Does the following claim hold?

Let $n\in \mathbb N_{\geq 2}$ and let $f:\mathbb R \to \mathbb R$ be an $n$-times differentiable function. Then the $n$-the derivative $f^{(n)}(x)\geq 0$ for all $x\in \mathbb R$ iff for any $a_1<a_2<\ldots<a_n$ $$ f(x) \leq p(x;a_1,\ldots,a_n) \quad \forall x \in [a_{n-1},a_n], \tag{*}\label{fp} $$ where $p(\cdot; a_1,\ldots,a_n)$ is the $n-1$ degree polynomial fitting the points $\big(a_1,f(a_1)\big),\ldots,\big(a_n,f(a_n)\big)$.

Note: The condition \eqref{fp} can be geometrically represented as follows: For any $n$ different points of the graph of $f$ consider the curve line segment between the last two points (those most on the right) of the polynomial curve fitting the graph of $f$ at the $n$ points. Then the curve line segment is contained in the epigraph of $f$.


For $n=2$ the condition \eqref{fp} simplifies into that the line segment between any two points of a graph of $f$ is in the epigraph of $f$ – the standard characterization of $\mathop{epi} f$ being a convex set.

In general the condition $f^{(n)}(x)\geq 0$ can be interpreted as that $f^{(n-1)}$ is increasing, and equivalently that $f^{(n-2)}$ is a convex function. The intuitive idea is that if $f^{(n)}$ was constant then $f$ would be identical to the polynomial $p$ that fits any $n$ points in the graph of $f$, and as $f^{(n)}$ is increasing it can be expected that $f$ would be above $p$ for $x$ large.



1 Answer 1


Yes, the equivalence does hold under the condition that $f$ is $n$-times continuously differentiable. This is a consequence of the formula for the interpolation error of polynomial interpolation: For $a_1 \le x \le a_n$ is $$ f(x) - p(x;a_1,\ldots,a_n) = \frac{f^{(n)}(\xi_x)}{n!} (x-a_1)(x-a_2)\cdots (x-a_n) $$ with some $\xi_x \in [a_1, a_n]$, depending on $x$.

If $f^{(n)}(x)\geq 0$ for all $x\in \Bbb R$ and $a_1 < a_2 < \cdots < a_n$ then the above formula shows that $$ f(x)-p(x) \le 0 \text{ for } a_{n-1}\le x \le a_n \, . $$

On the other hand, if $f^{(n)}(c) < 0$ for some $c \in \Bbb R$ then $f^{(n)}(x) < 0$ on some non-degenerated interval $[a, b]$. If we choose $$ a = a_1 < a_2 < \cdots < a_n = b $$ then the above formula shows that $$ f(x)-p(x) > 0 \text{ for } a_{n-1}< x < a_n \, . $$

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    $\begingroup$ thank you for a very nice solution! $\endgroup$ Commented Jan 12, 2023 at 21:44

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