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I have a question about the corollary to theorem 5.12 in Rudin's Principles of Mathematical Analysis (page 108):

Suppose $f$ is a real differentiable function on $[a,b]$ and suppose $f'(a)< \lambda < f'(b)$ then there is a point $x \in (a,b)$ such that $f'(x) = \lambda$

Corollary: If $f$ is differentiable on $[a,b]$ then $f'$ cannot have any simple discontinuities on $[a,b]$.

Can someone help me to show how he uses the result in the "main theorem" in the corollary?

(There are two cases of simple discontinuities $f(x+) = f(x-) \neq f(x)$ and $f(x +) \neq f(x-)$

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Danny
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2 Answers2

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The theorem in question (Darboux's theorem) basically states that the conclusion of the intermediate value theorem holds for the derivative of an everywhere differentiable function, even if the derivative is discontinuous.

For a simple discontinuity of either kind, it is true that either $f'(x-) \ne f'(x)$ or $f'(x+) \ne f'(x)$.

To be specific, let us treat the case where $f'(x)<f'(x+)$. Let $\lambda\in(f'(x),f'(x+))$, and pick $y>x$ with $f'(z)>\lambda$ whenever $z\in(x,y]$. Thus $f'(x)<\lambda<f'(x+)$, and $f'(z)\ne\lambda$ for all $z\in[x,y]$, contradicting Darboux's theorem.

The case $f'(x)>f'(x+)$ is treated similarly (or replace $f$ by $-f$ and use the case already treated). The case $f'(x-) \ne f'(x)$ is also treated the same way (or replace $f(x)$ by $f(-x)$).

Edited to fix a flaw pointed out in the comments, no less than seven years later!

Harald Hanche-Olsen
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  • thanks for your answer Harald. I think it should say that $z \in (x,y]$ in the second paragraph – Danny Dec 31 '13 at 01:40
  • @Danny You're welcome. And you're right, too. Fixed. – Harald Hanche-Olsen Dec 31 '13 at 09:57
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    @HaraldHanche-Olsen Why is there $y > x$ with $f'(z) < \lambda$ for $z \in (x,y]$? Thanks. –  Aug 06 '18 at 10:55
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    @Math_QED Because $f'(x-)<\lambda$ by choice. Note that $f'(x-)$ is short for the one-sided (from the left) limit of $f'$ at $x$, so the conclusion comes direct from the definition of one-sided limit. – Harald Hanche-Olsen Aug 06 '18 at 11:01
  • Thanks for the swift response. +1 –  Aug 06 '18 at 11:02
  • @HaraldHanche-Olsen I actually can't see how the statement is true for all z∈(x,y]. I see that it must be true for a certain z∈(x,y], but why for all z? –  Aug 06 '18 at 11:32
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    @Math_QED It's all $z\in[u,x)$ (and the opposite inequality in $(x,v]$). To take the first one, by the definition of the one-sided limit $f'(x-)$, for any $\varepsilon>0$ there is a $\delta>0$ so that $x-\delta<z<f'(x-)$ implies $|f'(z)-f'(x-)|<\varepsilon$. Now let $\varepsilon=\lambda-f'(x-)$, pick $\delta>0$ correspondingly, and take an arbitrary $u\in(x-\delta,x)$. Now for all $z\in[u,x)$, we have $x-\delta<z<f'(x-)$, therefore $|f'(z)-f'(x-)|<\varepsilon$, so $f'(z)<f'(x-)+\varepsilon=\lambda$, and we're done. – Harald Hanche-Olsen Aug 06 '18 at 12:00
  • Thanks. Clear now. –  Aug 06 '18 at 13:59
  • @HaraldHanche-Olsen One last question. How does the last statement lead to a contradiction? Since $f'(u) < \lambda < f'(v)$, there is $y \in (u,v)$ with $f'(y) = \lambda$. Then $y = x$, because the other options aren't possible, and $f'(x) = \lambda$. What's the contradiction? –  Aug 06 '18 at 16:02
  • @Math_QED I suppose the easiest way is to also ensure that $\lambda\ne f'(x)$. (After all, $\lambda$ can be freely chosen in an open interval.) Then $f'$ never takes the value $\lambda$ inside $[u,v]$, which would contradict the intermediate value theorem since $f'(u)<\lambda<f'(v)$. – Harald Hanche-Olsen Aug 06 '18 at 16:31
  • To consider a narrower interval than $[a, b]$ is a key. – tchappy ha Feb 16 '19 at 12:52
  • Thanks a lot for your answer! But let me ask you a question, please: when you considered the case of simple discontinuity of the first kind as far as I noticed you have never used the fact that $f'(x+)=f'(x-)$. Because you can take $\lambda \in (f'(x+),f'(x))$ and we can derive contradiction in the same way. And notice that in this process we did not use that $f'(x+)=f'(x-)$. Please correct if I am wrong – RFZ Apr 13 '21 at 01:41
  • Also in the case of second type I do not see how you can get contradiction. The function $f'(x)$ has Intermediate Value property on $[u,v]$, right? Since $f'(u)<\lambda<f'(v)$ then by Darboux's theorem exists $\xi \in (u,v)$ such that $f'(\xi)=\lambda$ then it follows that $\xi=x$, i.e. $f'(x)=\lambda$, right? So where is the contradiction? – RFZ Apr 13 '21 at 01:59
  • @ZFR You're right that I don't use $f'(x+)=f'(x-)$ for the first case. I did not need it! As for your second question, well spotted! The possibility that $f'(x)=\lambda$ is indeed problematic. I now see that the best way forward is to notice that for both cases, either $f'(x+) \ne f'(x)$ or $f'(x-) \ne f'(x)$, and then to proceed as I did for the first case. – Harald Hanche-Olsen Apr 16 '21 at 08:43
  • I edited the answer in accordance with the above comment. – Harald Hanche-Olsen Apr 16 '21 at 08:59
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Assume, on the contrary, that $f^{\prime}$ has a simple discontinuity of type (i), i.e., jump discontinuity such that $f^{\prime}(x-)\ne f^{\prime}(x+)$ in which case $f^{\prime}(x)$ is immaterial.
Then it follows that $f$ is not differentiable at that point, contrary to the fact that $f$ is differentiable on the interval $[a,b]$.
So $f^{\prime}$ cannot have simple discontinuity of type (i).
Since $f$ is continuous, the derivatives of $f$ need not have a simple discontinuity of type (ii).

In addition, suppose $x∈[a,b]$ is a point of discontinuity of $f^{\prime}$ such that $f^{\prime}(x-)\ne f^{\prime}(x+)$, since $f$ is differentiable on $[a,b]$, we see that $f$ is differentiable on $[x-δ,x+δ]$ for every $δ>0$, suppose w.l.o.g $f^{\prime} (x-δ)<f^{\prime} (x+δ)$, then Theorem 5.12 shows that $f^{\prime} (x-δ)<f^{\prime} (x)<f^{\prime} (x+δ)$. Since $δ$ was arbitrary, it follows that $f^{\prime} (x-)<f^{\prime} (x)<f^{\prime} (x+)$. To this case, $f^{\prime} (x)$ is material.
So, we conclude that the function $f^{\prime}$ cannot have any simple discontinuity.

Myo Nyunt
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    This answer is incorrect. It seems that you are not understanding the difference between $f'+(x)$ and $f'(x+)$ (similar for $f'-(x)$ and $f'(x-)$. – Mittens Oct 01 '22 at 01:26
  • You posted the same solution in another posting. That may raise eyebrows. As a general rule, do not post the same solution in different postings. – Mittens Oct 02 '22 at 04:47