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For a complex polynomial $p(x)\in \Bbb{C}[x]$ of degree $n$ such that $\int_{0}^{1}p(x)x^kdx=0$ for $1\le k\le n-1$, show that $p(\lambda)=0$ means $\lambda\in [0,1]$.

I haven't come by any theoretical direction so far. Since it is a question given in an Introduction to Operator Theory and Hilbert Spaces, I think that approaching Complex Analysis only will be very tedious. The problem is that I can't really tell what of the the course themes it is that I should use. I was thinking Min-Max theorem, Spectral Theory and things in that area of the course syllabus. I still can't see how it can be done; the question seems irrelevant. I would appreciate it if you could give any hint or guide me a little.

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    $\begingroup$ It may be relevant to notice that the integral constraint is equivalent to $$p(x)=\sum_{m\geq n}c_m P_m(2x-1) $$ by the orthogonality of shifted Legendre polynomials. Tricomi's inequalities about the zeroes of Legendre polynomials should solve the question. $\endgroup$ – Jack D'Aurizio Jul 19 '16 at 20:46
  • $\begingroup$ I don't understand how you arrive at it. Is there a Theorem you were using here? $\endgroup$ – Meitar Jul 19 '16 at 20:52
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    $\begingroup$ Shifted Legendre polynomials give a complete orthogonal base of $L^2(0,1)$ with respect to the usual inner product $$\langle f,g\rangle = \int_{0}^{1}f(x)\,g(x)\,dx.$$ $\endgroup$ – Jack D'Aurizio Jul 19 '16 at 20:54
  • $\begingroup$ Another approach may be to simply write the integral constraints in terms of the coefficients of $p(x)$: the Hilbert (en.wikipedia.org/wiki/Hilbert_matrix ) and Vandermonde matrices are so involved. $\endgroup$ – Jack D'Aurizio Jul 19 '16 at 20:57
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    $\begingroup$ As stated, the claim is wrong. $$p(x)=P_n(2ix-1)$$ with $P_n$ being a Legendre polynomial is a clear counter-example: orthogonality holds but not every root is real. $\endgroup$ – Jack D'Aurizio Jul 19 '16 at 21:12
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I think the intended question was something like:

Given $p(x)\in\mathbb{C}[x]$ with degree $n$, such that $$\int_{0}^{1}x^k p(x)\,dx=0 $$ for every $k\in\{1,2,\ldots,n-1\}$, show that the roots of $p(x)$ lie in the circle $|x|\leq 1$.

We may notice that $p(x)$ has $n+1$ coefficients and we have $n-1$ linear constraints on them, so the space of solutions has dimension $2$. We may also notice that $$ \int_{0}^{1} x^k\,P_n(2x-1)\,dx = 0 $$ for any $k\in\{0,1,2,\ldots,n-1\}$, with $P_n(x)$ being a Legendre polynomial.
Another solution is given by $$ p(x) = \frac{P_{n+1}(2x-1)-P_{n-1}(2x-1)}{x}, $$ so the integral constraints translate into: $$ p(x) = A\cdot P_n(2x-1)+B\cdot \frac{P_{n+1}(2x-1)-P_{n-1}(2x-1)}{x} $$ and the claim follows from Turan's inequality for Legendre polynomials, or by bounding the eigenvalues of a Sturm-Liouville problem associated with the Legendre differential equation.

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  • $\begingroup$ yo have $n$ constraints because you have forget the case $k=0$ $\endgroup$ – Hamza Jul 19 '16 at 21:59
  • $\begingroup$ @Hamza: the integral constraints are for $k\in\{1,2,\ldots,n-1\}$ and there are $n-1$ numbers in such a set, not $n$. $\endgroup$ – Jack D'Aurizio Jul 19 '16 at 22:02
  • $\begingroup$ I read in the title that $k=0$ is also a constraints $\endgroup$ – Hamza Jul 19 '16 at 22:04
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    $\begingroup$ @Hamza: then the OP modified the question, look at the comments. $\endgroup$ – Jack D'Aurizio Jul 19 '16 at 22:04
  • $\begingroup$ @JackD'Aurizio, Thank you very much for that. Is there, however, a way to avoid the Turan's inequality? It wasn't mentioned anywhere in the course notes... $\endgroup$ – Meitar Jul 20 '16 at 14:21

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