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Integral inequality of a function

Let n \geq 1 be an integer and let f : [0,1] \rightarrow \mathbb{R} be a continuous function. Suppose that \bigintsss_0^1 x^k f(x) \; \mathrm{d}x = 1 for all 0 \leq k \leq n-1. Show that

    \[\int_0^1 (f(x))^2 \mathrm{d}x \geq n^2\]


We begin by a lemma:

Lemma: For an integer n \geq 1 the n\times n  Hilbert matrix is defined by \mathcal{H}_n=[a_{ij}] where

    \[a_{ij}=\frac{1}{i+j-1} \quad , \quad 1 \leq i,j \leq n\]

It is known that \mathcal{H}_n is invertible and if \mathcal{H}_n^{-1}=[b_{ij}] then \sum \limits_{i,j}b_{ij}=n^2.

Since \mathcal{H}_n , the n\times n  Hilbert matrix ,  is invertible there exist real numbers p_0, p_1, \dots , p_{n-1} such that

    \[\sum_{i=1}^n\frac{p_{i-1}}{i+j-1}=1 \quad ,\quad 1 \leq j \leq n.\]

So the polynomial p(x)=\sum \limits_{k=0}^{n-1}p_k x^k satisfies the conditions

    \[\int_0^1x^k p(x) \, \mathrm{d} x =1 \quad , \quad 0 \leq k \leq n-1\]

Clearly \sum \limits_{k=0}^{n-1}p_k  is the sum of all the entries of \mathcal{H}_n^{-1} and so \sum \limits_{k=0}^{n-1}p_k=n^2. Now let f be a real-valued continuous function on [0,1] such that

    \[\int_0^1x^kf(x) \, \mathrm{d}x = 1 \quad , \quad 0 \leq k \leq n-1\]

Let p(x) be the above polynomial.Then since

    \[\left(f(x))^2-2f(x)p(x)+(p(x) \right)^2 =\left(f(x)-p(x) \right)^2 \geq 0\]

integrating gives:

    \begin{align*} \int_0^1 (f(x))^2 \; \mathrm{d}x &\geq 2\int_0^1f(x)p(x) \; \mathrm{d}x -\int_0^1(p(x))^2 \;\mathrm{d}x \\ &=2\sum_{k=0}^{n-1}p_k \int_0^1 x^kf(x) \;\mathrm{d}x- \sum_{k=0}^{n-1}p_k\int_0^1x^kp(x)\; \mathrm{d}x \\ &= 2\sum_{k=0}^{n-1}p_k-\sum_{k=0}^{n-1}p_k \\ &=\sum_{k=0}^{n-1}p_k \\ &=n^2 \end{align*}

and the result follows.

The exercise along with its solution was taken from here.

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