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

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

Solution

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:

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

Integral inequality of a function

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