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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

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  • $\begingroup$ 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, $\endgroup$
    – user284001
    Jan 22, 2020 at 9:43
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    $\begingroup$ 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? $\endgroup$ Jan 22, 2020 at 9:54
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    $\begingroup$ Are you assuming that $f$ is continuous everywhere except possibly at $c$? You have used this but have not mentioned it in the body. $\endgroup$ Jan 22, 2020 at 9:56
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    $\begingroup$ I would like to think that it does not hold, but I can't find a counterexample. $\endgroup$ Jan 22, 2020 at 12:16
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    $\begingroup$ I fixed a few typos. Hope you don't mind. $\endgroup$
    – Paramanand Singh
    Jan 22, 2020 at 12:38

1 Answer 1

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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 $0<t-c<\delta$. Thus we have $$L-\epsilon <f(t) <L+\epsilon$$ whenever $0<t-c<\delta$. Integrating the above inequality in interval $[c, c+h] $ where $0<h<\delta$ we get $$h(L-\epsilon) <\int_{c} ^{c+h} f(t) \, dt<h(L+\epsilon) $$ or $$L-\epsilon<\frac{F(c+h) - F(c)} {h} <L+\epsilon$$ whenever $0<h<\delta$. 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.

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  • $\begingroup$ But if $f$ has removable discontinuity then $F$ is differentiable at that point right ? $\endgroup$
    – manifold
    Jan 22, 2020 at 12:40
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    $\begingroup$ @viru: I added that also in my answer. See second last paragraph. $\endgroup$
    – Paramanand Singh
    Jan 22, 2020 at 12:47
  • $\begingroup$ 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 $\endgroup$
    – manifold
    Jan 22, 2020 at 14:16
  • $\begingroup$ @viru: you are correct. I have stated that in that case $F'(c) =L=\lim_{x\to c} f(x) $. $\endgroup$
    – Paramanand Singh
    Jan 22, 2020 at 15:32
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    $\begingroup$ @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 $0$. $\endgroup$
    – Paramanand Singh
    Jan 22, 2020 at 15:34

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