I've been grading the following problem for an undergraduate numerical analysis course:

Suppose $f \in C([a,b])$ and $f'(x)$ exists on $(a,b)$. Show that if $f'(x) \neq 0$ for all $x \in (a,b)$, then there can exist at most one number $p$ in $[a,b]$ with $f(p) = 0$.

Proof by contradiction plus the Mean Value Theorem does the trick just fine. But a few students tried proving it directly, as follows:

By the MVT, there exists $c \in (a,b)$ such that $$ f'(c) = \frac{f(b)-f(a)}{b-a}.$$ $f'(c) \neq 0$, hence $f(b) \neq f(a)$. Then either $f(a) < f(b)$ or $f(b) < f(a)$. But then $f$ is either strictly increasing or strictly decreasing; hence, there exists at most one zero in $(a,b)$.

Pedantic grading aside, there seems to be a leap of logic here. Due to the many examples of differentiable functions with discontinuous derivatives, I wonder if there is a counterexample to the students' final conclusion. Namely, is there a non-monotonic, everywhere differentiable function whose derivative is nowhere zero?

Note: I need the derivative to exist, so the Weierstrass function (as in this post and this post) doesn't work.

There may also be a simple way to interpret the students' proofs; they're generally poorly written and the above was my best guess at what they meant to say.

  • 3
    $\begingroup$ Darboux says "no". $\endgroup$ – David Mitra Sep 28 '14 at 14:06
  • $\begingroup$ As $f(a) < f(b)$ or $f(b) < f(a)$, doesn't $f$ need to be monotonic? $\endgroup$ – Clarinetist Sep 28 '14 at 14:10
  • 1
    $\begingroup$ @Clarinetist: That does not follow directly, i.e. without using Darboux' theorem. Note that the "sign" (i.e. wether $<$ or $>$ occurs) could change with different values of $a,b$. $\endgroup$ – PhoemueX Sep 28 '14 at 15:32

There are two arguments one can use here:

As David Mitra points out, Darboux' theorem (every derivative fulfils the intermediate value property) implies that either $f' >0$ or $f' <0$ on the whole interval, so that $f$ is strictly monotonous.

Another approach is to note that the argument given by the student proves that $f$ is injective. But it is well-known (and easy to see directly using the intermediate value theorem) that every continuous injective function on an interval is strictly monotonous.

  • $\begingroup$ For injectivity, one would need to show that $f(c) \neq f(d)$ for all $c \neq d \in [a,b]$, correct? Not just for $a$ and $b$. Regardless, I think you've affirmed for me that the students didn't know what they were doing. $\endgroup$ – artificial_moonlet Sep 28 '14 at 17:22
  • 1
    $\begingroup$ Yes, sure, fur $c\neq d$ and $c,d \in [a,b]$, but the proof using the mean value theorem is just the same. $\endgroup$ – PhoemueX Sep 28 '14 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.