Jump discontinuity implies that integral function is not differentiable

I am trying to prove following theorem . Is there any flaw in below proof

Assume that $$f$$ is integrable on $$[a,b]$$ and has a jump discontinuity at $$c \in (a,b)$$ this means that both one sided limits exist as $$x$$ approaches $$c$$ from the left and right but that $$\lim_{x \rightarrow c^{-}} f(x) \neq \lim_{x \rightarrow c^{+}} f(x)$$ then show that function $$F(x) = \int_a^x f(t) \, dt$$ is not differentiable at $$x=c$$.

Following is the proof strategy

• prove that $$\lim_{x \rightarrow c^{-}} \frac{F(x) - F(c)}{x-c} = \lim_{x \rightarrow c^{-}} f(x)$$
• prove that $$\lim_{x \rightarrow c^{+}} \frac{F(x) - F(c)}{ x- c} = \lim_{x \rightarrow c^{+}} f (x)$$
• from given hypothesis we immediately have that $$\lim_{x \rightarrow c^{-}} \frac{F(x) - F(c)}{x-c} \neq \lim_{x \rightarrow c^{+}} \frac{F(x) - F(c)}{x-c}$$
• conclude that function $$F$$ is not differentiable at $$c$$.

first part can proved as follows

we will show that $$\lim_{x \rightarrow c^{-}} \frac{F(c) - F(x)}{c-x} - f(x) = 0$$ consider arbitrary $$\epsilon > 0$$

$$|\frac{F(x) - F(c)}{x-c} - f(x) | \leq |\frac{F(x) - F(c)}{x-c} - f(c)| + |f(c) - f(x)| < \epsilon/2 + \epsilon/2 = \epsilon$$

sine we know that $$f$$ is left continuous we have that $$F$$ is left differetiable from Fundamental Theorem of Calculus then there exists $$\delta_1, \forall x$$ s.t $$c-x < \delta \implies |\frac{F(x) - F(c)}{x-c} -f(c)| < \epsilon/2$$ and again since $$f$$ is left continuous we have that there exists $$\delta_2, \forall x$$ s.t $$\forall c-x < \delta \implies |f(x) - f(c)| < \epsilon/2$$ required $$\delta = \min\{\delta_1 , \delta_2\}$$ and similarly we can show for right limits and we are done

• Looks good to me although I wonder if you should write there exists $\delta_1, \forall x$ s.t $c-x < \delta_1 \implies |\frac{F(x) - F(c)}{x-c} -f| < \varepsilon/2$ and similarly for $\delta_2$? Your picking of a $\delta$ which is the minimum of these two then holds up the proof,
– user284001
Jan 22 '20 at 9:43
• Are you assuming that $f$ is continuous everywhere except possibly at $c$? You have used this but have not mentioned it in the body, and you cannot use FTC in parts where you haven't been given it, right? Jan 22 '20 at 9:54
• Are you assuming that $f$ is continuous everywhere except possibly at $c$? You have used this but have not mentioned it in the body. Jan 22 '20 at 9:56
• I would like to think that it does not hold, but I can't find a counterexample. Jan 22 '20 at 12:16
• I fixed a few typos. Hope you don't mind. Jan 22 '20 at 12:38

You can't assume left or right continuity of $$f$$. Use the same approach as in proof of FTC and show the following.

Lemma: If $$f$$ is Riemann integrable on $$[a, b]$$ and if for some $$c\in[a, b)$$ the limit $$\lim_{x\to c^{+}}f(x)$$ exists then the function $$F$$ defined by $$F(x) =\int_{a} ^{x} f(t) \, dt$$ has right derivative at $$c$$ and we have $$D^{+} F(c)=\lim_{x\to c^{+}} f(x)$$.

Let $$f(x) \to L$$ as $$x\to c^{+}$$ and $$\epsilon >0$$ be arbitrary. Then there is a $$\delta >0$$ such that $$|f(t) - L|<\epsilon$$ whenever $$0. Thus we have $$L-\epsilon whenever $$0. Integrating the above inequality in interval $$[c, c+h]$$ where $$0 we get $$h(L-\epsilon) <\int_{c} ^{c+h} f(t) \, dt or $$L-\epsilon<\frac{F(c+h) - F(c)} {h} whenever $$0. Thus $$D^{+} F(c) =\lim_{h\to 0^{+}}\frac {F(c+h) - F(c)} {h} =L=\lim_{x\to c^{+}} f(x)$$ A similar lemma can be proved in same manner for left limit and left derivative. Going further if $$f(x) \to L$$ as $$x\to c$$ then $$F$$ is differentiable at $$c$$ with $$F'(c) =L$$ and if $$f$$ is continuous at $$c$$ then $$L=f(c)$$ so that $$F'(c) =f(c)$$ which brings us to the traditional form of FTC.

It is now obvious that if $$f$$ has a jump discontinuity at some point then $$F$$ is not differentiable at that point. However if $$f$$ has essential discontinuity at some point then $$F$$ may or may not be differentiable at that point.

• But if $f$ has removable discontinuity then $F$ is differentiable at that point right ? Jan 22 '20 at 12:40
• @viru: I added that also in my answer. See second last paragraph. Jan 22 '20 at 12:47
• Yes . That is what made me ask about remaining case of discontinuity which is removable discontinuity.I was thinking removable discontinuity is different from Essential Discontinuity. Removable discontinuity is when $\Big[ \lim_{x \rightarrow c^{+}} f = \lim_{x \rightarrow c^{-}}f \Big] \neq f(c)$ which readily implies from your proof that it is differentiable at c . right ? or is there anything that I am missing Jan 22 '20 at 14:16
• @viru: you are correct. I have stated that in that case $F'(c) =L=\lim_{x\to c} f(x)$. Jan 22 '20 at 15:32
• @viru: essential discontinuity is when one or both of $\lim_{x\to c^{+}} f(x), \lim_{x\to c^{-}} f(x)$ don't exist. As an example let $f(x) =\cos(1/x),x\neq 0,f(0)=0$. Then $f$ has essential discontinuity at $0$ but $F'(0)$ exists and equals $0$. Jan 22 '20 at 15:34