# A proof with Legendre polynomials and an integral minimum value

I need to prove that, over the monic polynomials $f$ of degree $n$, the integral $$\int_{-1}^1\bigl(f(x)\bigr)^2\,dx$$ takes its minimum value when $$f(x)=\frac{2^n}{\binom{2n}n}L_n(x),$$ where $L_n(x)$ is the $n$th Legendre polynomial.

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What do you mean by "an is the reciprocal of the leading coefficient of Ln(x)"?, in the proof it self? Your result is not what I want to prove, it's upside down, it should be the inverse value right? – user68315 Mar 24 '13 at 19:44
In my answer, I am using $a_n$ to denote the coefficient in front of $L_n(x)$ in the expansion of $f(x)$ as I write in the first bullet point in my answer. In order for $a_nL_n(x)$ to be monic, $a_n$ must be the reciprocal of the leading coefficient of $L_n(x)$. – anon Mar 24 '13 at 23:18

• The Legendre polynomials form a basis for the space of polynomials, so we may write $$f(x)=\sum_{k=0}^n a_kL_k(x)$$ for some real numbers $a_k\in\bf R$, for any $f(x)$ of degree $n$.
• If $f(x)$ is monic, then $a_n$ is the reciprocal of the leading coefficient of $L_n(x)$. Looks like that means you have to verify that the leading coefficient of $L_n(x)$ is $2^{-n}{2n\choose n}$.
• The Legendre polynomials are orthogonal, in the sense that $$\langle L_r,L_s\rangle=\int_{-1}^{+1}L_r(x)L_s(x)dx=0$$ when $r\ne s$. Therefore (with $\|\cdot\|$ induced from $\langle\cdot,\cdot\rangle$ defined directly above) $$\|f(x)\|^2=\sum_{k=0}^na_k^2\|L_k(x)\|^2\ge a_n^2\|L_n(x)\|^2$$ with equality if and only if $a_0=a_1=a_2=\cdots=a_{n-1}=0$.
@deiota Indeed. If you expand $(x^2-1)^n$ with the binomial expansion the leading term is $x^{2n}$; if you differentiate that $n$ times the leading coefficient is ... ? – anon Mar 24 '13 at 10:50