A function with weakly positive $n$-th derivative has at most $n$ roots – an inequality version of the Fundamental theorem of Algebra.

Question

This is an attempt to generalize the result in [1].

Claim: Let $n\in \mathbb N$, and let $f:\mathbb R \to \mathbb R$ be such that its $n$-th derivative $f^{(n)}(x)\geq 0, \ \forall x\in \mathbb R$. Then the set $R$ of roots of $f$ either consists of at most of $n$ isolated points, or it is a (non-degenerate) closed interval.

Note: As argued in [1], under the stronger assumption $f^{(n)}(x)> 0$ the case of $R$ being a closed interval can be excluded.

The analogy of this claim to the fundamental theorem of algebra (restricted to real numbers) is shown in [1].

Special cases:

• for $n=1$ the claim states that if a function has nonnegative derivative, it has a single root or a closed interval of roots, e.g. consider the function $f(x) = \min\{x+1,0\} + \max\{x-1,0\}$ for which $R = [-1,1]$;
• for $n=2$ the condition $f^{(n)}(x)\geq 0$ implies that $f$ is (weakly) convex, and example could be the function $f(x)=x^2-1$ which has two roots, or $$ f(x) = \begin{cases} 0, &\text{ if } x\in [-1,1] \\ (|x|-1)^2, &\text{otherwise}, \end{cases} $$ in which case $R= [-1,1]$.

The reason why I think that $[a,b]\subset R$ for some $a<b$ implies that $f$ has no isolated roots is that this implies that all the derivatives of $f$ are zero on $(a,b)$, and so (assuming that the interval $[a,b]$ is maximal among the closed intervals contained in $R$) the derivatives $f^{(n-1)}(x),f^{(n-2)}(x),f'(x),f(x)$ are all positive for $x>b$ as $f^{(n)}(x)\geq 0$ for $x>b$ with inequality strict at $x=b+\varepsilon$ for $\varepsilon>0$ arbitrarily small.

Does the claim hold?

Answer 1

Let $f:\Bbb R \to \Bbb R$ be $n$ times differentiable with $f^{(n)}(x) \ge 0$ for all $x \in \Bbb R$. First we show:

If $f$ has (at least) $n+1$ distinct zeros $a_1 < a_2 < \cdots < a_{n+1}$ then $f$ is identically zero on $[a_1, a_{n+1}]$.

Proof: $p(x) \equiv 0$ interpolates $f$ both on $(a_1, \ldots, a_n)$ and on $(a_2, \ldots, a_{n+1})$. The formula for the interpolation error of polynomial interpolation gives that for every $x \in \Bbb R$, $$ f(x) = \frac{f^{(n)}(\xi_x)}{n!} (x-a_1)(x-a_2)\cdots (x-a_n) \\ = \frac{f^{(n)}(\eta_x)}{n!} (x-a_2)(x-a_3)\cdots (x-a_{n+1}) $$ with some $\xi_x$ and $\eta_x$, depending on $x$. It follows that $f(x)$ is either zero or has the same sign as both the products $$ (x-a_1)(x-a_2)\cdots (x-a_n) \\ (x-a_2)(x-a_3)\cdots (x-a_{n+1}) $$ But those products have different sign on each interval $(a_k, a_{k+1})$, $1 \le k \le n$. Therefore is $f$ identically zero on each interval $(a_k, a_{k+1})$, and consequently zero on $[a_1, a_{n+1}]$.

Remark: The interpolation error formula as stated on the Wikipedia page requires $f^{(n)}$ to be continuous, but that is not needed. It suffices that $f$ is $n$-times differentiable, so that Rolle's theorem can be applied to $f, f', \ldots, f^{(n-1)}$.

Now we can show

If $f$ has (at least) $n+1$ distinct zeros then the set $Z = \{ x \in \Bbb R \mid f(x) = 0 \}$ is a closed interval.

Proof: Let $a_1 < a_2 < \cdots < a_{n+1}$ be zeros of $f$. If $a, b\in Z$ with $a \le a_1$ and $b \ge a_{n+1}$ then the first part shows that $[a, b]$ is contained in $ Z$. So $Z$ is connected and therefore an interval. And it is closed because $f$ is continuous.