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

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 $$a implies that $$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?

• Let me mention that was motivated by the closely related question. Commented Jan 13, 2023 at 22:18

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.

• Thank you kindly for this very nice proof! Commented Jan 13, 2023 at 6:23
• @PavelKocourek: You are welcome! – Btw, contrary to what is said on the Wikipedia page on polynomial interpolation, it suffices that $f$ is n-times differentiable, but not necessarily n-times continuously differentiable. I have edited the answer to clarify that. Commented Jan 13, 2023 at 7:46