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! :-)


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 $a = b=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.