Norms and orthogonal Polynomials

[Tomer]

Homework Statement

Thanks very much for reading.
I actually have two problems, I hope it's ok to state both of the in the same thread.

1. Let V_n be the space of all functions having the n'th derivitve in the point x₀.
I've been given the semi-norm (holds all the norm axioms other than ||v|| = 0 => v = 0) defined by:
||f|| = $\sum^{n}_{k=0}\frac{1}{k!}|f^{(k)}(x_{0})|$

I need to show that $||fg|| \leq ||f|| ||g||$

2. I need to prove that if the set of polynomials {P_i}, i=0,...,n is orthonormal on [a,b] in respect to the inner product defined by: $<f,g> = \int ^{b}_{a}f(x)g(x)dx$
then they hold the recurrance relation:
P_k+1(x) = (a_kx + b_k)P_k(x) + c_kP_k-1(x)

The Attempt at a Solution

1. This one's been a nightmare for me. I've tried proving it straightforward, and then using induction, but I just get lost in a sea of indexes. I'm thinking there might be a more elegant way, but can't see it.
I'm of course trying to use the triangle inequality, which works great for the cases n=1 and n=2, but I just cannot generlize it.
I could write the whole development I've done here, it's like one whole page, but it'll take me ages and therefore think it's rather unnescesarry. I'll greatly appreciate a hint. If there's no other way but an ugly induction I think I'll skip it :-)
(I've used the fact that $(fg)^{(k)}(x_{0}) = \sum^{k}_{i=0}(\frac{k!}{i!(k-i)!})f^{(k-i)}g^{(i)}(x_{0})$

2. Well, I've been thinking polynomial division. If I assume the set of given P_i's is of ascending degrees (0,1,2...), I can always divide P_k+1(x) in P_k(x) and write:
$P_{k+1}(x) = (a_{k}x + b_{k})P_{k}(x) + L(x)$
Where deg(L(x)) < deg(P_k(x))
I don't know how to take it from here. How do I use the orthogonality? I would greatly appreciate hints :-)

 

[micromass]

OK, for (a). You've come to the conclusion already that

$$\begin{eqnarray*} \|fg\| & = & \sum_{k=0}^n{\frac{1}{k!}\left|\sum_{i=0}^k\frac{k!}{i!(k-i)!}f^{(k-i)}(x_0)g^{(i)}(x_0)\right|}\\ & \leq & \sum_{k=0}^n{\sum_{i=0}^k\frac{1}{i!(k-i)!}|f^{(k-i)}(x_0)g^{(i)}(x_0)|} \end{eqnarray*}$$

The thing to do now is to switch those two sums appropriately.

For (b). The convention for orthogonal polynomials is that $P_k$ always has degree k. So I shall follow that convention here. (I'm pretty sure the result is false if the convention is not followed).

Anyway. If you pick a suitable $a_k$, then $P_{k+1}-a_kxP_k$ has degree $\leq k$. So we can write

$$P_{k+1}-a_k xP_k=b_kP_k+c_kP_{k-1}+\sum_{j=1}^{k-2}{d_jP_{k-2}}$$

Our job is to prove $d_j=0$. Do this by taking appropriate inner products.

 

[Tomer]

Thanks, I'll try working on that.