Hermite-type convex interpolation

Question

Let $f : [0, 1] \to \mathbb{R}$ be a smooth strictly convex function. Let $0 \leq x_1 < \dots < x_n \leq 1.$ I am interested in whether there exists a polynomial $p$ such that it is convex on $[0, 1]$ and satisfies $p (x_i) = f (x_i)$ and $p^\prime (x_i) = f^\prime (x_i)$ for $i = 1, \dots, n.$

I know that the Hermite interpolation satisfies the zeroth and first order derivative matching, but it is not necessarily convex. My question is whether the convexity can be imposed by allowing for any order of polynomial.

This is written as a feasibility problem of a conic optimization problem. Let $P_d$ be the set of polynomials of degree up to $d.$ Let $C_d \subset P_d$ be the set of polynomials that are convex on $[0, 1]$ and of degree up to $d,$ and $$ K_d = \left\{ p \in P_d \mid \text{$p (x_i) = f (x_i)$ and $p^\prime (x_i) = f^\prime (x_i)$ for $i = 1, \dots, n$} \right\} . $$ Then $P_d$ is a $d$-dimensional vector space, $C_d$ is a (regular) convex cone in $P_d,$ and $K_d$ is an affine subspace in $P_d.$ My question can be written as whether $C_d \cap K_d \neq \emptyset$ for large $d.$

Fedor Petrov · Accepted Answer · 2025-05-08 09:31:08Z

Yes, it exists. The strict convexity yields that the points $P_i=(x_i,f(x_i))$ and on the graph of $f$ and the tangents $\ell_i$ at these points form a strictly convex picture (every chord $P_iP_{i+1}$ lies above the lines $\ell_i$ and $\ell_{i+1}$). These points and tangents may be realized by a convex smooth function $f$ satisfying $f''>0$ on $[0,1]$ (ask if you need details).

Now we consider the case when $f''>0$ on $[0,1]$.

Let $g_0$ be arbitrary polynomial satisfying $g_0(x_i)=f(x_i)$ and $g_0'(x_i)=f'(x_i)$ for all $i=1,2,\ldots,n$. Denote $H(x)=\prod_i (x-x_i)^2$. Then the function $w:=(f-g_0)/H$ is smooth on $[0,1]$. We look at a polynomial of the form $p(x)=g_0(x)+H(x)\cdot u(x)$ for the unknown polynomial $u(x)$. It automatically enjoys the conditions $p(x_i)=g_0(x_i)=f(x_i)$ and $p'(x_i)=g_0'(x_i)=f'(x_i)$ for all $i=1,\ldots,n$, and it remains to achieve convexity. Well, $p''=g_0''+Hu''+2H'u'+H''u$, and if it were $w$ instead of $u$, this would be $f''$, which is bounded away from 0 on $[0,1]$. It remains to uniformly approximate $w''$ by a polynomial denoted by $u''$ (that is, $u$ itself is determined up to adding a polynomial of degree at most 1) using Weierstrass theorem, and choose $u$ so that $u(0)=w(0)$, $u'(0)=w'(0)$. Then all $u,u',u''$ are uniformly on $[0,1]$ close to $w,w',w''$ respectively, and this $u$ works

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