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 $f(x +) \neq f(x-)$


2 Answers 2


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!

  • $\begingroup$ thanks for your answer Harald. I think it should say that $z \in (x,y]$ in the second paragraph $\endgroup$
    – Danny
    Commented Dec 31, 2013 at 1:40
  • $\begingroup$ @Danny You're welcome. And you're right, too. Fixed. $\endgroup$ Commented Dec 31, 2013 at 9:57
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    $\begingroup$ @HaraldHanche-Olsen Why is there $y > x$ with $f'(z) < \lambda$ for $z \in (x,y]$? Thanks. $\endgroup$
    – user370967
    Commented Aug 6, 2018 at 10:55
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    $\begingroup$ @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. $\endgroup$ Commented Aug 6, 2018 at 11:01
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    $\begingroup$ @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. $\endgroup$ Commented Aug 6, 2018 at 12:00

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.

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    $\begingroup$ This answer is incorrect. It seems that you are not understanding the difference between $f'_+(x)$ and $f'(x+)$ (similar for $f'_-(x)$ and $f'(x-)$. $\endgroup$
    – Mittens
    Commented Oct 1, 2022 at 1:26
  • $\begingroup$ 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. $\endgroup$
    – Mittens
    Commented Oct 2, 2022 at 4:47

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