Spivak Calculus: $f$ continuous on $[a,b]$, $n$-times differentiable on $(a,b)$, with $n+1$ roots in $(a,b)$. $f^{(n)}(x)=0$ for some $x$ in $(a,b)$. The following is a problem from chapter 10, "Significance of the Derivative", in Spivak's Calculus



*Suppose that $f$ is continuous on $[a,b]$, that it is $n$-times differentiable on $(a,b)$, and that $f(x)=0$ for $n+1$ different $x$
in $[a,b]$. Prove that $f^{(n)}(x)=0$ for some $x$ in $(a,b)$.


It is relatively straightforward to prove this:
Let $a_1<a_2<...<a_n$ be the $n+1$ roots of $f in $[a,b]$.
Applying Rolle's Theorem to each consecutive pair of roots tells us that $f'$ has $n$ roots, one between each pair of consecutive roots of $f$.
Applying Rolle's Theorem to each consecutive pair of roots of $f'$ tells us that $f''$ has $n-1$ roots in $[a,b]$. And so on.
In particular, after $n$ applications of Rolle's Theorem in this way, we reach the result that $f^{(n)}$ has exactly one root in $[a,b]$.
The solution manual solution is similar, but at the end it says that "we are all set up for a proof by induction".

If $f(x_i)=0$ for $x_1<x_2<...<x_{n+1}$ then $f'(x)=0$ for some $x$ in
each of the $n$ intervals $(x_i,x_{i+1})$. Consequently $f''(x)=0$ for
$n-1$ numbers $x$, etc. (In other words, we are all set up for a proof
by induction)

What exactly is the induction argument here?
 A: It's an induction argument in terms of $n$. By using Rolle's theorem, we see that the function $g = f'$ has $n$ roots (instead of $n + 1$ roots), and is $(n-1)$-times differentiable (as $g^{(n-1)} = f^{(n)}$ and $f$ is $n$-times differentiable). That is, $g$ satisfies the premises of the question, but with $n$ replaced by $n - 1$. This is indeed the basis for an induction proof. However, there are some slight wrinkles that need navigating, which the text doesn't highlight, so let's go through a proof.
I'm going to write out a predicate in $n \in \Bbb{N}$ for us to show. Note, there's already a small wrinkle at the end.

$P(n)$ : Given any closed, bounded interval $[a, b]$ and continuous function $f : [a, b] \to \Bbb{R}$ with the following properties:

*

*$f$ is $n$-times differentiable on $(a, b)$, and

*$f$ has at least $n+1$ zeros,

we have $f^{(n)}(x) = 0$ for some $x \in (a, b)$ if $n > 0$, or $x \in [a, b]$ if $n = 0$.

Let's prove the base case: $P(0)$, taking the convention that $f^{(0)} = f$. As any function (continuous or not) can be considered $0$-times differentiable, then having $0 + 1 = 1$ roots of $f$ implies $f^{(0)} = f$ has a root in $[a, b]$ trivially. Thus, $P(0)$ is true. (Note: we cannot conclude there is a root in $(a, b)$; the question does not assume the roots must exist in the open interval.)
Now, suppose that $P(k)$ is true for some $k \ge 0$. We wish to prove $P(k+1)$ is true.
Suppose $f : [a, b] \to \Bbb{R}$ is continuous, and $(k + 1)$-times differentiable on $(a, b)$. Further, suppose
$$a \le x_1 < x_2 < \ldots < x_{k+2} \le b$$
are roots of $f$ (i.e. $f$ has at least $(k + 1) + 1$ roots). Then, by Rolle's theorem, applied to each $[x_i, x_{i+1}]$ interval, there must be some $y_i \in (x_i, x_{i+1})$ such that $f'(y_i) = 0$. Since each $y_i$ lies in one of $k + 1$ pairwise-disjoint open intervals, we see that $f'$ has at least $k + 1$ roots. We also know that $f'$ is $k$-times differentiable on $(a, b)$ and hence continuous on $(a, b)$.
It's not clear (and indeed false in general) that $f'$ can be made continuous on the closed interval (e.g. in the case of a semicircle, the derivative tend to $\pm\infty$ as we approach the endpoints). So, what we do is consider $f'$ on the restricted domain to $[y_1, y_k]$. This domain is contained in $(a, b)$, so $f'$ is continuous on this closed interval, while still be $k$-times differentiable on $(y_1, y_k)$. We also still have $k$ roots of $f'$ on $[y_1, y_k]$.
So, we let $g = f'|_{[y_1, y_k]}$. Using our assumption $P(k)$, we have at least one root $x \in [y_1, y_k]$ of $g^{(k)}$ (note: I put such a root in the closed interval, in case $k = 0$). Thus,
$$0 = g^{(k)}(x) = f^{(k+1)}(x)$$
for some $x \in [y_1, y_k] \subseteq (a, b)$, as required. Hence $P(k+1)$ holds as well.
