# For $p(x)\in \Bbb{C}[x]$ such that $\int_{0}^{1}p(x)x^kdx=0$ for $0\le k\le n-1$, show that $p(\lambda)=0\Rightarrow \lambda\in [0,1]$

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.

• 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. – Jack D'Aurizio Jul 19 '16 at 20:46
• I don't understand how you arrive at it. Is there a Theorem you were using here? – Meitar Jul 19 '16 at 20:52
• Shifted Legendre polynomials give a complete orthogonal base of $L^2(0,1)$ with respect to the usual inner product $$\langle f,g\rangle = \int_{0}^{1}f(x)\,g(x)\,dx.$$ – Jack D'Aurizio Jul 19 '16 at 20:54
• 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. – Jack D'Aurizio Jul 19 '16 at 20:57
• 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. – Jack D'Aurizio Jul 19 '16 at 21:12

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.
• yo have $n$ constraints because you have forget the case $k=0$ – Hamza Jul 19 '16 at 21:59
• @Hamza: the integral constraints are for $k\in\{1,2,\ldots,n-1\}$ and there are $n-1$ numbers in such a set, not $n$. – Jack D'Aurizio Jul 19 '16 at 22:02
• I read in the title that $k=0$ is also a constraints – Hamza Jul 19 '16 at 22:04