# Orthogonal polynomials and inner product space

• Mar 5th 2010, 12:50 PM
alice34
Orthogonal polynomials and inner product space
Say we have a set {v1, v2, ..., vk, ...} of orthogonal polynomials in some real inner product space < , >, and assume they're normalised to be monic. If q is a polynomial with deg(q) = k, how would I show that

|| vk || <= || q ||, for all q?

(||x|| = <x,x>^(1/2).)
• Mar 5th 2010, 05:53 PM
NonCommAlg
Quote:

Originally Posted by alice34
Say we have a set {v1, v2, ..., vk, ...} of orthogonal polynomials in some real inner product space < , >, and assume they're normalised to be monic. If q is a polynomial with deg(q) = k, how would I show that

|| vk || <= || q ||, for all q?

(||x|| = <x,x>^(1/2).)

the problem in this form is just non-sense! first of all we must have $\deg v_k = k$ and i think we also need a $v_0.$ we also don't need them to be monic. finally if they are monic, then the absolute

value of the leading coefficient of $q$ should be at least 1 because otherwise $q=\frac{1}{2}v_k$ would be a trivial counter-example.

anyway, your problem is a quick result of this simple fact that $q$ is a linear combinaton of $v_j, \ 0 \leq j \leq k$: if $q=\sum_{j=0}^kc_jv_j,$ then $\langle q, q \rangle = \sum_{j=0}^k c_j^2 \langle v_j, v_j \rangle \geq c_k^2 \langle v_k,v_k \rangle \geq \langle v_k,v_k \rangle.$