Let $z_1,z_2,\dots,z_n\in\Bbb{C}$ be distinct and $w_1,w_2,\dots,w_n\in\Bbb{C}$ be arbitrary. Suppose $f, g$ are two polynomials of degree less than $n$ such that $$f(z_j)=w_j,\qquad g(z_j)=\bar{w}_j \qquad\text{for $1\leq j\leq n$}.$$ Define $\Omega(z)=\prod_{j=1}^n(z-z_j)$. The following puzzles me.

Question 1. Is this true? $$\sum_{k=1}^n\frac{\vert f^{\prime}(z_k)\vert^2}{\vert \Omega^{\prime}(z_k)\vert^2}\leq \sum_{k=1}^n\frac{\vert g^{\prime}(z_k)\vert^2}{\vert\Omega^{\prime}(z_k)\vert^2}.$$

EDIT. I'm sorry, one of the conditions was missing. To give credit, I'll leave the question as it stands and ask the correct one below.

Question 2. What if we insist that $f$ and $g$ have equal degrees?

  • $\begingroup$ Any further background where's this coming from? thanks. $\endgroup$
    – Suvrit
    Feb 5, 2017 at 4:28
  • $\begingroup$ It was a technical need that occurred in some Monge-Ampere equations. $\endgroup$ Feb 5, 2017 at 4:53
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    $\begingroup$ I am puzzled. Polynomials $\ f\ g\ $ appear in a symmetric way. Then why inequality and not equality? Indeed, if the theorem held for all $\ f\ g\ $ then there would be equality. $\endgroup$ Feb 5, 2017 at 5:45
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    $\begingroup$ Well $f,g$ are uniquely defined by interpolation given the $z_j$'s and $w_j$'s, so $f=\bar{g}$ necessarily... $\endgroup$ Feb 5, 2017 at 7:55
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    $\begingroup$ @PhilippeNadeau: it's not as easy as that; the $w_i$ are replaced by their complex conjugates but not the $z_i$, so the change in coefficients is a bit more mysterious. $\endgroup$ Feb 5, 2017 at 9:41

2 Answers 2


For $n\ge 3$, let's take $z_j$, for $1\le j\le n$, be the $n$-th roots of unity, and $w_j:=z_j+4\bar{z_j}$, so that the assumption are satisfied by $$f(z):=z+4z^{n-1}$$ $$g(z):=4z+z^{n-1}.$$ However, $$ |g'(z_k)|=|4+(n-1) z_k^{-2}| \le n+3<\phantom{Z} $$ $$ \phantom{ZZZZ} <4n-5\le|1+4(n-1) z_k^{-2}|= |f'(z_k)| $$ for any index $k$ in the sums.

Rmk. For $n=2$, $f$ and $g$ are affine functions, with $|f'|=|g'|={|w_1-w_2|\over |z_1-z_2|}$, so that the claim is true.

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    $\begingroup$ Modified to fit question 2 (the customer is always right) $\endgroup$ Feb 5, 2017 at 13:49
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    $\begingroup$ That was funny, Pietro! $\endgroup$ Feb 5, 2017 at 14:22

This Matlab program shows that your conjecture is in general not true for $n\geq 3$

n = 3;

z = rand(1, n) + 1i * rand(1, n)
w = rand(1, n) + 1i * rand(1, n)

f = polyfit(z, w, n-1);
g = polyfit(z, conj(w), n - 1);

% note that Omega(z_i) = 0 and Omega(z) = z^n+q(z) where deg(q)<=n-1 and
% hence q(z_i)=-z_i^n
Omega = [1 -polyfit(z, z.^n, n - 1)];

df = polyder(f);
dg = polyder(g);
dOmega = polyder(Omega);

adf = abs(polyval(df, z));
adg = abs(polyval(dg, z));
adO = abs(polyval(dOmega, z));

  • $\begingroup$ I don't use mathlab. Do you mind giving the specific complex numbers that generate the example? $\endgroup$ Feb 5, 2017 at 12:49
  • $\begingroup$ Pasting this into the Matlab console a couple of times gives a positive final value (11.1599) for z = [0.8143 + 0.3500i, 0.2435 + 0.1966i, 0.9293 + 0.2511i], w = [0.6160 + 0.8308i, 0.4733 + 0.5853i, 0.3517 + 0.5497i] $\endgroup$
    – J.J. Green
    Feb 5, 2017 at 23:13

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