# Interpolation with nonvanishing constraint

Let $x_1,x_2,\ldots,x_n$ be distinct complex numbers. Let $y_1,y_2,\ldots,y_n$ be nonzero complex numbers, and let $K$ be a bounded subset of $\mathbb C$. Does there always exist a polynomial $P$ such that $P(x_i)=y_i$ for every $i$ and $P$ is everywhere nonzero on $K$ ?

This is obviously false when $\mathbb C$ is replaced by $\mathbb R$.

## 1 Answer

Yes, such a polynomial always exists:

By scaling and replacing $K$ with a superset, we may assume that $K = \overline{\mathbb{D}} = \{ z : \lvert z\rvert \leqslant 1\}$ and that $\lvert x_i\rvert \leqslant 1$ for all $i$.

For each $i$, consider the interpolation polynomial

$$P_i(z) = \prod_{\substack{k=1\\k\neq i}}^n \frac{z-x_k}{x_i-x_k}$$

and define $C_i = \max \{ \lvert P_i(z)\rvert : z\in K\}$, and $C = \max \{ C_i : 1 \leqslant i \leqslant n\}$.

Lagrange interpolation gives us a polynomial $L$ with $L(x_i) = \log y_i$ for $1 \leqslant i \leqslant n$, whichever logarithm of $y_i$ we choose.

Then $f(z) = e^{L(z)}$ is an entire function with $f(x_i) = y_i$ for $1\leqslant i \leqslant n$, and $f(z) \neq 0$ for all $z\in \mathbb{C}$. Thus $m := \min \{ \lvert f(z)\rvert : z\in K\} > 0$. Since the Taylor series of $f$ converges uniformly to $f$ on $K$, there is a Taylor polynomial $T$ of $f$ with

$$\lvert T(z) - f(z)\rvert < \frac{m}{n(C+1)}$$

for $z \in K$. Finally, consider the polynomial

$$P(z) = T(z) + \sum_{j=1}^n \bigl(y_j - T(x_j)\bigr)P_j(z).$$

Then we have

$$P(x_i) = T(x_i) + \sum_{j=1}^n \bigl(y_j - T(x_j)\bigr)P_j(x_i) = T(x_i) + \bigl(y_i - T(x_i)\bigr) = y_i$$

for $1 \leqslant i \leqslant n$, and

\begin{align} \lvert P(z)\rvert &\geqslant \lvert T(z)\rvert - \sum_{j=1}^n \lvert y_j - T(x_j)\rvert\cdot \lvert P_j(z)\rvert\\ &\geqslant m - \sum_{j=1}^n \lvert y_j - T(x_j)\rvert C_j\\ &\geqslant m - nC\frac{m}{n(C+1)}\\ &= \frac{m}{C+1}\\ &> 0 \end{align}

for $z\in K$.