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I'm trying to deduce from the following theorem that $f'(x)$ cannot have jump discontinuities.

Theorem :

Let $f(x)$ be continuous in a point $c$ and differentiable in a (right) neighbourhood $I_{+}(c)-\big \{c \big \}$. If the limit $lim_{x\to > c^{+}} f'(x)=l \in \mathbb{R}$ exists and it's finite, than $f(x)$ is differentiable (from the right) in $c$ and $f'_{+}(c)=l$. The same is valid for $f'_{-}(c)$ (from the left).

Is it correct to deduce the following, just using the contronominal of that implication (denying thesis and hypothesis)?

Let $f(x)$ differentiable in all $I_{+}(c)$, if $ lim_{x\to c^{+}} f'(x)\neq f'_{+}(c)(=f'(c))$ (which means $f'(c)$ is not continuous in $c$) than the limit $ lim_{x\to c^{+}} f'(x)$ either does not exists or it is equal to $\infty$ (which is indeed an essential discontinuity).

If this is correct, does it also prove that $f(x)$ cannot have removable discontinuities?

Edit

With $I_{+}(c)$ I mean a neighbourhood "from the right" of the point $c$. If I choose a particular length $\delta$ than $I_{+}(c)=(c,c+\delta)$

Similarly $f'_{+}(c)=$ means the derivative from the right, thus $f'_{+}(c)=lim_{x\to c^{+}} \frac{f(x)-f(c)}{x-c}$

tinlyx
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midu10
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1 Answers1

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I am confused slightly by your notation/wording, but here is a hint I believe you'll find useful:

Let $h(x)=\begin{cases}0&x<0\\1&x\geq0\end{cases}$ be the Heaviside function and consider the function $$f(x)=\int_0^xh(t)dt.$$

charlestoncrabb
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    What does this show? The function $f$ is not differentiable at $0$ so the question of continuity of $f'$ at $0$ does not arise. Perhaps you were under the impression that via FTC $f'(x) =h(x) $ everywhere. This is not correct and the equation $f'(x) =h(x) $ may not hold at point where where $h$ is discontinuous. – Paramanand Singh Mar 11 '18 at 05:01