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I would greatly appreciate any hint for proving the following.

Question: Let $f:[0, 1] \to {\bf R}$. Can it be proved that if $[0, 1/(N+m),\dots, (N+m)/(N+m) ; f ]=0$ for all $m=1,2, 3,\dots$, then $f$ is necessarily a polynomial of degree less than or equal to $N$?

(Here $[0, 1/(N+m),\dots, (N+m)/(N+m) ; f]$ denotes the divided difference of $f$ on the nodes $0, 1/(N+m), \dots, 1$.)

Remark: In order to make it easier to obtain a proof, it can be supposed that $f$ satisfies some smoothness properties, as for example to be infinitely differentiable or analytic on $[0, 1]$.

Thank you,

G

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  • $\begingroup$ In your question, it is not clear to me whether or not continuity of $f$ on $[0,1]$ is a requirement ... is it? Otherwise $f:[0,1]\to\mathbb{R}$ given by, $$ f(x) = \begin{cases} 0 &;x\in\mathbb{Q}\cap [0,1] \newline 1 &;x\in[0,1]\backslash \mathbb{Q} \end{cases}$$ may be relevant. $\endgroup$
    – JCM
    Commented May 27, 2013 at 11:12
  • $\begingroup$ Thank you for your remark, in the question of course that f must be supposed at least continuous on [0, 1], but as I said in the Remark, it could be supposed to be differentiable too, if that hypothesis would be helpful for the proof. Irene $\endgroup$
    – George
    Commented May 27, 2013 at 11:16
  • $\begingroup$ Cheers for confirming. $\endgroup$
    – JCM
    Commented May 27, 2013 at 12:14
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    $\begingroup$ I don't know enough about the background of this question to determine whether it is "off-topic" -- but I ask people who vote to close on such or similar grounds to make sure they do know. $\endgroup$
    – Todd Trimble
    Commented May 27, 2013 at 13:09
  • $\begingroup$ Actually this is the point of the question, that in the class of all continuous functions on [0, 1], only the polynomials of degree < or = N satisfy that condition. Intuitively, this is suggested by the fact that the set of knots 1/(N+m), ..., (N+m)/(N+m), when m takes all the natural numbers, is dense in [0, 1]. $\endgroup$
    – George
    Commented May 27, 2013 at 15:33

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