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I'm stuck with a homework problem from my functional analysis class. The question is:

"Show that the Gram-Schmidt orthonomalization procedure in $L^2(-1,1)$, starting from the series $(x^n)^{\infty}_{n=0}$ provides an orthonomal basis, given by $(b^n)^{\infty}_{n=0}$, $b_n = \sqrt{n + \frac{1}{2}}P_n$ with $P_n(x)=\frac{1}{2^n n!} \frac{d^n}{dx^n} [(x^2-1)^n]$ beeing the n-th Legendre polynomial."

I tried two different approaches:

$ p_n(x) = x p_{n-1}(x) - \frac{(y p_{n-1}(y),p_{n-1}(y))}{(p_{n-1}(y),p_{n-1}(y))} p_{n-1}(x) - \frac{(y p_{n-1}(y),p_{n-2}(y))}{(p_{n-2}(y),p_{n-2}(y))} p_{n-2}(x) $

with $P_n(x) = \frac{p_n(x)}{p_n(1)}$ is equivalent to

$ (n+1) P_{n+1}(x) = (2n+1)x P_n(x) - n P_{n-1}(x), n=1,2,...;P_0 = 1; P_1 = x$.

And I just don't know how to do that.

  • The second approach was to show explicitely for $P_0, P_1$ and $P_2$ that the equivalence

$b_n = \sqrt{n + \frac{1}{2}}P_n = \frac{v_n}{|v_n|} $ with $ v_n = (x^n - \sum_{k=0}^{n-1} \frac{(x^n,b_{n-1})}{(b_{n-1},b_{n-1})} b_{n-1} )$

holds and then use an inductive argument, which goes like: "The linear subspace of polynomials of degree $n$ has dimension $n+1$. The orthogonal complement of the polynomials of degree $n − 1$ in the space of polynomials of degree $n$ is equal to $1$, and therefore ${Pn}$ is a basis of the orthogonal complement. The Gram-Schmidt orthogonalization of the monomials gives a polynomial of degree $n$ in this complement, so it gives the Legendre polynomials up to normalization."

But by that argument I would assume, that in the space of polynomials of degree $\le n$, given $n$ orthogonal polynomials, an (n+1)'th orthogonal polynomial must be unique, up to scaling. But can I just assume that?

I would appreciate any help, thank you! :-)

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For any inner product there is indeed a unique (up to scaling) polynomial $f\in P_{n+1}(x)$ orthogonal to $P_n(x)$ (the space of polynomials of degree less than or equal to n).

To see this assume (seeking contradiction) that $f, g$ are linearly independent polynomials in the orthogonal compliment of $P_n(x) \leq P_{n+1}(x)$. Note $\{1,x,x^2...x^{n+1}\}$ form a basis of $P_{n+1}(x)$ so this is an n+2 dimensional vector space over your field.

Suppose $f,g$ are linearly independent. By Gram Schmidt WLOG they are orthogonal. Then $\{1,x,x^2...x^n,f,g\}$ are $n+2$ linearly independent. This is a contradiction since it implies $dim (P_{n+1}(x))=n+2$.

To see the above set is linearly independent note $1,...,x^n$ are manifestly independent. f and g are independent of each other by assumption. Suppose: $$\sum_{i=0}^n{\lambda_i x^i} = a f(x) +bg(x)$$

Then taking inner products on both sides with f and g implies $a = b=0$.

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