$ and differentiable on $\left({a \,.\,.\, b}\right)$, the Mean Value Theorem gives that a point $c$ in $\left({a \,.\,.\, a+\delta}\right)$ exists such that


 * $f\left({a+\delta}\right) - f\left({a}\right) = f'\left({c}\right)\delta$

We use this equation in the expression for $f'_+\left({a}\right)$:

Since $c$ is in $\left({a \,.\,.\, a+\delta}\right)$, we have $\left\vert{c - a}\right\vert < \delta$.

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

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

Since $\displaystyle \lim_{x \to a^+} f'\left({x}\right)$ exists, so does $\displaystyle \lim_{\delta \to 0^+} f'\left({c}\right)$ because $c \to a^+$ when $\delta \to 0^+$.

Furthermore, they are equal.

Hence, since we proved above that $\displaystyle \lim_{\delta \to 0^+} f'\left({c}\right)$ equals $f'_+\left({a}\right)$


 * $\displaystyle f'_+\left({a}\right) = \lim_{x \to a^+} f'\left({x}\right)$

Since $\displaystyle \lim_{x \to a^+} f'\left({x}\right)$ exists per presupposition, we conclude that $f'_+\left({a}\right)$ exists, and that their values are equal.

Hence $f'\left({a}\right)$ exists and equals $\displaystyle \lim_{x \to a^+} f'\left({x}\right)$.

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 $\displaystyle f'\left({a}\right) = \lim_{x \to a^+} f'\left({x}\right)$.

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

This finishes the "if" part of the proof.

The presupposition of the "only if" part of the proof is that $f$ is continuously differentiable on $\left[{a \,.\,.\, b}\right]$.

We need to show that


 * $(1): \quad f$ is continuously differentiable on $\left({a \,.\,.\, b}\right)$

and:


 * $(2): \quad \displaystyle \lim_{x \to a^+} f'\left({x}\right)$ and $\displaystyle \lim_{x \to b^-} f'\left({x}\right)$ exist.

$f$ is continuously differentiable on $\left({a \,.\,.\, b}\right)$ because $f$ is continuously differentiable on $\left[{a \,.\,.\, b}\right]$ and $\left({a \,.\,.\, b}\right)$ is a subset of $\left[{a \,.\,.\, b}\right]$.

What remains to show is that $\displaystyle \lim_{x \to a^+} f'\left({x}\right)$ and $\displaystyle \lim_{x \to b^-} f'\left({x}\right)$ exist.

We shall only make a proof of the existence of $\displaystyle \lim_{x \to a^+} f'\left({x}\right)$ as the proof of the existence of $\displaystyle \lim_{x \to b^-} f'\left({x}\right)$ is similar.

Since $f$ is differentiable at $a$, $f'\left({a}\right)$ exists.

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


 * $\displaystyle f'\left({a}\right) = \lim_{x \to a^+} f'\left({x}\right)$

Therefore, the right-hand side of this equation exists.

This finishes the proof.